What is electrostatic potential energy?

Electrostatic potential energy is the work done in assembling a system of charges from infinity without changing kinetic energy.

What is the formula for potential energy of two point charges?

The formula is U = k q1 q2 / r, where k = 1/(4 pi epsilon0), q1 and q2 are charges, and r is their separation.

How many interaction terms are present for n charges?

The number of distinct pair interactions is n(n - 1)/2.

Why can electrostatic potential energy be negative?

It is negative for attractive unlike charges when zero energy is chosen at infinity, meaning energy must be supplied to separate the charges to infinity.

Physics notes

Electrostatic Potential Energy and System of Charges

Complete derivations, diagrams, formulas, numerical problems and exam questions.

CBSENEETJEE MainJEE AdvancedIB PhysicsIGCSEICSEA-Level

Electrostatics Master Resource

1. Introduction

Electrostatic potential energy is the energy stored in a system of charges because electric forces have arranged those charges in a particular configuration. If two charges repel, external work is needed to push them close together; that work becomes positive electrostatic potential energy. If two charges attract, the field can pull them together and the system has negative potential energy relative to the reference state at infinity.

Potential energy exists because the electrostatic force is conservative. Work done by or against a conservative force depends only on initial and final positions, not on the path. Therefore a system of charges can be assigned a single energy value for a given geometry.

The connection between force and potential energy is central: when potential energy changes with separation, a force appears. In one dimension, F = -dU/dr. A system naturally tends to move toward lower potential energy if it is free to move.

Electrostatic potential energy is important in every exam syllabus because it links Coulomb's law, electric potential, work, stability, field energy, capacitors, atomic binding, ionic crystals and numerical problem solving.

Physical Meaning

Positive U means external work has been stored in the charge system.
Negative U means the system can release energy while the charges come together.
Zero U is normally chosen when the charges are infinitely far apart.
Only energy differences directly determine work done.

Electrostatics Master Resource

2. Potential Energy of Two Point Charges

Consider a fixed charge q₁. A second charge q₂ is slowly brought from infinity to a point at distance r from q₁. Slow motion means there is no change in kinetic energy, so the external work done is stored as electrostatic potential energy.

Complete derivation from work done

Electric potential due to q₁ at distance x is V(x) = kq₁/x, where k = 1/(4πε₀).
Potential energy of q₂ placed at that point is U = q₂V.
At infinity, V(∞) = 0, so U(∞) = 0.
At distance r, V(r) = kq₁/r.
Therefore U = q₂[kq₁/r] = kq₁q₂/r.

U = (1/4πε₀) (q₁q₂/r) = kq₁q₂/r

If q₁q₂ is positive, U is positive. If q₁q₂ is negative, U is negative. This single sign decision often decides the whole question.

Electrostatics Master Resource

3. Sign of Potential Energy

The sign of electrostatic potential energy is controlled by the product q₁q₂. Distance r is always positive, and k is positive, so the charge signs do all the work.

Repulsive pair: U > 0

Attractive pair: U < 0

Case	Product	Sign of U	Physical meaning
+q and +q	positive	U > 0	External work is required to assemble because the charges repel.
-q and -q	positive	U > 0	External work is required to assemble because the charges repel.
+q and -q	negative	U < 0	The pair is bound relative to infinity because the charges attract.

A positive value does not mean the system is automatically stable. Like charges placed close together have high energy and tend to fly apart. A negative value usually represents an energetically favorable attractive configuration, although full stability still depends on all forces and constraints.

Electrostatics Master Resource

4. Potential Energy of Three Charges

For three charges, the total electrostatic potential energy is the sum of all distinct pair energies. There are three pairs: (q₁, q₂), (q₂, q₃) and (q₃, q₁).

Pairwise derivation

Bring q₁ from infinity. No other charge is present, so W₁ = 0.
Bring q₂ to distance r₁₂ from q₁. Work stored is W₂ = kq₁q₂/r₁₂.
Bring q₃. It interacts with q₁ and q₂, so W₃ = kq₃q₁/r₃₁ + kq₃q₂/r₂₃.
Total energy is U = W₁ + W₂ + W₃.

U = (1/4πε₀) [q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁]

Pairwise addition is used because electrostatic forces obey superposition. Each pair stores its own interaction energy, and the total is obtained by adding every distinct pair once.

Electrostatics Master Resource

5. Potential Energy of Four Charges

Four charges have six distinct pairs. The safest method is to list every pair systematically and stop only when each pair has appeared exactly once.

Complete expression

Pairs involving q₁: q₁q₂, q₁q₃, q₁q₄.
Pairs involving q₂ that have not already appeared: q₂q₃, q₂q₄.
Remaining pair involving q₃: q₃q₄.
Add the six terms with their own separation distances.

U = k[q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₁q₄/r₁₄ + q₂q₃/r₂₃ + q₂q₄/r₂₄ + q₃q₄/r₃₄]

Every missing term means one interaction has been ignored. Every repeated term means double counting. In square problems, the four side pairs and two diagonal pairs must be treated with different distances.

Electrostatics Master Resource

6. General Formula for N Charges

For N charges, the total energy is the sum of the energy of every distinct pair. The notation i < j is used so each pair is counted once.

U = (1/4πε₀) ∑_i<j (q_iq_j/r_ij)

Why the number of interactions is n(n - 1)/2

Each charge can pair with n - 1 other charges.
This gives n(n - 1) ordered choices if q_iq_j and q_jq_i are both counted.
But each physical pair has been counted twice.
Therefore the number of distinct interactions is n(n - 1)/2.

Number of charges	Distinct pairs	Example of pair count
2	1	Only 12
3	3	12, 23, 31
4	6	Four sides and two diagonals in a square
5	10	Five nearest-neighbor pairs plus five diagonal pairs in a pentagon
6	15	Six sides, six short diagonals and three long diagonals in a hexagon

Electrostatics Master Resource

7. Charges at Vertices of an Equilateral Triangle

Let three equal charges +q be placed at the vertices of an equilateral triangle of side a. All three separations are equal to a.

Equal positive charges

Pair 12 contributes kq²/a.
Pair 23 contributes kq²/a.
Pair 31 contributes kq²/a.
Total potential energy is the sum of the three equal terms.

U = 3(kq²/a)

One charge negative

For +q, +q, -q: U = kq²/a - kq²/a - kq²/a = -kq²/a.

Two charges negative

For +q, -q, -q: U = -kq²/a + kq²/a - kq²/a = -kq²/a.

Electrostatics Master Resource

8. Charges at Vertices of a Square

Let four equal charges +q be placed at the vertices of a square of side a. The square has four side pairs and two diagonal pairs. Side distance is a, and diagonal distance is √2 a.

Complete square derivation

Four side pairs each contribute kq²/a, so side contribution is 4kq²/a.
Two diagonal pairs each contribute kq²/(√2 a), so diagonal contribution is 2kq²/(√2 a).
Add side and diagonal contributions.

U = kq²[4/a + 2/(√2 a)]

U = (kq²/a)(4 + √2)

The common mistake is to use a for the diagonals. The diagonal of a square is longer, so diagonal terms are smaller.

Electrostatics Master Resource

9. Charges at Vertices of Regular Polygons

Regular polygon questions are pair counting questions. The charges may all be equal, but the separations are not all equal once diagonals appear.

Polygon	Pairs	Useful grouping	Method
Pentagon	10	5 sides and 5 diagonals	U = 5kq²/a + 5kq²/d, where d is the pentagon diagonal.
Hexagon	15	6 sides, 6 short diagonals, 3 opposite pairs	Use distances a, √3a and 2a for a regular hexagon.
General n-gon	n(n - 1)/2	Group pairs by equal chord length	Add number of pairs at each distance multiplied by kq²/distance.

For a regular polygon on a circle of radius R, the distance between vertices separated by m sides is chord length 2R sin(mπ/n). This is often the fastest route for JEE Advanced and Olympiad-style extensions.

U = kq² ∑ (number of pairs at a chord length)/(that chord length)

Electrostatics Master Resource

10. Work Done in Assembling a System of Charges

Assembling a system means bringing charges from infinity one by one and placing them at their final positions without changing kinetic energy. The total external work done equals the final electrostatic potential energy of the system.

Step-by-step assembly logic

Choose the zero of potential energy when all charges are infinitely separated.
Bring the first charge. There is no electric field due to other placed charges, so no work is required.
Bring the second charge. Work is required only against the field of the first charge.
Bring the third charge. Work is required against the fields of the first and second charges.
Continue this pattern. Every new charge forms pair interactions with all previously placed charges.
When all charges are placed, each distinct pair has appeared exactly once, so W_external = U.

Exam Shortcut

You do not have to simulate the actual path if the final geometry is known. Write all distinct pair terms directly, preserving signs.

Electrostatics Master Resource

11. Potential Energy Curves

As r increases, positive potential energy falls toward zero.

Attraction gives negative potential energy that approaches zero from below.

Like charges

For like charges, q₁q₂ > 0, so U = kq₁q₂/r is positive. As r approaches infinity, U approaches zero from above. As r approaches zero, the mathematical model predicts U approaches +infinity.

Unlike charges

For unlike charges, q₁q₂ < 0, so U is negative. As r approaches infinity, U approaches zero from below. As r approaches zero, U becomes very large negative in the point-charge model.

Electrostatics Master Resource

12. Stability of Charge Configurations

Stability is decided by how potential energy changes near an equilibrium position. A system is stable if a small displacement increases potential energy and produces a restoring tendency.

Minimum of potential energy.

Maximum of potential energy.

Constant of potential energy.

Type	Energy shape	Meaning	Charge-system interpretation
Stable equilibrium	Local minimum	Small displacement raises U	The system tends to return if constraints permit.
Unstable equilibrium	Local maximum	Small displacement lowers U	The system moves away after disturbance.
Neutral equilibrium	Flat U	Small displacement does not change U	No restoring or separating tendency from energy change.

Electrostatics Master Resource

13. Electrostatic Potential Energy of a Charged Conducting Sphere

For an isolated conducting sphere of radius R and charge Q, charge resides on the surface. To charge the sphere gradually, bring small charge dq from infinity to the sphere when it already has charge q.

Derivation of sphere energy

Potential of a conducting sphere carrying charge q is V = kq/R.
Small work done to bring dq is dW = Vdq = (kq/R)dq.
Total work is integral from q = 0 to Q: W = ∫₀^Q (kq/R)dq.
W = (k/R)[Q²/2].
Since k = 1/(4πε₀), U = Q²/(8πε₀R).

U = Q²/(8πε₀R) = kQ²/(2R)

This energy is physically stored in the electric field surrounding the sphere. A smaller sphere with the same charge has greater energy because the field is more intense near the surface.

