Why is there a negative sign in E = -dV/dr?

The negative sign shows that electric field points in the direction in which electric potential decreases.

If potential is constant, what is electric field?

If potential is constant throughout a region, its spatial derivative is zero, so electric field in that region is zero.

Can potential be zero while electric field is non-zero?

Yes. Potential value and potential gradient are different. A point may have zero potential while nearby potential changes, producing a non-zero electric field.

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Potential vs Field Complete Resource

Relationship Between Electric Field and Electric Potential

Q: What is the relationship between electric field and electric potential?

In one dimension, electric field is the negative rate of change of potential with distance: E = -dV/dr. In three dimensions, vector electric field is the negative gradient of potential: E = -grad V.

Understand the most important concept connecting Electric Field and Electric Potential.

CBSENEETJEE MainJEE AdvancedIB PhysicsIGCSEICSEA-LevelUP Board

E = -dV/dr

One-dimensional relation: field equals negative potential slope.

E = -∇V

Vector form: field is the negative gradient of potential.

Electric field and potential

1. Introduction

What is electric field?

Electric field is the force experienced by a unit positive test charge placed at a point. It tells us how strongly and in what direction a charge would be pushed. Mathematically, E = F/q.

Because force has direction, electric field is a vector. The direction of electric field is defined as the direction of force on a positive test charge.

What is electric potential?

Electric potential is electric potential energy per unit charge. It tells us how much work per unit charge is associated with bringing a small positive test charge to a point.

Potential is a scalar. It has a value but no direction. The direction information appears only when we examine how potential changes from point to point.

Why are they related?

A force field does work when a charge moves. Work changes potential energy. Since electric potential is potential energy per unit charge, the electric field must be connected to how potential changes with position.

A steep potential change means the electric force per unit charge is large. A gentle potential change means the field is small.

Why students find it difficult

Students often memorize E = -dV/dr without seeing that it is simply a slope rule. They also confuse the value of potential with the rate of change of potential. Zero potential is not the same thing as zero electric field.

The negative sign also causes errors because it is a direction statement, not a decoration.

This topic is important in NEET and JEE because it connects electrostatic force, energy, work, graphs, equipotential surfaces, conductors, capacitors and calculus. In JEE Advanced and IB Physics, it becomes the gateway to gradient ideas in more than one dimension.

Potential basics

2. Revision of Electric Potential

V = W/q

Potential at a point is work done per unit charge.

ΔV = W/q

Potential difference is work per unit charge between two points.

W = qΔV

Work or energy change for charge q across potential difference ΔV.

Physical meaning

If a point is at high electric potential, a positive charge placed there has high electric potential energy per coulomb. Potential is like an electric height. Moving a positive charge to a higher potential requires external work; moving it to lower potential releases energy through the electric field.

Unit of potential

The SI unit of potential is volt. One volt means one joule of work per coulomb of charge.

1 V = 1 J/C.

Field basics

3. Revision of Electric Field

E = F/q

Electric field intensity is force per unit positive test charge.

Unit: N/C

Equivalent unit: V/m.

Direction: force on +q

A positive test charge defines the field direction.

Electric field intensity

Electric field intensity measures force per coulomb. A field of 500 N/C means every coulomb of positive charge would experience 500 N of force in the field direction.

Direction

Field direction is away from positive source charges and toward negative source charges. At any point, a positive test charge accelerates along the field if released from rest.

Field lines

Field lines are closer where field is stronger. They never cross because the field at a point cannot have two directions. Electric field is perpendicular to equipotential surfaces.

Step-by-step derivation

4. Complete Derivation of E = -dV/dr

Consider a positive test charge q moving a very small distance dr along the direction chosen as positive r. The work done by the electric force during this displacement is connected to the change in potential energy. We use a very small displacement so that the field can be treated as nearly constant over that interval.