Electrostatics Master Resource

14. Energy Stored in Electric Field

The charge-system formula is one way to calculate electrostatic energy. A deeper view says the energy is stored in the electric field itself. Wherever an electric field exists, space contains energy density.

u = (1/2)ε₀E²

This idea prepares students for capacitors. In a parallel-plate capacitor, the energy can be written as U = (1/2)CV², and it is stored mainly in the uniform electric field between the plates.

For point-charge systems, field-energy calculations require integration over space and can be advanced. For school and entrance exams, pairwise potential energy is usually the fastest method for discrete charges.

Electrostatics Master Resource

15. Important Formula Sheet

Two charges

U = kq₁q₂/r

Three charges

U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁)

Four charges

U = k sum of all six pair terms

N charges

U = k ∑_i<j q_iq_j/r_ij

Number of interactions

n(n - 1)/2

Equilateral triangle

U = 3kq²/a for three +q charges

Square

U = (kq²/a)(4 + √2) for four +q charges

Conducting sphere

U = Q²/(8πε₀R) = kQ²/2R

Energy density

u = (1/2)ε₀E²

Electrostatics Master Resource

16. Solved Examples

The following 120 solved examples are original practice problems arranged by exam style. Each solution uses the same reliable strategy: identify pairs, preserve signs, substitute distances in metres and add algebraically.

CBSE Solved Examples (20)

CBSE Example 1: Work needed to change separation

Two equal charges 7 µC and 7 µC are moved slowly from 32 cm separation to 16 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 1.378 x 10^+0 J.
Final energy U_f = kq²/r_f = 2.756 x 10^+0 J.
W_external = U_f - U_i = 1.378 x 10^+0 J.
The work is positive because like charges are being pushed closer.

CBSE Example 2: Three-charge system with unequal distances

Charges q₁ = 2 µC, q₂ = -4 µC and q₃ = 2 µC have separations r₁₂ = 10 cm, r₂₃ = 18 cm and r₃₁ = 17 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -7.200 x 10^-1 J.
Term 23 = -4.000 x 10^-1 J.
Term 31 = 2.118 x 10^-1 J.
Total U = -9.082 x 10^-1 J after preserving all charge signs.

CBSE Example 3: Equilateral triangle configuration

An equilateral triangle of side 12 cm carries two positive charges and one negative charge, each of magnitude 90 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.12 m.
For the given signs, the algebraic pair multiplier is -1.
U = -1kq²/a.
U = -1(9 x 10⁹)(90 x 10^-9)²/0.12.
U = -6.075 x 10^-4 J.

CBSE Example 4: Four equal charges on a square

Four equal charges 20 nC are placed at the corners of a square of side 10 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.1 m and q = 20 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.1 + 2/(0.141)].
U = 1.949 x 10^-4 J.

CBSE Example 5: Charged conducting sphere

A conducting sphere of radius 28 cm is charged to 2 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 2 x 10^-6 C and R = 0.28 m.
U = (9 x 10⁹)(2 x 10^-6)²/(2 x 0.28).
U = 6.429 x 10^-2 J.
The energy is stored in the electric field outside the conducting sphere.

CBSE Example 6: Pair counting for identical separations

6 equal charges of 1 µC are arranged so every pair is separated by 40 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 6(5)/2 = 15.
Energy of each pair = kq²/r.
Total U = 15kq²/r.
Substitution gives U = 15(9 x 10⁹)(1 x 10^-6)²/0.4.
U = 3.375 x 10^-1 J.

CBSE Example 7: Energy ratio from graph behavior

For charges 6 µC and 6 µC, potential energy at separation r = 6 cm is U₀. What is the potential energy when the separation becomes 5r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 5r, the energy becomes U₀/5.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/5.

CBSE Example 8: All six terms for four charges

Four charges 2 µC, -2 µC, 6 µC, -5 µC have separations r₁₂ = 12 cm, r₁₃ = 15 cm, r₁₄ = 18 cm, r₂₃ = 16 cm, r₂₄ = 19 cm, r₃₄ = 22 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -3.000 x 10^-1 J, 7.200 x 10^-1 J, -5.000 x 10^-1 J, -6.750 x 10^-1 J, 4.737 x 10^-1 J, -1.227 x 10^+0 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -1.509 x 10^+0 J.

CBSE Example 9: Assembly order with three charges

Charges 5 µC, -2 µC and 2 µC are assembled at the corners of an equilateral triangle of side 15 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -6.000 x 10^-1 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = 3.600 x 10^-1 J.
Total work equals total potential energy: U = -2.400 x 10^-1 J.

CBSE Example 10: Two-charge energy with sign control

Two point charges 2 µC and 6 µC are separated by 27 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 2 x 10^-6 C and q₂ = 6 x 10^-6 C.
Convert distance: r = 0.27 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(2 x 10^-6)(6 x 10^-6)/0.27.
U = 4.000 x 10^-1 J. The sign is positive because the charges are like charges.

CBSE Example 11: Work needed to change separation

Two equal charges 9 µC and 9 µC are moved slowly from 16 cm separation to 8 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 4.556 x 10^+0 J.
Final energy U_f = kq²/r_f = 9.112 x 10^+0 J.
W_external = U_f - U_i = 4.556 x 10^+0 J.
The work is positive because like charges are being pushed closer.

CBSE Example 12: Three-charge system with unequal distances

Charges q₁ = 2 µC, q₂ = -2 µC and q₃ = 6 µC have separations r₁₂ = 10 cm, r₂₃ = 12 cm and r₃₁ = 17 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -3.600 x 10^-1 J.
Term 23 = -9.000 x 10^-1 J.
Term 31 = 6.353 x 10^-1 J.
Total U = -6.247 x 10^-1 J after preserving all charge signs.

CBSE Example 13: Equilateral triangle configuration

An equilateral triangle of side 6 cm carries one positive charge and two negative charges, each of magnitude 100 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.06 m.
For the given signs, the algebraic pair multiplier is -1.
U = -1kq²/a.
U = -1(9 x 10⁹)(100 x 10^-9)²/0.06.
U = -1.500 x 10^-3 J.

CBSE Example 14: Four equal charges on a square

Four equal charges 40 nC are placed at the corners of a square of side 16 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.16 m and q = 40 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.16 + 2/(0.226)].
U = 4.873 x 10^-4 J.

CBSE Example 15: Charged conducting sphere

A conducting sphere of radius 28 cm is charged to 3 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 3 x 10^-6 C and R = 0.28 m.
U = (9 x 10⁹)(3 x 10^-6)²/(2 x 0.28).
U = 1.446 x 10^-1 J.
The energy is stored in the electric field outside the conducting sphere.

CBSE Example 16: Pair counting for identical separations

8 equal charges of 1 µC are arranged so every pair is separated by 30 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 8(7)/2 = 28.
Energy of each pair = kq²/r.
Total U = 28kq²/r.
Substitution gives U = 28(9 x 10⁹)(1 x 10^-6)²/0.3.
U = 8.400 x 10^-1 J.

CBSE Example 17: Energy ratio from graph behavior

For charges 4 µC and 2 µC, potential energy at separation r = 6 cm is U₀. What is the potential energy when the separation becomes 3r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 3r, the energy becomes U₀/3.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/3.

CBSE Example 18: All six terms for four charges

Four charges 3 µC, -1 µC, 4 µC, -4 µC have separations r₁₂ = 12 cm, r₁₃ = 15 cm, r₁₄ = 18 cm, r₂₃ = 16 cm, r₂₄ = 19 cm, r₃₄ = 22 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -2.250 x 10^-1 J, 7.200 x 10^-1 J, -6.000 x 10^-1 J, -2.250 x 10^-1 J, 1.895 x 10^-1 J, -6.545 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -7.951 x 10^-1 J.

CBSE Example 19: Assembly order with three charges

Charges 5 µC, -6 µC and 2 µC are assembled at the corners of an equilateral triangle of side 17 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -1.588 x 10^+0 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = -1.059 x 10^-1 J.
Total work equals total potential energy: U = -1.694 x 10^+0 J.

CBSE Example 20: Two-charge energy with sign control

Two point charges 5 µC and 4 µC are separated by 32 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 5 x 10^-6 C and q₂ = 4 x 10^-6 C.
Convert distance: r = 0.32 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(5 x 10^-6)(4 x 10^-6)/0.32.
U = 5.625 x 10^-1 J. The sign is positive because the charges are like charges.

NEET Solved Examples (20)

NEET Example 1: Four equal charges on a square

Four equal charges 20 nC are placed at the corners of a square of side 10 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.1 m and q = 20 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.1 + 2/(0.141)].
U = 1.949 x 10^-4 J.

NEET Example 2: Charged conducting sphere

A conducting sphere of radius 28 cm is charged to 2 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 2 x 10^-6 C and R = 0.28 m.
U = (9 x 10⁹)(2 x 10^-6)²/(2 x 0.28).
U = 6.429 x 10^-2 J.
The energy is stored in the electric field outside the conducting sphere.

NEET Example 3: Pair counting for identical separations

6 equal charges of 1 µC are arranged so every pair is separated by 40 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 6(5)/2 = 15.
Energy of each pair = kq²/r.
Total U = 15kq²/r.
Substitution gives U = 15(9 x 10⁹)(1 x 10^-6)²/0.4.
U = 3.375 x 10^-1 J.

NEET Example 4: Energy ratio from graph behavior

For charges 6 µC and 6 µC, potential energy at separation r = 6 cm is U₀. What is the potential energy when the separation becomes 5r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 5r, the energy becomes U₀/5.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/5.

NEET Example 5: All six terms for four charges

Four charges 2 µC, -2 µC, 6 µC, -5 µC have separations r₁₂ = 12 cm, r₁₃ = 15 cm, r₁₄ = 18 cm, r₂₃ = 16 cm, r₂₄ = 19 cm, r₃₄ = 22 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -3.000 x 10^-1 J, 7.200 x 10^-1 J, -5.000 x 10^-1 J, -6.750 x 10^-1 J, 4.737 x 10^-1 J, -1.227 x 10^+0 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -1.509 x 10^+0 J.

NEET Example 6: Assembly order with three charges

Charges 5 µC, -2 µC and 2 µC are assembled at the corners of an equilateral triangle of side 15 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -6.000 x 10^-1 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = 3.600 x 10^-1 J.
Total work equals total potential energy: U = -2.400 x 10^-1 J.

NEET Example 7: Two-charge energy with sign control

Two point charges 2 µC and 6 µC are separated by 27 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 2 x 10^-6 C and q₂ = 6 x 10^-6 C.
Convert distance: r = 0.27 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(2 x 10^-6)(6 x 10^-6)/0.27.
U = 4.000 x 10^-1 J. The sign is positive because the charges are like charges.

NEET Example 8: Work needed to change separation

Two equal charges 9 µC and 9 µC are moved slowly from 16 cm separation to 8 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 4.556 x 10^+0 J.
Final energy U_f = kq²/r_f = 9.112 x 10^+0 J.
W_external = U_f - U_i = 4.556 x 10^+0 J.
The work is positive because like charges are being pushed closer.