Step	Mathematics	Meaning
1	dW = F dr	For a very small displacement dr along the field line, work done by electric force is force times displacement.
2	F = qE	Electric field intensity is force per unit positive test charge, so the force on charge q is qE.
3	dW = qE dr	Substitute F = qE in the work expression.
4	dV = -dW/q	Potential difference is negative work done by electric field per unit charge, because external work against the field raises potential.
5	dV = -(qE dr)/q	Replace dW with qEdr and divide by q.
6	dV = -E dr	The charge q cancels; the relation is a property of the field, not of the test charge.
7	E = -dV/dr	Divide both sides by dr and rearrange.

Important interpretation: The equation E = -dV/dr is not saying field equals potential. It says field equals the negative spatial rate of change of potential. Therefore the same value of V can produce different fields depending on how V changes nearby.

Direction of decreasing V

5. Physical Meaning of the Negative Sign

Why does the sign appear?

The electric field does positive work when a positive charge moves in the field direction. During that natural motion, electric potential energy decreases. Therefore potential decreases in the direction of electric field.

Mathematically, if dr is along E, work by field is positive, dW > 0. Since dV = -dW/q, dV is negative. Hence E and increasing V point in opposite directions.

Three statements to remember

Electric field always points toward decreasing potential.
Potential decreases in the direction of electric field.
Positive charges naturally move toward lower electric potential when only electric force acts.

The hill analogy is useful: height represents potential. A ball rolls downhill because gravitational potential energy decreases. A positive charge released in an electric field moves in the direction where electric potential decreases.

Slope controls field

6. Conceptual Interpretation

Case 1: V decreases rapidly

A large magnitude of dV/dr means potential changes a lot in a small distance. Since E = -dV/dr, the electric field is large. Field lines would be crowded, and a charge would experience a strong force.

Case 2: V changes slowly

A small magnitude of dV/dr means potential changes gently with distance. The field is weak because the force per unit charge is small.

Case 3: V is constant

If V is constant, dV/dr = 0. Therefore E = 0. There is no potential slope to push a charge.

Very important: If V is constant in a region, then E = 0 in that region. If E = 0 in a connected region, potential may be constant throughout that region because there is no spatial variation of V. Examples include the inside of a conductor in electrostatic equilibrium and a flat segment of a potential-distance graph.

Uniform field

7. Constant Electric Field

If electric field is constant, then E = constant and dV/dr = -E.

Step	Working
1	E = -dV/dr
2	dV/dr = -E
3	dV = -E dr
4	Integrate both sides: ∫dV = ∫-E dr
5	V = -Er + C

Parallel plate example

Between ideal large parallel plates, E is nearly constant. Therefore potential decreases linearly from the positive plate to the negative plate.

Equal distances correspond to equal drops in potential. This is why the V-r graph is a straight line and the slope is constant.

Zero gradient

8. Constant Potential

V = constant

Potential has the same value everywhere in the region.

dV/dr = 0

Slope of the potential graph is zero.

E = 0

No electric field exists in that region.

Conductors in electrostatic equilibrium

Inside a conductor in electrostatic equilibrium, free charges rearrange until internal electric field becomes zero. Since E = 0, potential is constant inside and on the conductor surface.

From 1/r to 1/r squared

9. Point Charge Analysis

For a point charge Q, potential at distance r is V = kQ/r. To find the field, differentiate with respect to r.

Step	Mathematics	Explanation
1	V = kQ/r = kQr^-1	Write in power form for differentiation.
2	dV/dr = -kQr^-2	Derivative of r^-1 is -r^-2.
3	E = -dV/dr	Use the field-potential relation.
4	E = -(-kQ/r²) = kQ/r²	For Q positive, the field is radially outward.

Potential variation

V varies as 1/r. Doubling r halves the potential.

Field variation

E varies as 1/r². Doubling r makes the field one-fourth.

Seven required graphs

10. Graph Analysis

Graph questions are slope questions. On a V-r graph, the slope is dV/dr and electric field is the negative of that slope. On an E-r graph, the graph directly gives field magnitude or component.

For a positive point charge, potential is large near the charge and decreases toward zero as r increases.