NEET Example 9: Three-charge system with unequal distances

Charges q₁ = 2 µC, q₂ = -2 µC and q₃ = 6 µC have separations r₁₂ = 10 cm, r₂₃ = 12 cm and r₃₁ = 17 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -3.600 x 10^-1 J.
Term 23 = -9.000 x 10^-1 J.
Term 31 = 6.353 x 10^-1 J.
Total U = -6.247 x 10^-1 J after preserving all charge signs.

NEET Example 10: Equilateral triangle configuration

An equilateral triangle of side 6 cm carries one positive charge and two negative charges, each of magnitude 100 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.06 m.
For the given signs, the algebraic pair multiplier is -1.
U = -1kq²/a.
U = -1(9 x 10⁹)(100 x 10^-9)²/0.06.
U = -1.500 x 10^-3 J.

NEET Example 11: Four equal charges on a square

Four equal charges 40 nC are placed at the corners of a square of side 16 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.16 m and q = 40 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.16 + 2/(0.226)].
U = 4.873 x 10^-4 J.

NEET Example 12: Charged conducting sphere

A conducting sphere of radius 28 cm is charged to 3 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 3 x 10^-6 C and R = 0.28 m.
U = (9 x 10⁹)(3 x 10^-6)²/(2 x 0.28).
U = 1.446 x 10^-1 J.
The energy is stored in the electric field outside the conducting sphere.

NEET Example 13: Pair counting for identical separations

8 equal charges of 1 µC are arranged so every pair is separated by 30 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 8(7)/2 = 28.
Energy of each pair = kq²/r.
Total U = 28kq²/r.
Substitution gives U = 28(9 x 10⁹)(1 x 10^-6)²/0.3.
U = 8.400 x 10^-1 J.

NEET Example 14: Energy ratio from graph behavior

For charges 4 µC and 2 µC, potential energy at separation r = 6 cm is U₀. What is the potential energy when the separation becomes 3r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 3r, the energy becomes U₀/3.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/3.

NEET Example 15: All six terms for four charges

Four charges 3 µC, -1 µC, 4 µC, -4 µC have separations r₁₂ = 12 cm, r₁₃ = 15 cm, r₁₄ = 18 cm, r₂₃ = 16 cm, r₂₄ = 19 cm, r₃₄ = 22 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -2.250 x 10^-1 J, 7.200 x 10^-1 J, -6.000 x 10^-1 J, -2.250 x 10^-1 J, 1.895 x 10^-1 J, -6.545 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -7.951 x 10^-1 J.

NEET Example 16: Assembly order with three charges

Charges 5 µC, -6 µC and 2 µC are assembled at the corners of an equilateral triangle of side 17 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -1.588 x 10^+0 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = -1.059 x 10^-1 J.
Total work equals total potential energy: U = -1.694 x 10^+0 J.

NEET Example 17: Two-charge energy with sign control

Two point charges 5 µC and 4 µC are separated by 32 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 5 x 10^-6 C and q₂ = 4 x 10^-6 C.
Convert distance: r = 0.32 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(5 x 10^-6)(4 x 10^-6)/0.32.
U = 5.625 x 10^-1 J. The sign is positive because the charges are like charges.

NEET Example 18: Work needed to change separation

Two equal charges 3 µC and 3 µC are moved slowly from 28 cm separation to 14 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 2.893 x 10^-1 J.
Final energy U_f = kq²/r_f = 5.786 x 10^-1 J.
W_external = U_f - U_i = 2.893 x 10^-1 J.
The work is positive because like charges are being pushed closer.

NEET Example 19: Three-charge system with unequal distances

Charges q₁ = 2 µC, q₂ = -4 µC and q₃ = 4 µC have separations r₁₂ = 10 cm, r₂₃ = 18 cm and r₃₁ = 17 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -7.200 x 10^-1 J.
Term 23 = -8.000 x 10^-1 J.
Term 31 = 4.235 x 10^-1 J.
Total U = -1.096 x 10^+0 J after preserving all charge signs.

NEET Example 20: Equilateral triangle configuration

An equilateral triangle of side 8 cm carries three equal positive charges, each of magnitude 20 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.08 m.
For the given signs, the algebraic pair multiplier is 3.
U = 3kq²/a.
U = 3(9 x 10⁹)(20 x 10^-9)²/0.08.
U = 1.350 x 10^-4 J.

JEE Main Solved Examples (20)

JEE Main Example 1: Energy ratio from graph behavior

For charges 4 µC and 3 µC, potential energy at separation r = 10 cm is U₀. What is the potential energy when the separation becomes 5r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 5r, the energy becomes U₀/5.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/5.

JEE Main Example 2: All six terms for four charges

Four charges 3 µC, -1 µC, 4 µC, -4 µC have separations r₁₂ = 11 cm, r₁₃ = 14 cm, r₁₄ = 17 cm, r₂₃ = 20 cm, r₂₄ = 18 cm, r₃₄ = 21 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -2.455 x 10^-1 J, 7.714 x 10^-1 J, -6.353 x 10^-1 J, -1.800 x 10^-1 J, 2.000 x 10^-1 J, -6.857 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -7.750 x 10^-1 J.

JEE Main Example 3: Assembly order with three charges

Charges 4 µC, -6 µC and 6 µC are assembled at the corners of an equilateral triangle of side 11 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -1.964 x 10^+0 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = -9.818 x 10^-1 J.
Total work equals total potential energy: U = -2.945 x 10^+0 J.

JEE Main Example 4: Two-charge energy with sign control

Two point charges 6 µC and 4 µC are separated by 14 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 6 x 10^-6 C and q₂ = 4 x 10^-6 C.
Convert distance: r = 0.14 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(6 x 10^-6)(4 x 10^-6)/0.14.
U = 1.543 x 10^+0 J. The sign is positive because the charges are like charges.

JEE Main Example 5: Work needed to change separation

Two equal charges 5 µC and 5 µC are moved slowly from 32 cm separation to 16 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 7.031 x 10^-1 J.
Final energy U_f = kq²/r_f = 1.406 x 10^+0 J.
W_external = U_f - U_i = 7.031 x 10^-1 J.
The work is positive because like charges are being pushed closer.

JEE Main Example 6: Three-charge system with unequal distances

Charges q₁ = 1 µC, q₂ = -2 µC and q₃ = 4 µC have separations r₁₂ = 8 cm, r₂₃ = 12 cm and r₃₁ = 15 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -2.250 x 10^-1 J.
Term 23 = -6.000 x 10^-1 J.
Term 31 = 2.400 x 10^-1 J.
Total U = -5.850 x 10^-1 J after preserving all charge signs.

JEE Main Example 7: Equilateral triangle configuration

An equilateral triangle of side 10 cm carries three equal positive charges, each of magnitude 50 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.1 m.
For the given signs, the algebraic pair multiplier is 3.
U = 3kq²/a.
U = 3(9 x 10⁹)(50 x 10^-9)²/0.1.
U = 6.750 x 10^-4 J.

JEE Main Example 8: Four equal charges on a square

Four equal charges 80 nC are placed at the corners of a square of side 10 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.1 m and q = 80 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.1 + 2/(0.141)].
U = 3.119 x 10^-3 J.

JEE Main Example 9: Charged conducting sphere

A conducting sphere of radius 16 cm is charged to 7 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 7 x 10^-6 C and R = 0.16 m.
U = (9 x 10⁹)(7 x 10^-6)²/(2 x 0.16).
U = 1.378 x 10^+0 J.
The energy is stored in the electric field outside the conducting sphere.

JEE Main Example 10: Pair counting for identical separations

4 equal charges of 5 µC are arranged so every pair is separated by 20 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 4(3)/2 = 6.
Energy of each pair = kq²/r.
Total U = 6kq²/r.
Substitution gives U = 6(9 x 10⁹)(5 x 10^-6)²/0.2.
U = 6.750 x 10^+0 J.

JEE Main Example 11: Energy ratio from graph behavior

For charges 2 µC and 6 µC, potential energy at separation r = 10 cm is U₀. What is the potential energy when the separation becomes 3r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 3r, the energy becomes U₀/3.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/3.

JEE Main Example 12: All six terms for four charges

Four charges 4 µC, -3 µC, 5 µC, -3 µC have separations r₁₂ = 11 cm, r₁₃ = 14 cm, r₁₄ = 17 cm, r₂₃ = 20 cm, r₂₄ = 18 cm, r₃₄ = 21 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -9.818 x 10^-1 J, 1.286 x 10^+0 J, -6.353 x 10^-1 J, -6.750 x 10^-1 J, 4.500 x 10^-1 J, -6.429 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -1.199 x 10^+0 J.

JEE Main Example 13: Assembly order with three charges

Charges 4 µC, -4 µC and 6 µC are assembled at the corners of an equilateral triangle of side 13 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -1.108 x 10^+0 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = 0 J.
Total work equals total potential energy: U = -1.108 x 10^+0 J.

JEE Main Example 14: Two-charge energy with sign control

Two point charges 2 µC and 2 µC are separated by 19 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 2 x 10^-6 C and q₂ = 2 x 10^-6 C.
Convert distance: r = 0.19 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(2 x 10^-6)(2 x 10^-6)/0.19.
U = 1.895 x 10^-1 J. The sign is positive because the charges are like charges.

JEE Main Example 15: Work needed to change separation

Two equal charges 7 µC and 7 µC are moved slowly from 16 cm separation to 8 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 2.756 x 10^+0 J.
Final energy U_f = kq²/r_f = 5.513 x 10^+0 J.
W_external = U_f - U_i = 2.756 x 10^+0 J.
The work is positive because like charges are being pushed closer.

JEE Main Example 16: Three-charge system with unequal distances

Charges q₁ = 1 µC, q₂ = -4 µC and q₃ = 2 µC have separations r₁₂ = 8 cm, r₂₃ = 18 cm and r₃₁ = 15 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -4.500 x 10^-1 J.
Term 23 = -4.000 x 10^-1 J.
Term 31 = 1.200 x 10^-1 J.
Total U = -7.300 x 10^-1 J after preserving all charge signs.

JEE Main Example 17: Equilateral triangle configuration

An equilateral triangle of side 12 cm carries two positive charges and one negative charge, each of magnitude 60 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.12 m.
For the given signs, the algebraic pair multiplier is -1.
U = -1kq²/a.
U = -1(9 x 10⁹)(60 x 10^-9)²/0.12.
U = -2.700 x 10^-4 J.

JEE Main Example 18: Four equal charges on a square

Four equal charges 20 nC are placed at the corners of a square of side 16 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.16 m and q = 20 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.16 + 2/(0.226)].
U = 1.218 x 10^-4 J.

JEE Main Example 19: Charged conducting sphere

A conducting sphere of radius 16 cm is charged to 8 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 8 x 10^-6 C and R = 0.16 m.
U = (9 x 10⁹)(8 x 10^-6)²/(2 x 0.16).
U = 1.800 x 10^+0 J.
The energy is stored in the electric field outside the conducting sphere.