The field of a point charge falls faster than potential because it follows an inverse-square law.

A constant field graph is horizontal because E has the same value at every point.

A constant potential graph is horizontal; its slope is zero, so E is zero.

A straight falling V-r line has constant negative slope, so the field is constant and positive along r.

Positive charge gives positive potential that decreases with distance; field points outward.

Negative charge gives negative potential that rises toward zero; field points inward.

Slope method

11. Electric Field from Potential Graph

Positive slope

If dV/dr is positive, then E = -dV/dr is negative. The field points opposite to increasing r.

Negative slope

If dV/dr is negative, then E is positive. The field points along increasing r.

Zero slope

If dV/dr = 0, then E = 0. A flat potential graph means no field in that region.

Graph feature	Slope	Electric field	Meaning
Steep downward V-r graph	Large negative	Large positive	Strong field along +r
Gentle downward V-r graph	Small negative	Small positive	Weak field along +r
Flat graph	Zero	Zero	No electric field
Upward graph	Positive	Negative	Field opposite +r

No work along surface

12. Equipotential Surfaces

Definition

An equipotential surface is a surface on which every point has the same electric potential. Moving a charge along this surface gives ΔV = 0.

No work done

Since W = qΔV, motion along an equipotential surface requires zero work against the electric field. The electric force also does no work along the surface.

Relation with field

Electric field is perpendicular to equipotential surfaces. If it had a tangential component, charges would move along the surface and potential would change, contradicting the definition.

Gradient form

13. Multi-Dimensional Form

E = -∇V

Vector field equals negative gradient of scalar potential.

E_x = -∂V/∂x

x-component from x-slope of potential.

E_y = -∂V/∂y

y-component from y-slope of potential.

E_z = -∂V/∂z

z-component from z-slope of potential.

For JEE Advanced and IB Physics, the gradient is the direction in which potential increases fastest. Since electric field is negative gradient, field points in the direction of fastest decrease of potential. If V depends on x, y and z, each component of E is found by differentiating V with respect to that coordinate while keeping the other coordinates constant.

Concept checks

14. Common Confusions

Is potential zero when electric field is zero?

Show Answer

Not necessarily. If E = 0 in a region, potential is constant there, but that constant may be 0 V, 10 V or any other value depending on reference.

Is electric field zero when potential is constant?

Show Answer

Yes, in that region. Constant potential means dV/dr = 0, so E = 0.

Can potential be zero but electric field non-zero?

Show Answer

Yes. At a zero-potential point between charges, potential may cancel algebraically while the field components do not cancel. Field depends on gradient, not absolute value.

Can electric field be non-zero while potential is positive?

Show Answer

Yes. Around a positive point charge, V is positive and E is non-zero.

Can electric field exist on equipotential surfaces?

Show Answer

Yes, but it is perpendicular to the surface. There is no field component along the equipotential surface.

Does larger potential always mean larger field?

Show Answer

No. A high constant potential has zero field. Field depends on how rapidly potential changes with distance.

Side-by-side comparison

15. Comparison Table: Electric Field vs Electric Potential

Feature	Electric Field	Electric Potential
Definition	Force per unit positive test charge	Work or potential energy per unit charge
Formula	E = F/q and E = -dV/dr	V = W/q and V = kQ/r
Unit	N/C or V/m	Volt or J/C
Vector or scalar	Vector	Scalar
Direction	Direction of force on positive test charge	No direction
Sign	Sign can represent component direction	Positive or negative depending on source charge and reference
Measurement	Measured by force per unit charge or graph slope	Measured by work per unit charge
Graph interpretation	Negative slope of V-r graph	Height/value of V-r graph

Formula cards

16. Important Formulas

E = F/q

Definition of electric field intensity.

V = W/q

Definition of electric potential.

ΔV = W/q

Potential difference from work per charge.

W = qΔV

Work-energy relation in electrostatics.

E = -dV/dr

One-dimensional field-potential relation.

E = -∇V

Three-dimensional vector form.

V = kQ/r

Potential due to point charge.

E = kQ/r²

Field due to point charge.

V = -Er + C

Potential in a constant electric field.

120 step-by-step solved questions

17. Solved Examples

20 CBSE Solved Examples

CBSE Example 1: The potential changes from 273 V to 138 V when moving from r = 0.16 m to r = 0.4 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 138 - 273 = -135 V.

Step 2: dr = 0.4 - 0.16 = 0.24 m.

Step 3: dV/dr = -135/0.24 = -562.5 V m^-1.

Step 4: E = -dV/dr = 562.5 N C^-1.

The field points in the direction of decreasing potential.

CBSE Example 2: A charge of 6 microcoulomb moves through a potential difference of 40 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 6 microcoulomb = 6 x 10^-6 C.

Step 2: ΔV = 40 V.

Step 3: W = (6 x 10^-6)(40) = 2.40e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

CBSE Example 3: A point charge of +7 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 7 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(7 x 10^-9)/0.16 = 393.75 V.

Step 3: E = kQ/r² = (9 x 10⁹)(7 x 10^-9)/(0.16)² = 2460.938 N C^-1.

Since Q is positive, E is radially outward.

CBSE Example 4: The electric field between two parallel plates is 350 N/C and plate separation is 0.02 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (350)(0.02) = 7 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

CBSE Example 5: In a region, V(r) = -50r + 135 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -50.

Step 2: E = -dV/dr = -(-50) = 50 N C^-1.

The positive sign shows the field is along increasing r.

CBSE Example 6: In a small region, V is constant at 96 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

CBSE Example 7: The potential is V(x) = 60x² + 9x + 17 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(60)x + 9.

Step 2: At x = 3, dV/dx = 360 + 9 = 369 V/m.

Step 3: E_x = -369 N/C.

The negative sign means field points opposite to increasing x at that point.

CBSE Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

CBSE Example 9: The slope of a V-r graph at a point is +70 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +70 V/m.

Step 2: E = -dV/dr = -70 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

CBSE Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 300 V. Find the work done by electric force on a 6 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 300 - 300 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

CBSE Example 11: The potential changes from 303 V to 198 V when moving from r = 0.12 m to r = 0.44 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 198 - 303 = -105 V.

Step 2: dr = 0.44 - 0.12 = 0.32 m.

Step 3: dV/dr = -105/0.32 = -328.125 V m^-1.

Step 4: E = -dV/dr = 328.125 N C^-1.

The field points in the direction of decreasing potential.

CBSE Example 12: A charge of 8 microcoulomb moves through a potential difference of 140 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 8 microcoulomb = 8 x 10^-6 C.

Step 2: ΔV = 140 V.

Step 3: W = (8 x 10^-6)(140) = 1.12e-3 J.

The positive answer means external work is required to move a positive charge to higher potential.

CBSE Example 13: A point charge of +8 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 8 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(8 x 10^-9)/0.16 = 450 V.

Step 3: E = kQ/r² = (9 x 10⁹)(8 x 10^-9)/(0.16)² = 2812.5 N C^-1.

Since Q is positive, E is radially outward.

CBSE Example 14: The electric field between two parallel plates is 250 N/C and plate separation is 0.05 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (250)(0.05) = 12.5 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

CBSE Example 15: In a region, V(r) = -45r + 145 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -45.

Step 2: E = -dV/dr = -(-45) = 45 N C^-1.

The positive sign shows the field is along increasing r.

CBSE Example 16: In a small region, V is constant at 106 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

CBSE Example 17: The potential is V(x) = 55x² + 7x + 27 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(55)x + 7.

Step 2: At x = 3, dV/dx = 330 + 7 = 337 V/m.

Step 3: E_x = -337 N/C.

The negative sign means field points opposite to increasing x at that point.

CBSE Example 18: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

CBSE Example 19: The slope of a V-r graph at a point is +65 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +65 V/m.