JEE Main Example 20: Pair counting for identical separations

6 equal charges of 5 µC are arranged so every pair is separated by 40 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 6(5)/2 = 15.
Energy of each pair = kq²/r.
Total U = 15kq²/r.
Substitution gives U = 15(9 x 10⁹)(5 x 10^-6)²/0.4.
U = 8.437 x 10^+0 J.

JEE Advanced Solved Examples (20)

JEE Advanced Example 1: Two-charge energy with sign control

Two point charges 3 µC and 2 µC are separated by 26 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 3 x 10^-6 C and q₂ = 2 x 10^-6 C.
Convert distance: r = 0.26 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(3 x 10^-6)(2 x 10^-6)/0.26.
U = 2.077 x 10^-1 J. The sign is positive because the charges are like charges.

JEE Advanced Example 2: Work needed to change separation

Two equal charges 9 µC and 9 µC are moved slowly from 20 cm separation to 10 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 3.645 x 10^+0 J.
Final energy U_f = kq²/r_f = 7.290 x 10^+0 J.
W_external = U_f - U_i = 3.645 x 10^+0 J.
The work is positive because like charges are being pushed closer.

JEE Advanced Example 3: Three-charge system with unequal distances

Charges q₁ = 5 µC, q₂ = -2 µC and q₃ = 2 µC have separations r₁₂ = 16 cm, r₂₃ = 12 cm and r₃₁ = 23 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -5.625 x 10^-1 J.
Term 23 = -3.000 x 10^-1 J.
Term 31 = 3.913 x 10^-1 J.
Total U = -4.712 x 10^-1 J after preserving all charge signs.

JEE Advanced Example 4: Equilateral triangle configuration

An equilateral triangle of side 6 cm carries two positive charges and one negative charge, each of magnitude 90 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.06 m.
For the given signs, the algebraic pair multiplier is -1.
U = -1kq²/a.
U = -1(9 x 10⁹)(90 x 10^-9)²/0.06.
U = -1.215 x 10^-3 J.

JEE Advanced Example 5: Four equal charges on a square

Four equal charges 40 nC are placed at the corners of a square of side 18 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.18 m and q = 40 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.18 + 2/(0.255)].
U = 4.331 x 10^-4 J.

JEE Advanced Example 6: Charged conducting sphere

A conducting sphere of radius 24 cm is charged to 2 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 2 x 10^-6 C and R = 0.24 m.
U = (9 x 10⁹)(2 x 10^-6)²/(2 x 0.24).
U = 7.500 x 10^-2 J.
The energy is stored in the electric field outside the conducting sphere.

JEE Advanced Example 7: Pair counting for identical separations

8 equal charges of 4 µC are arranged so every pair is separated by 40 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 8(7)/2 = 28.
Energy of each pair = kq²/r.
Total U = 28kq²/r.
Substitution gives U = 28(9 x 10⁹)(4 x 10^-6)²/0.4.
U = 1.008 x 10^+1 J.

JEE Advanced Example 8: Energy ratio from graph behavior

For charges 6 µC and 3 µC, potential energy at separation r = 14 cm is U₀. What is the potential energy when the separation becomes 3r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 3r, the energy becomes U₀/3.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/3.

JEE Advanced Example 9: All six terms for four charges

Four charges 2 µC, -2 µC, 6 µC, -5 µC have separations r₁₂ = 10 cm, r₁₃ = 13 cm, r₁₄ = 16 cm, r₂₃ = 19 cm, r₂₄ = 22 cm, r₃₄ = 20 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -3.600 x 10^-1 J, 8.308 x 10^-1 J, -5.625 x 10^-1 J, -5.684 x 10^-1 J, 4.091 x 10^-1 J, -1.350 x 10^+0 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -1.601 x 10^+0 J.

JEE Advanced Example 10: Assembly order with three charges

Charges 3 µC, -2 µC and 5 µC are assembled at the corners of an equilateral triangle of side 17 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -3.176 x 10^-1 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = 2.647 x 10^-1 J.
Total work equals total potential energy: U = -5.294 x 10^-2 J.

JEE Advanced Example 11: Two-charge energy with sign control

Two point charges 6 µC and 6 µC are separated by 31 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 6 x 10^-6 C and q₂ = 6 x 10^-6 C.
Convert distance: r = 0.31 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(6 x 10^-6)(6 x 10^-6)/0.31.
U = 1.045 x 10^+0 J. The sign is positive because the charges are like charges.

JEE Advanced Example 12: Work needed to change separation

Two equal charges 3 µC and 3 µC are moved slowly from 32 cm separation to 16 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 2.531 x 10^-1 J.
Final energy U_f = kq²/r_f = 5.062 x 10^-1 J.
W_external = U_f - U_i = 2.531 x 10^-1 J.
The work is positive because like charges are being pushed closer.

JEE Advanced Example 13: Three-charge system with unequal distances

Charges q₁ = 5 µC, q₂ = -4 µC and q₃ = 6 µC have separations r₁₂ = 16 cm, r₂₃ = 18 cm and r₃₁ = 23 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -1.125 x 10^+0 J.
Term 23 = -1.200 x 10^+0 J.
Term 31 = 1.174 x 10^+0 J.
Total U = -1.151 x 10^+0 J after preserving all charge signs.

JEE Advanced Example 14: Equilateral triangle configuration

An equilateral triangle of side 8 cm carries one positive charge and two negative charges, each of magnitude 100 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.08 m.
For the given signs, the algebraic pair multiplier is -1.
U = -1kq²/a.
U = -1(9 x 10⁹)(100 x 10^-9)²/0.08.
U = -1.125 x 10^-3 J.

JEE Advanced Example 15: Four equal charges on a square

Four equal charges 60 nC are placed at the corners of a square of side 10 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.1 m and q = 60 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.1 + 2/(0.141)].
U = 1.754 x 10^-3 J.

JEE Advanced Example 16: Charged conducting sphere

A conducting sphere of radius 24 cm is charged to 3 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 3 x 10^-6 C and R = 0.24 m.
U = (9 x 10⁹)(3 x 10^-6)²/(2 x 0.24).
U = 1.688 x 10^-1 J.
The energy is stored in the electric field outside the conducting sphere.

JEE Advanced Example 17: Pair counting for identical separations

10 equal charges of 4 µC are arranged so every pair is separated by 30 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 10(9)/2 = 45.
Energy of each pair = kq²/r.
Total U = 45kq²/r.
Substitution gives U = 45(9 x 10⁹)(4 x 10^-6)²/0.3.
U = 2.160 x 10^+1 J.

JEE Advanced Example 18: Energy ratio from graph behavior

For charges 4 µC and 6 µC, potential energy at separation r = 14 cm is U₀. What is the potential energy when the separation becomes 5r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 5r, the energy becomes U₀/5.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/5.

JEE Advanced Example 19: All six terms for four charges

Four charges 3 µC, -1 µC, 4 µC, -4 µC have separations r₁₂ = 10 cm, r₁₃ = 13 cm, r₁₄ = 16 cm, r₂₃ = 19 cm, r₂₄ = 22 cm, r₃₄ = 20 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -2.700 x 10^-1 J, 8.308 x 10^-1 J, -6.750 x 10^-1 J, -1.895 x 10^-1 J, 1.636 x 10^-1 J, -7.200 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -8.601 x 10^-1 J.

JEE Advanced Example 20: Assembly order with three charges

Charges 3 µC, -6 µC and 5 µC are assembled at the corners of an equilateral triangle of side 11 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -1.473 x 10^+0 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = -1.227 x 10^+0 J.
Total work equals total potential energy: U = -2.700 x 10^+0 J.

IB Physics Solved Examples (10)

IB Physics Example 1: Equilateral triangle configuration

An equilateral triangle of side 12 cm carries one positive charge and two negative charges, each of magnitude 70 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.12 m.
For the given signs, the algebraic pair multiplier is -1.
U = -1kq²/a.
U = -1(9 x 10⁹)(70 x 10^-9)²/0.12.
U = -3.675 x 10^-4 J.

IB Physics Example 2: Four equal charges on a square

Four equal charges 20 nC are placed at the corners of a square of side 14 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.14 m and q = 20 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.14 + 2/(0.198)].
U = 1.392 x 10^-4 J.

IB Physics Example 3: Charged conducting sphere

A conducting sphere of radius 20 cm is charged to 9 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 9 x 10^-6 C and R = 0.2 m.
U = (9 x 10⁹)(9 x 10^-6)²/(2 x 0.2).
U = 1.822 x 10^+0 J.
The energy is stored in the electric field outside the conducting sphere.

IB Physics Example 4: Pair counting for identical separations

6 equal charges of 2 µC are arranged so every pair is separated by 30 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 6(5)/2 = 15.
Energy of each pair = kq²/r.
Total U = 15kq²/r.
Substitution gives U = 15(9 x 10⁹)(2 x 10^-6)²/0.3.
U = 1.800 x 10^+0 J.

IB Physics Example 5: Energy ratio from graph behavior

For charges 4 µC and 8 µC, potential energy at separation r = 12 cm is U₀. What is the potential energy when the separation becomes 5r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 5r, the energy becomes U₀/5.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/5.

IB Physics Example 6: All six terms for four charges

Four charges 3 µC, -1 µC, 4 µC, -4 µC have separations r₁₂ = 13 cm, r₁₃ = 16 cm, r₁₄ = 14 cm, r₂₃ = 17 cm, r₂₄ = 20 cm, r₃₄ = 23 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -2.077 x 10^-1 J, 6.750 x 10^-1 J, -7.714 x 10^-1 J, -2.118 x 10^-1 J, 1.800 x 10^-1 J, -6.261 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -9.620 x 10^-1 J.

IB Physics Example 7: Assembly order with three charges

Charges 6 µC, -6 µC and 3 µC are assembled at the corners of an equilateral triangle of side 15 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -2.160 x 10^+0 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = 0 J.
Total work equals total potential energy: U = -2.160 x 10^+0 J.

IB Physics Example 8: Two-charge energy with sign control

Two point charges 4 µC and 4 µC are separated by 25 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 4 x 10^-6 C and q₂ = 4 x 10^-6 C.
Convert distance: r = 0.25 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(4 x 10^-6)(4 x 10^-6)/0.25.
U = 5.760 x 10^-1 J. The sign is positive because the charges are like charges.

IB Physics Example 9: Work needed to change separation

Two equal charges 9 µC and 9 µC are moved slowly from 24 cm separation to 12 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 3.038 x 10^+0 J.
Final energy U_f = kq²/r_f = 6.075 x 10^+0 J.
W_external = U_f - U_i = 3.038 x 10^+0 J.
The work is positive because like charges are being pushed closer.

IB Physics Example 10: Three-charge system with unequal distances

Charges q₁ = 3 µC, q₂ = -2 µC and q₃ = 4 µC have separations r₁₂ = 12 cm, r₂₃ = 12 cm and r₃₁ = 19 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -4.500 x 10^-1 J.
Term 23 = -6.000 x 10^-1 J.
Term 31 = 5.684 x 10^-1 J.
Total U = -4.816 x 10^-1 J after preserving all charge signs.

IGCSE Solved Examples (10)

IGCSE Example 1: Pair counting for identical separations

9 equal charges of 2 µC are arranged so every pair is separated by 35 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 9(8)/2 = 36.
Energy of each pair = kq²/r.
Total U = 36kq²/r.
Substitution gives U = 36(9 x 10⁹)(2 x 10^-6)²/0.35.
U = 3.703 x 10^+0 J.

IGCSE Example 2: Energy ratio from graph behavior

For charges 5 µC and 3 µC, potential energy at separation r = 7 cm is U₀. What is the potential energy when the separation becomes 4r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 4r, the energy becomes U₀/4.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/4.

IGCSE Example 3: All six terms for four charges

Four charges 4 µC, -3 µC, 5 µC, -3 µC have separations r₁₂ = 13 cm, r₁₃ = 16 cm, r₁₄ = 14 cm, r₂₃ = 17 cm, r₂₄ = 20 cm, r₃₄ = 23 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -8.308 x 10^-1 J, 1.125 x 10^+0 J, -7.714 x 10^-1 J, -7.941 x 10^-1 J, 4.050 x 10^-1 J, -5.870 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -1.453 x 10^+0 J.

IGCSE Example 4: Assembly order with three charges

Charges 6 µC, -1 µC and 3 µC are assembled at the corners of an equilateral triangle of side 10 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -5.400 x 10^-1 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = 1.350 x 10^+0 J.
Total work equals total potential energy: U = 8.100 x 10^-1 J.

IGCSE Example 5: Two-charge energy with sign control

Two point charges 6 µC and -5 µC are separated by 10 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 6 x 10^-6 C and q₂ = -5 x 10^-6 C.
Convert distance: r = 0.1 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(6 x 10^-6)(-5 x 10^-6)/0.1.
U = -2.700 x 10^+0 J. The sign is negative because the charges are unlike charges.

IGCSE Example 6: Work needed to change separation

Two equal charges 4 µC and 4 µC are moved slowly from 32 cm separation to 16 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 4.500 x 10^-1 J.
Final energy U_f = kq²/r_f = 9.000 x 10^-1 J.
W_external = U_f - U_i = 4.500 x 10^-1 J.
The work is positive because like charges are being pushed closer.

IGCSE Example 7: Three-charge system with unequal distances

Charges q₁ = 3 µC, q₂ = -5 µC and q₃ = 5 µC have separations r₁₂ = 12 cm, r₂₃ = 21 cm and r₃₁ = 19 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -1.125 x 10^+0 J.
Term 23 = -1.071 x 10^+0 J.
Term 31 = 7.105 x 10^-1 J.
Total U = -1.486 x 10^+0 J after preserving all charge signs.

IGCSE Example 8: Equilateral triangle configuration

An equilateral triangle of side 9 cm carries two positive charges and one negative charge, each of magnitude 30 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.09 m.
For the given signs, the algebraic pair multiplier is -1.
U = -1kq²/a.
U = -1(9 x 10⁹)(30 x 10^-9)²/0.09.
U = -9.000 x 10^-5 J.

IGCSE Example 9: Four equal charges on a square

Four equal charges 70 nC are placed at the corners of a square of side 10 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.1 m and q = 70 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.1 + 2/(0.141)].
U = 2.388 x 10^-3 J.

IGCSE Example 10: Charged conducting sphere

A conducting sphere of radius 10 cm is charged to 5 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 5 x 10^-6 C and R = 0.1 m.
U = (9 x 10⁹)(5 x 10^-6)²/(2 x 0.1).
U = 1.125 x 10^+0 J.
The energy is stored in the electric field outside the conducting sphere.

ICSE Solved Examples (10)

ICSE Example 1: Assembly order with three charges

Charges 5 µC, -6 µC and 2 µC are assembled at the corners of an equilateral triangle of side 17 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -1.588 x 10^+0 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = -1.059 x 10^-1 J.
Total work equals total potential energy: U = -1.694 x 10^+0 J.

ICSE Example 2: Two-charge energy with sign control

Two point charges 5 µC and 4 µC are separated by 32 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 5 x 10^-6 C and q₂ = 4 x 10^-6 C.
Convert distance: r = 0.32 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(5 x 10^-6)(4 x 10^-6)/0.32.
U = 5.625 x 10^-1 J. The sign is positive because the charges are like charges.

ICSE Example 3: Work needed to change separation

Two equal charges 3 µC and 3 µC are moved slowly from 28 cm separation to 14 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 2.893 x 10^-1 J.
Final energy U_f = kq²/r_f = 5.786 x 10^-1 J.
W_external = U_f - U_i = 2.893 x 10^-1 J.
The work is positive because like charges are being pushed closer.

ICSE Example 4: Three-charge system with unequal distances

Charges q₁ = 2 µC, q₂ = -4 µC and q₃ = 4 µC have separations r₁₂ = 10 cm, r₂₃ = 18 cm and r₃₁ = 17 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -7.200 x 10^-1 J.
Term 23 = -8.000 x 10^-1 J.
Term 31 = 4.235 x 10^-1 J.
Total U = -1.096 x 10^+0 J after preserving all charge signs.

ICSE Example 5: Equilateral triangle configuration

An equilateral triangle of side 8 cm carries three equal positive charges, each of magnitude 20 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.08 m.
For the given signs, the algebraic pair multiplier is 3.
U = 3kq²/a.
U = 3(9 x 10⁹)(20 x 10^-9)²/0.08.
U = 1.350 x 10^-4 J.

ICSE Example 6: Four equal charges on a square

Four equal charges 60 nC are placed at the corners of a square of side 8 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.08 m and q = 60 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.08 + 2/(0.113)].
U = 2.193 x 10^-3 J.

ICSE Example 7: Charged conducting sphere

A conducting sphere of radius 28 cm is charged to 4 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 4 x 10^-6 C and R = 0.28 m.
U = (9 x 10⁹)(4 x 10^-6)²/(2 x 0.28).
U = 2.571 x 10^-1 J.
The energy is stored in the electric field outside the conducting sphere.

ICSE Example 8: Pair counting for identical separations

10 equal charges of 1 µC are arranged so every pair is separated by 20 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 10(9)/2 = 45.
Energy of each pair = kq²/r.
Total U = 45kq²/r.
Substitution gives U = 45(9 x 10⁹)(1 x 10^-6)²/0.2.
U = 2.025 x 10^+0 J.

ICSE Example 9: Energy ratio from graph behavior

For charges 2 µC and 5 µC, potential energy at separation r = 6 cm is U₀. What is the potential energy when the separation becomes 5r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 5r, the energy becomes U₀/5.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/5.

ICSE Example 10: All six terms for four charges

Four charges 4 µC, -3 µC, 5 µC, -3 µC have separations r₁₂ = 12 cm, r₁₃ = 15 cm, r₁₄ = 18 cm, r₂₃ = 16 cm, r₂₄ = 19 cm, r₃₄ = 22 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -9.000 x 10^-1 J, 1.200 x 10^+0 J, -6.000 x 10^-1 J, -8.437 x 10^-1 J, 4.263 x 10^-1 J, -6.136 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -1.331 x 10^+0 J.

A-Level Solved Examples (10)

A-Level Example 1: Three-charge system with unequal distances

Charges q₁ = 5 µC, q₂ = -3 µC and q₃ = 1 µC have separations r₁₂ = 16 cm, r₂₃ = 15 cm and r₃₁ = 23 cm. Calculate total potential energy.

Show Answer

Use U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₃q₁/r₃₁).
Term 12 = -8.437 x 10^-1 J.
Term 23 = -1.800 x 10^-1 J.
Term 31 = 1.957 x 10^-1 J.
Total U = -8.281 x 10^-1 J after preserving all charge signs.

A-Level Example 2: Equilateral triangle configuration

An equilateral triangle of side 11 cm carries three equal positive charges, each of magnitude 50 nC. Find the total potential energy.

Show Answer

All pair distances are a = 0.11 m.
For the given signs, the algebraic pair multiplier is 3.
U = 3kq²/a.
U = 3(9 x 10⁹)(50 x 10^-9)²/0.11.
U = 6.136 x 10^-4 J.

A-Level Example 3: Four equal charges on a square

Four equal charges 90 nC are placed at the corners of a square of side 14 cm. Find the total electrostatic potential energy.

Show Answer

There are four side pairs at distance a and two diagonal pairs at distance √2a.
Use U = kq²[4/a + 2/(√2a)].
Here a = 0.14 m and q = 90 x 10^-9 C.
U = (9 x 10⁹)q²[4/0.14 + 2/(0.198)].
U = 2.819 x 10^-3 J.

A-Level Example 4: Charged conducting sphere

A conducting sphere of radius 14 cm is charged to 7 µC. Find the electrostatic energy stored in the sphere.

Show Answer

For an isolated conducting sphere, U = kQ²/(2R).
Convert Q = 7 x 10^-6 C and R = 0.14 m.
U = (9 x 10⁹)(7 x 10^-6)²/(2 x 0.14).
U = 1.575 x 10^+0 J.
The energy is stored in the electric field outside the conducting sphere.

A-Level Example 5: Pair counting for identical separations

5 equal charges of 4 µC are arranged so every pair is separated by 35 cm. Find the number of interaction terms and total potential energy.

Show Answer

Number of distinct pairs = n(n - 1)/2 = 5(4)/2 = 10.
Energy of each pair = kq²/r.
Total U = 10kq²/r.
Substitution gives U = 10(9 x 10⁹)(4 x 10^-6)²/0.35.
U = 4.114 x 10^+0 J.

A-Level Example 6: Energy ratio from graph behavior

For charges 5 µC and 8 µC, potential energy at separation r = 9 cm is U₀. What is the potential energy when the separation becomes 4r?

Show Answer

For fixed charges, U is inversely proportional to r.
If separation becomes 4r, the energy becomes U₀/4.
The sign remains unchanged because the signs of the charges did not change.
Final answer: U = U₀/4.

A-Level Example 7: All six terms for four charges

Four charges 4 µC, -3 µC, 5 µC, -3 µC have separations r₁₂ = 10 cm, r₁₃ = 13 cm, r₁₄ = 16 cm, r₂₃ = 19 cm, r₂₄ = 22 cm, r₃₄ = 20 cm. Find U.

Show Answer

Write all six terms once: 12, 13, 14, 23, 24, 34.
The six numerical terms are -1.080 x 10^+0 J, 1.385 x 10^+0 J, -6.750 x 10^-1 J, -7.105 x 10^-1 J, 3.682 x 10^-1 J, -6.750 x 10^-1 J.
Add algebraically, preserving the signs of the charges.
Total electrostatic potential energy U = -1.388 x 10^+0 J.

A-Level Example 8: Assembly order with three charges

Charges 3 µC, -1 µC and 5 µC are assembled at the corners of an equilateral triangle of side 14 cm. Calculate the total external work needed.

Show Answer

Bring the first charge: W₁ = 0.
Bring the second charge: W₂ = kq_Aq_B/a = -1.929 x 10^-1 J.
Bring the third charge: W₃ = kq_Aq_C/a + kq_Bq_C/a = 6.429 x 10^-1 J.
Total work equals total potential energy: U = 4.500 x 10^-1 J.

A-Level Example 9: Two-charge energy with sign control

Two point charges 4 µC and -5 µC are separated by 21 cm. Find the electrostatic potential energy of the pair.

Show Answer

Convert charges: q₁ = 4 x 10^-6 C and q₂ = -5 x 10^-6 C.
Convert distance: r = 0.21 m.
Use U = kq₁q₂/r.
U = (9 x 10⁹)(4 x 10^-6)(-5 x 10^-6)/0.21.
U = -8.571 x 10^-1 J. The sign is negative because the charges are unlike charges.

A-Level Example 10: Work needed to change separation

Two equal charges 8 µC and 8 µC are moved slowly from 24 cm separation to 12 cm separation. Find the external work done.

Show Answer

For slow motion, W_external = ΔU.
Initial energy U_i = kq²/r_i = 2.400 x 10^+0 J.
Final energy U_f = kq²/r_f = 4.800 x 10^+0 J.
W_external = U_f - U_i = 2.400 x 10^+0 J.
The work is positive because like charges are being pushed closer.

Electrostatics Master Resource

17. Question Bank

This question bank contains 300 original exam-style questions across MCQ, Assertion Reason, Numerical Value, Match the Following, Graph Interpretation and Conceptual formats.

NEET Question Bank (50+)

NEET Q1 Numerical Value

Two equal charges of 8 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.152 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.5 = 1.152 x 10^+0 J.

NEET Q2 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-2.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

NEET Q3 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

NEET Q4 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

NEET Q5 MCQ

Two charges 1 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

NEET Q6 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

NEET Q7 Numerical Value

Two equal charges of 7 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.103 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(7 x 10^-6)²/0.4 = 1.103 x 10^+0 J.

NEET Q8 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-8.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

NEET Q9 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

NEET Q10 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

NEET Q11 MCQ

Two charges 2 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

NEET Q12 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

NEET Q13 Numerical Value

Two equal charges of 6 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.080 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(6 x 10^-6)²/0.3 = 1.080 x 10^+0 J.

NEET Q14 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-14.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

NEET Q15 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

NEET Q16 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

NEET Q17 MCQ

Two charges 3 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

NEET Q18 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

NEET Q19 Numerical Value

Two equal charges of 5 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.125 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(5 x 10^-6)²/0.2 = 1.125 x 10^+0 J.

NEET Q20 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-20.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

NEET Q21 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

NEET Q22 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

NEET Q23 MCQ

Two charges 4 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

NEET Q24 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

NEET Q25 Numerical Value

Two equal charges of 4 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.880 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(4 x 10^-6)²/0.5 = 2.880 x 10^-1 J.

NEET Q26 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-26.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

NEET Q27 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

NEET Q28 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

NEET Q29 MCQ

Two charges 5 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

NEET Q30 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

NEET Q31 Numerical Value

Two equal charges of 3 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.025 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.4 = 2.025 x 10^-1 J.

NEET Q32 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-32.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

NEET Q33 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

NEET Q34 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

NEET Q35 MCQ

Two charges 1 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

NEET Q36 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

NEET Q37 Numerical Value

Two equal charges of 2 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.200 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(2 x 10^-6)²/0.3 = 1.200 x 10^-1 J.

NEET Q38 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-38.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

NEET Q39 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

NEET Q40 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

NEET Q41 MCQ

Two charges 2 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

NEET Q42 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

NEET Q43 Numerical Value

Two equal charges of 8 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.880 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.2 = 2.880 x 10^+0 J.

NEET Q44 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-44.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

NEET Q45 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

NEET Q46 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

NEET Q47 MCQ

Two charges 3 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

NEET Q48 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

NEET Q49 Numerical Value

Two equal charges of 7 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 8.820 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(7 x 10^-6)²/0.5 = 8.820 x 10^-1 J.

NEET Q50 Match the Following

Match the configuration with its total pair count or sign feature for set NEET-50.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Question Bank (50+)

JEE Main Q1 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q2 Numerical Value

Two equal charges of 4 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 7.200 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(4 x 10^-6)²/0.2 = 7.200 x 10^-1 J.

JEE Main Q3 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Main-3.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Q4 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

JEE Main Q5 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Main Q6 MCQ

Two charges 1 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Main Q7 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q8 Numerical Value

Two equal charges of 3 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.620 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.5 = 1.620 x 10^-1 J.

JEE Main Q9 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Main-9.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Q10 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

JEE Main Q11 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Main Q12 MCQ

Two charges 2 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Main Q13 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q14 Numerical Value

Two equal charges of 2 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 9.000 x 10^-2 J

Explanation: U = kq²/r = (9 x 10⁹)(2 x 10^-6)²/0.4 = 9.000 x 10^-2 J.

JEE Main Q15 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Main-15.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Q16 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

JEE Main Q17 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Main Q18 MCQ

Two charges 3 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Main Q19 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q20 Numerical Value

Two equal charges of 8 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.920 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.3 = 1.920 x 10^+0 J.

JEE Main Q21 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Main-21.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Q22 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

JEE Main Q23 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Main Q24 MCQ

Two charges 4 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Main Q25 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q26 Numerical Value

Two equal charges of 7 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.205 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(7 x 10^-6)²/0.2 = 2.205 x 10^+0 J.

JEE Main Q27 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Main-27.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Q28 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

JEE Main Q29 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Main Q30 MCQ

Two charges 5 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Main Q31 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q32 Numerical Value

Two equal charges of 6 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 6.480 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(6 x 10^-6)²/0.5 = 6.480 x 10^-1 J.

JEE Main Q33 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Main-33.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Q34 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

JEE Main Q35 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Main Q36 MCQ

Two charges 1 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Main Q37 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q38 Numerical Value

Two equal charges of 5 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 5.625 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(5 x 10^-6)²/0.4 = 5.625 x 10^-1 J.

JEE Main Q39 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Main-39.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Q40 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

JEE Main Q41 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Main Q42 MCQ

Two charges 2 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Main Q43 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q44 Numerical Value

Two equal charges of 4 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 4.800 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(4 x 10^-6)²/0.3 = 4.800 x 10^-1 J.

JEE Main Q45 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Main-45.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Main Q46 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

JEE Main Q47 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Main Q48 MCQ

Two charges 3 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Main Q49 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Main Q50 Numerical Value

Two equal charges of 3 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 4.050 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.2 = 4.050 x 10^-1 J.

JEE Advanced Question Bank (40+)

JEE Advanced Q1 MCQ

Two charges 5 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Advanced Q2 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Advanced Q3 Numerical Value

Two equal charges of 7 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.470 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(7 x 10^-6)²/0.3 = 1.470 x 10^+0 J.

JEE Advanced Q4 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Advanced-4.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Advanced Q5 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

JEE Advanced Q6 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Advanced Q7 MCQ

Two charges 1 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Advanced Q8 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Advanced Q9 Numerical Value

Two equal charges of 6 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.620 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(6 x 10^-6)²/0.2 = 1.620 x 10^+0 J.

JEE Advanced Q10 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Advanced-10.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Advanced Q11 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

JEE Advanced Q12 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Advanced Q13 MCQ

Two charges 2 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Advanced Q14 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Advanced Q15 Numerical Value

Two equal charges of 5 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 4.500 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(5 x 10^-6)²/0.5 = 4.500 x 10^-1 J.

JEE Advanced Q16 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Advanced-16.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Advanced Q17 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

JEE Advanced Q18 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Advanced Q19 MCQ

Two charges 3 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Advanced Q20 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Advanced Q21 Numerical Value

Two equal charges of 4 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 3.600 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(4 x 10^-6)²/0.4 = 3.600 x 10^-1 J.

JEE Advanced Q22 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Advanced-22.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Advanced Q23 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

JEE Advanced Q24 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Advanced Q25 MCQ

Two charges 4 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Advanced Q26 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Advanced Q27 Numerical Value

Two equal charges of 3 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.700 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.3 = 2.700 x 10^-1 J.

JEE Advanced Q28 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Advanced-28.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Advanced Q29 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

JEE Advanced Q30 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Advanced Q31 MCQ

Two charges 5 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Advanced Q32 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Advanced Q33 Numerical Value

Two equal charges of 2 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.800 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(2 x 10^-6)²/0.2 = 1.800 x 10^-1 J.

JEE Advanced Q34 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Advanced-34.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

JEE Advanced Q35 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

JEE Advanced Q36 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

JEE Advanced Q37 MCQ

Two charges 1 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

JEE Advanced Q38 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

JEE Advanced Q39 Numerical Value

Two equal charges of 8 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.152 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.5 = 1.152 x 10^+0 J.

JEE Advanced Q40 Match the Following

Match the configuration with its total pair count or sign feature for set JEE Advanced-40.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

CBSE Question Bank (40+)

CBSE Q1 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

CBSE Q2 MCQ

Two charges 3 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

CBSE Q3 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

CBSE Q4 Numerical Value

Two equal charges of 5 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.125 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(5 x 10^-6)²/0.2 = 1.125 x 10^+0 J.

CBSE Q5 Match the Following

Match the configuration with its total pair count or sign feature for set CBSE-5.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

CBSE Q6 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

CBSE Q7 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

CBSE Q8 MCQ

Two charges 4 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

CBSE Q9 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

CBSE Q10 Numerical Value

Two equal charges of 4 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.880 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(4 x 10^-6)²/0.5 = 2.880 x 10^-1 J.

CBSE Q11 Match the Following

Match the configuration with its total pair count or sign feature for set CBSE-11.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

CBSE Q12 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

CBSE Q13 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

CBSE Q14 MCQ

Two charges 5 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

CBSE Q15 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

CBSE Q16 Numerical Value

Two equal charges of 3 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.025 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.4 = 2.025 x 10^-1 J.

CBSE Q17 Match the Following

Match the configuration with its total pair count or sign feature for set CBSE-17.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

CBSE Q18 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

CBSE Q19 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

CBSE Q20 MCQ

Two charges 1 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

CBSE Q21 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

CBSE Q22 Numerical Value

Two equal charges of 2 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.200 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(2 x 10^-6)²/0.3 = 1.200 x 10^-1 J.

CBSE Q23 Match the Following

Match the configuration with its total pair count or sign feature for set CBSE-23.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

CBSE Q24 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

CBSE Q25 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

CBSE Q26 MCQ

Two charges 2 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

CBSE Q27 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

CBSE Q28 Numerical Value

Two equal charges of 8 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.880 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.2 = 2.880 x 10^+0 J.

CBSE Q29 Match the Following

Match the configuration with its total pair count or sign feature for set CBSE-29.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

CBSE Q30 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

CBSE Q31 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

CBSE Q32 MCQ

Two charges 3 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

CBSE Q33 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

CBSE Q34 Numerical Value

Two equal charges of 7 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 8.820 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(7 x 10^-6)²/0.5 = 8.820 x 10^-1 J.

CBSE Q35 Match the Following

Match the configuration with its total pair count or sign feature for set CBSE-35.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

CBSE Q36 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

CBSE Q37 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

CBSE Q38 MCQ

Two charges 4 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

CBSE Q39 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

CBSE Q40 Numerical Value

Two equal charges of 6 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 8.100 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(6 x 10^-6)²/0.4 = 8.100 x 10^-1 J.

IB Physics Question Bank (30+)

IB Physics Q1 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

IB Physics Q2 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IB Physics Q3 MCQ

Two charges 5 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

IB Physics Q4 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IB Physics Q5 Numerical Value

Two equal charges of 3 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.025 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.4 = 2.025 x 10^-1 J.

IB Physics Q6 Match the Following

Match the configuration with its total pair count or sign feature for set IB Physics-6.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IB Physics Q7 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

IB Physics Q8 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IB Physics Q9 MCQ

Two charges 1 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

IB Physics Q10 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IB Physics Q11 Numerical Value

Two equal charges of 2 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.200 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(2 x 10^-6)²/0.3 = 1.200 x 10^-1 J.

IB Physics Q12 Match the Following

Match the configuration with its total pair count or sign feature for set IB Physics-12.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IB Physics Q13 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

IB Physics Q14 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IB Physics Q15 MCQ

Two charges 2 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

IB Physics Q16 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IB Physics Q17 Numerical Value

Two equal charges of 8 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.880 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.2 = 2.880 x 10^+0 J.

IB Physics Q18 Match the Following

Match the configuration with its total pair count or sign feature for set IB Physics-18.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IB Physics Q19 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

IB Physics Q20 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IB Physics Q21 MCQ

Two charges 3 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

IB Physics Q22 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IB Physics Q23 Numerical Value

Two equal charges of 7 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 8.820 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(7 x 10^-6)²/0.5 = 8.820 x 10^-1 J.

IB Physics Q24 Match the Following

Match the configuration with its total pair count or sign feature for set IB Physics-24.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IB Physics Q25 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

IB Physics Q26 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IB Physics Q27 MCQ

Two charges 4 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

IB Physics Q28 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IB Physics Q29 Numerical Value

Two equal charges of 6 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 8.100 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(6 x 10^-6)²/0.4 = 8.100 x 10^-1 J.

IB Physics Q30 Match the Following

Match the configuration with its total pair count or sign feature for set IB Physics-30.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IGCSE Question Bank (30+)

IGCSE Q1 Match the Following

Match the configuration with its total pair count or sign feature for set IGCSE-1.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IGCSE Q2 Graph Interpretation

For like charges, if the separation becomes 3 times its original value, what happens to U on the U-r graph?

A. U becomes 3U
B. U becomes U/3
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 3 divides U by 3. The sign does not change.

IGCSE Q3 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IGCSE Q4 MCQ

Two charges 1 µC and -5 µC are separated by 25 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

IGCSE Q5 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IGCSE Q6 Numerical Value

Two equal charges of 4 µC are 45 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 3.200 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(4 x 10^-6)²/0.45 = 3.200 x 10^-1 J.

IGCSE Q7 Match the Following

Match the configuration with its total pair count or sign feature for set IGCSE-7.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IGCSE Q8 Graph Interpretation

For like charges, if the separation becomes 5 times its original value, what happens to U on the U-r graph?

A. U becomes 5U
B. U becomes U/5
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 5 divides U by 5. The sign does not change.

IGCSE Q9 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IGCSE Q10 MCQ

Two charges 2 µC and -3 µC are separated by 15 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

IGCSE Q11 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IGCSE Q12 Numerical Value

Two equal charges of 3 µC are 35 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.314 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.35 = 2.314 x 10^-1 J.

IGCSE Q13 Match the Following

Match the configuration with its total pair count or sign feature for set IGCSE-13.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IGCSE Q14 Graph Interpretation

For like charges, if the separation becomes 3 times its original value, what happens to U on the U-r graph?

A. U becomes 3U
B. U becomes U/3
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 3 divides U by 3. The sign does not change.

IGCSE Q15 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IGCSE Q16 MCQ

Two charges 3 µC and -5 µC are separated by 45 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

IGCSE Q17 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IGCSE Q18 Numerical Value

Two equal charges of 2 µC are 25 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.440 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(2 x 10^-6)²/0.25 = 1.440 x 10^-1 J.

IGCSE Q19 Match the Following

Match the configuration with its total pair count or sign feature for set IGCSE-19.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IGCSE Q20 Graph Interpretation

For like charges, if the separation becomes 5 times its original value, what happens to U on the U-r graph?

A. U becomes 5U
B. U becomes U/5
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 5 divides U by 5. The sign does not change.

IGCSE Q21 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IGCSE Q22 MCQ

Two charges 4 µC and -3 µC are separated by 35 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

IGCSE Q23 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IGCSE Q24 Numerical Value

Two equal charges of 8 µC are 55 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.047 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.55 = 1.047 x 10^+0 J.

IGCSE Q25 Match the Following

Match the configuration with its total pair count or sign feature for set IGCSE-25.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

IGCSE Q26 Graph Interpretation

For like charges, if the separation becomes 3 times its original value, what happens to U on the U-r graph?

A. U becomes 3U
B. U becomes U/3
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 3 divides U by 3. The sign does not change.

IGCSE Q27 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

IGCSE Q28 MCQ

Two charges 5 µC and -5 µC are separated by 25 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

IGCSE Q29 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

IGCSE Q30 Numerical Value

Two equal charges of 7 µC are 45 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 9.800 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(7 x 10^-6)²/0.45 = 9.800 x 10^-1 J.

ICSE Question Bank (30+)

ICSE Q1 Numerical Value

Two equal charges of 3 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.025 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.4 = 2.025 x 10^-1 J.

ICSE Q2 Match the Following

Match the configuration with its total pair count or sign feature for set ICSE-2.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

ICSE Q3 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

ICSE Q4 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

ICSE Q5 MCQ

Two charges 1 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

ICSE Q6 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

ICSE Q7 Numerical Value

Two equal charges of 2 µC are 30 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.200 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(2 x 10^-6)²/0.3 = 1.200 x 10^-1 J.

ICSE Q8 Match the Following

Match the configuration with its total pair count or sign feature for set ICSE-8.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

ICSE Q9 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

ICSE Q10 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

ICSE Q11 MCQ

Two charges 2 µC and 4 µC are separated by 40 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

ICSE Q12 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

ICSE Q13 Numerical Value

Two equal charges of 8 µC are 20 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 2.880 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.2 = 2.880 x 10^+0 J.

ICSE Q14 Match the Following

Match the configuration with its total pair count or sign feature for set ICSE-14.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

ICSE Q15 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

ICSE Q16 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

ICSE Q17 MCQ

Two charges 3 µC and 2 µC are separated by 30 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

ICSE Q18 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

ICSE Q19 Numerical Value

Two equal charges of 7 µC are 50 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 8.820 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(7 x 10^-6)²/0.5 = 8.820 x 10^-1 J.

ICSE Q20 Match the Following

Match the configuration with its total pair count or sign feature for set ICSE-20.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

ICSE Q21 Graph Interpretation

For like charges, if the separation becomes 2 times its original value, what happens to U on the U-r graph?

A. U becomes 2U
B. U becomes U/2
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 2 divides U by 2. The sign does not change.

ICSE Q22 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

ICSE Q23 MCQ

Two charges 4 µC and 4 µC are separated by 20 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

ICSE Q24 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

ICSE Q25 Numerical Value

Two equal charges of 6 µC are 40 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 8.100 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(6 x 10^-6)²/0.4 = 8.100 x 10^-1 J.

ICSE Q26 Match the Following

Match the configuration with its total pair count or sign feature for set ICSE-26.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

ICSE Q27 Graph Interpretation

For like charges, if the separation becomes 4 times its original value, what happens to U on the U-r graph?

A. U becomes 4U
B. U becomes U/4
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 4 divides U by 4. The sign does not change.

ICSE Q28 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

ICSE Q29 MCQ

Two charges 5 µC and 2 µC are separated by 10 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: A

Explanation: U = kq₁q₂/r. Since q₁q₂ is positive, U is positive.

ICSE Q30 Assertion Reason

Assertion: The potential energy of two unlike charges is negative if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

A-Level Question Bank (30+)

A-Level Q1 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

A-Level Q2 Numerical Value

Two equal charges of 5 µC are 45 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 5.000 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(5 x 10^-6)²/0.45 = 5.000 x 10^-1 J.

A-Level Q3 Match the Following

Match the configuration with its total pair count or sign feature for set A-Level-3.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

A-Level Q4 Graph Interpretation

For like charges, if the separation becomes 5 times its original value, what happens to U on the U-r graph?

A. U becomes 5U
B. U becomes U/5
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 5 divides U by 5. The sign does not change.

A-Level Q5 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

A-Level Q6 MCQ

Two charges 5 µC and -3 µC are separated by 15 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

A-Level Q7 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

A-Level Q8 Numerical Value

Two equal charges of 4 µC are 35 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 4.114 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(4 x 10^-6)²/0.35 = 4.114 x 10^-1 J.

A-Level Q9 Match the Following

Match the configuration with its total pair count or sign feature for set A-Level-9.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

A-Level Q10 Graph Interpretation

For like charges, if the separation becomes 3 times its original value, what happens to U on the U-r graph?

A. U becomes 3U
B. U becomes U/3
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 3 divides U by 3. The sign does not change.

A-Level Q11 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

A-Level Q12 MCQ

Two charges 1 µC and -5 µC are separated by 45 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

A-Level Q13 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

A-Level Q14 Numerical Value

Two equal charges of 3 µC are 25 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 3.240 x 10^-1 J

Explanation: U = kq²/r = (9 x 10⁹)(3 x 10^-6)²/0.25 = 3.240 x 10^-1 J.

A-Level Q15 Match the Following

Match the configuration with its total pair count or sign feature for set A-Level-15.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

A-Level Q16 Graph Interpretation

For like charges, if the separation becomes 5 times its original value, what happens to U on the U-r graph?

A. U becomes 5U
B. U becomes U/5
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 5 divides U by 5. The sign does not change.

A-Level Q17 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

A-Level Q18 MCQ

Two charges 2 µC and -3 µC are separated by 35 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

A-Level Q19 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

A-Level Q20 Numerical Value

Two equal charges of 2 µC are 55 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 6.545 x 10^-2 J

Explanation: U = kq²/r = (9 x 10⁹)(2 x 10^-6)²/0.55 = 6.545 x 10^-2 J.

A-Level Q21 Match the Following

Match the configuration with its total pair count or sign feature for set A-Level-21.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

A-Level Q22 Graph Interpretation

For like charges, if the separation becomes 3 times its original value, what happens to U on the U-r graph?

A. U becomes 3U
B. U becomes U/3
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 3 divides U by 3. The sign does not change.

A-Level Q23 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

A-Level Q24 MCQ

Two charges 3 µC and -5 µC are separated by 25 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

A-Level Q25 Assertion Reason

Assertion: The potential energy of two like charges is positive if zero is chosen at infinity. Reason: The sign of U depends on q₁q₂.

A. Both are true and Reason explains Assertion
B. Both are true but Reason does not explain Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Show Answer

Answer: A

Explanation: The product q₁q₂ directly controls the sign of U, so the reason correctly explains the assertion.

A-Level Q26 Numerical Value

Two equal charges of 8 µC are 45 cm apart. Find U in joule. Take k = 9 x 10⁹ N m²C^-2.

Show Answer

Answer: 1.280 x 10^+0 J

Explanation: U = kq²/r = (9 x 10⁹)(8 x 10^-6)²/0.45 = 1.280 x 10^+0 J.

A-Level Q27 Match the Following

Match the configuration with its total pair count or sign feature for set A-Level-27.

Column I: (P) 3 charges, (Q) 4 charges, (R) +q and -q, (S) four +q at square corners
Column II: (1) six pairs, (2) three pairs, (3) negative pair energy, (4) four side terms and two diagonal terms

Show Answer

Answer: P-2, Q-1, R-3, S-4

Explanation: Three charges give 3 pairs, four charges give 6 pairs, unlike charges give negative energy and square geometry has four side plus two diagonal interactions.

A-Level Q28 Graph Interpretation

For like charges, if the separation becomes 5 times its original value, what happens to U on the U-r graph?

A. U becomes 5U
B. U becomes U/5
C. U becomes zero immediately
D. U changes sign

Show Answer

Answer: B

Explanation: For fixed charges, U is proportional to 1/r. Multiplying r by 5 divides U by 5. The sign does not change.

A-Level Q29 Conceptual Question

Why should the pair q₂q₁ not be added separately after adding q₁q₂ in a system of charges?

Show Answer

Answer: Because it represents the same physical interaction and would double count the pair energy.

Explanation: The formula uses i < j to count each pair once. q₁q₂/r₁₂ and q₂q₁/r₂₁ are identical terms.

A-Level Q30 MCQ

Two charges 4 µC and -3 µC are separated by 15 cm. What is the sign of their electrostatic potential energy?

A. Positive
B. Negative
C. Zero
D. Independent of charge sign

Show Answer

Answer: B

Explanation: U = kq₁q₂/r. Since q₁q₂ is negative, U is negative.

Electrostatics Master Resource

18. Case Studies

Each CBSE-style case study has a passage followed by five option-based questions with answers and explanations.

Case Study 1: Two charges

A laboratory demonstration uses two small charged beads separated by 12 cm. The teacher asks students to predict whether external work is needed when the separation is reduced and how the sign of potential energy changes for like and unlike charges.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Two charges"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 2 charges, how many distinct pair interactions exist?

A. 2
B. 1
C. 1
D. 4

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 2, the count is 1.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 2: Three charges

Three point charges are fixed at the corners of a scalene triangle. Students are told that total energy must be calculated from all distinct pairs rather than from the net force on one charge.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Three charges"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 3 charges, how many distinct pair interactions exist?

A. 3
B. 2
C. 3
D. 9

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 3, the count is 3.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 3: Triangle configuration

Three equal charges are placed at the vertices of an equilateral triangle of side a. The class then replaces one charge by an equal negative charge and compares the new energy with the all-positive case.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Triangle configuration"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 4 charges, how many distinct pair interactions exist?

A. 4
B. 3
C. 6
D. 16

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 4, the count is 6.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 4: Square configuration

Four equal charges are placed on a square frame. Measurements show that the side distance is a and each diagonal distance is √2a. The question is to identify all side and diagonal contributions.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Square configuration"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 5 charges, how many distinct pair interactions exist?

A. 5
B. 4
C. 10
D. 25

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 5, the count is 10.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 5: Potential energy graph

A graph of U against r is drawn for two charges. One curve lies above the r-axis and approaches zero from above, while the other lies below the r-axis and approaches zero from below.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Potential energy graph"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 6 charges, how many distinct pair interactions exist?

A. 6
B. 5
C. 15
D. 36

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 6, the count is 15.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 6: System of charges

A group of n charges is arranged in space. The teacher warns the students that q₁q₂ and q₂q₁ must not both be counted.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "System of charges"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 7 charges, how many distinct pair interactions exist?

A. 7
B. 6
C. 21
D. 49

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 7, the count is 21.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 7: Assembly of charges

Charges are brought from infinity one by one and placed at fixed points. The process is slow enough that kinetic energy remains zero, so the work done externally becomes stored energy.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Assembly of charges"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 8 charges, how many distinct pair interactions exist?

A. 8
B. 7
C. 28
D. 64

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 8, the count is 28.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 8: Charged sphere

An isolated conducting sphere of radius R is charged gradually from zero to Q. At an intermediate charge q, its potential is kq/R.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Charged sphere"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 9 charges, how many distinct pair interactions exist?

A. 9
B. 8
C. 36
D. 81

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 9, the count is 36.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 9: Field energy

A capacitor chapter begins by introducing the idea that energy can be stored in the electric field, with energy density proportional to E².

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Field energy"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 10 charges, how many distinct pair interactions exist?

A. 10
B. 9
C. 45
D. 100

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 10, the count is 45.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Case Study 10: Mixed electrostatics

A mixed electrostatics problem combines potential energy, electric potential and pair counting. Students must decide whether the problem asks for energy of a system or potential at a point.

Question 1

Which reference state is normally used for electrostatic potential energy of point charges?

A. Charges at zero separation
B. Charges at infinity
C. Charges on Earth only
D. Charges moving at constant speed

Show Answer

Answer: B

Explanation: For point-charge systems, U is normally chosen as zero when the charges are infinitely separated.

Question 2

Which formula is most directly useful for the case study topic "Mixed electrostatics"?

A. U = kq1q2/r for every distinct pair
B. v = u + at
C. pV = nRT
D. F = ma only

Show Answer

Answer: A

Explanation: Electrostatic energy problems begin with pair energy and then add all relevant distinct pairs.

Question 3

If two like charges are pushed closer slowly, what happens to external work?

A. It is positive
B. It is always zero
C. It is negative for all cases
D. It cannot be related to U

Show Answer

Answer: A

Explanation: Like charges repel, so external work is required to reduce their separation. Potential energy increases.

Question 4

For 11 charges, how many distinct pair interactions exist?

A. 11
B. 10
C. 55
D. 121

Show Answer

Answer: C

Explanation: The number of distinct pairs is n(n - 1)/2. With n = 11, the count is 55.

Question 5

What is the most common exam trap in this case study?

A. Ignoring signs or missing a pair
B. Writing units
C. Drawing a diagram
D. Choosing infinity as reference

Show Answer

Answer: A

Explanation: Most errors come from missing interaction terms, double counting or losing the sign of a negative charge.

Electrostatics Master Resource

19. Common Student Mistakes

Forgetting sign of charge

Always substitute negative charges with their sign. Do not calculate using magnitudes unless the question asks only for magnitude.

Missing interaction terms

For n charges, count n(n - 1)/2 terms before solving.

Double counting pairs

q1q2 and q2q1 are the same pair and must not both appear.

Wrong diagonal distance in square

Use √2a for square diagonals, not a.

Wrong triangle calculations

In an equilateral triangle all side distances are a; in a general triangle use r12, r23 and r31 separately.

Sign errors in mixed charges

Mixed signs can produce cancellation. Add algebraically, not by magnitude.

Confusing potential and potential energy

Potential V is energy per unit charge. Potential energy U is qV or pair energy.

Electrostatics Master Resource

20. Exam Strategy

Exam	Most important patterns	Fast solving tricks
CBSE	Definitions, derivation of two-charge formula, sign interpretation, three-charge and square numericals.	Draw the geometry first, list pairs, write units and explain signs in words.
NEET	Direct formula use, signs, quick graph interpretation, dimensional checks.	Use k = 9 x 10^9, convert microcoulomb and centimetre quickly, eliminate options by sign.
JEE Main	Multi-charge numericals, square and triangle configurations, assembly work.	Count pairs before substituting. Keep diagonal terms separate.
JEE Advanced	Regular polygons, symbolic derivations, stability and constrained motion.	Group equal distances and use chord lengths for polygons.
IB Physics	Energy conservation, field energy concept and explanation-based reasoning.	Write assumptions clearly and connect work done with change in potential energy.
IGCSE	Conceptual meaning, simple two-charge calculations and sign of energy.	Use clear sentence explanations and avoid overcomplicated algebra.
ICSE	Formula recall, derivation steps and standard numerical substitution.	Show every step in sequence: formula, conversion, substitution, answer.
A-Level	Energy curves, conservative force relation and conducting sphere extension.	Use U-r graph behavior and connect force with slope of potential energy.

Electrostatics Master Resource

21. Final Quick Revision

Definitions

Electrostatic potential energy is the work done in assembling a charge configuration from infinity without changing kinetic energy.

Formula Summary

Two charges: U = kq₁q₂/r
N charges: U = k ∑_i<j q_iq_j/r_ij
Pairs: n(n - 1)/2
Conducting sphere: U = kQ²/2R

Graph Summary

Like charges: U > 0 and tends to zero from above. Unlike charges: U < 0 and tends to zero from below.

Exam Traps

Do not ignore signs, do not miss diagonal terms, do not double count pairs and do not confuse potential V with potential energy U.

Frequently Used Results

Triangle of three +q charges: 3kq²/a. Square of four +q charges: (kq²/a)(4 + √2).

Electrostatics Master Resource

SEO Publishing Details

SEO Title

Electrostatic Potential Energy and System of Charges - Complete Physics Guide

Meta Description

Learn electrostatic potential energy with full derivations, diagrams, formula sheet, solved examples, question bank, case studies and exam strategy for CBSE, NEET, JEE, IB, IGCSE, ICSE and A-Level.

Focus Keyword

electrostatic potential energy

Slug

electrostatic-energy

Internal Linking Suggestions

Electric Potential, Coulomb's Law, Capacitors, Electric Field.