Step 2: E = -dV/dr = -65 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

CBSE Example 20: A charge moves along an equipotential surface from A to B. The potential at both points is 330 V. Find the work done by electric force on a 8 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 330 - 330 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

20 NEET Solved Examples

NEET Example 1: The potential changes from 483 V to 378 V when moving from r = 0.12 m to r = 0.28 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 378 - 483 = -105 V.

Step 2: dr = 0.28 - 0.12 = 0.16 m.

Step 3: dV/dr = -105/0.16 = -656.25 V m^-1.

Step 4: E = -dV/dr = 656.25 N C^-1.

The field points in the direction of decreasing potential.

NEET Example 2: A charge of 4 microcoulomb moves through a potential difference of 140 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 4 microcoulomb = 4 x 10^-6 C.

Step 2: ΔV = 140 V.

Step 3: W = (4 x 10^-6)(140) = 5.60e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

NEET Example 3: A point charge of +5 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 5 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(5 x 10^-9)/0.16 = 281.25 V.

Step 3: E = kQ/r² = (9 x 10⁹)(5 x 10^-9)/(0.16)² = 1757.813 N C^-1.

Since Q is positive, E is radially outward.

NEET Example 4: The electric field between two parallel plates is 250 N/C and plate separation is 0.02 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (250)(0.02) = 5 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

NEET Example 5: In a region, V(r) = -70r + 205 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -70.

Step 2: E = -dV/dr = -(-70) = 70 N C^-1.

The positive sign shows the field is along increasing r.

NEET Example 6: In a small region, V is constant at 166 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

NEET Example 7: The potential is V(x) = 80x² + 7x + 87 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(80)x + 7.

Step 2: At x = 3, dV/dx = 480 + 7 = 487 V/m.

Step 3: E_x = -487 N/C.

The negative sign means field points opposite to increasing x at that point.

NEET Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

NEET Example 9: The slope of a V-r graph at a point is +35 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +35 V/m.

Step 2: E = -dV/dr = -35 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

NEET Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 510 V. Find the work done by electric force on a 4 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 510 - 510 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

NEET Example 11: The potential changes from 513 V to 438 V when moving from r = 0.16 m to r = 0.4 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 438 - 513 = -75 V.

Step 2: dr = 0.4 - 0.16 = 0.24 m.

Step 3: dV/dr = -75/0.24 = -312.5 V m^-1.

Step 4: E = -dV/dr = 312.5 N C^-1.

The field points in the direction of decreasing potential.

NEET Example 12: A charge of 6 microcoulomb moves through a potential difference of 120 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 6 microcoulomb = 6 x 10^-6 C.

Step 2: ΔV = 120 V.

Step 3: W = (6 x 10^-6)(120) = 7.20e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

NEET Example 13: A point charge of +6 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 6 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(6 x 10^-9)/0.16 = 337.5 V.

Step 3: E = kQ/r² = (9 x 10⁹)(6 x 10^-9)/(0.16)² = 2109.375 N C^-1.

Since Q is positive, E is radially outward.

NEET Example 14: The electric field between two parallel plates is 750 N/C and plate separation is 0.05 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (750)(0.05) = 37.5 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

NEET Example 15: In a region, V(r) = -65r + 215 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -65.

Step 2: E = -dV/dr = -(-65) = 65 N C^-1.

The positive sign shows the field is along increasing r.

NEET Example 16: In a small region, V is constant at 176 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

NEET Example 17: The potential is V(x) = 75x² + 5x + 97 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(75)x + 5.

Step 2: At x = 3, dV/dx = 450 + 5 = 455 V/m.

Step 3: E_x = -455 N/C.

The negative sign means field points opposite to increasing x at that point.

NEET Example 18: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

NEET Example 19: The slope of a V-r graph at a point is +30 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +30 V/m.

Step 2: E = -dV/dr = -30 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

NEET Example 20: A charge moves along an equipotential surface from A to B. The potential at both points is 540 V. Find the work done by electric force on a 6 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 540 - 540 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

20 JEE Main Solved Examples

JEE Main Example 1: The potential changes from 693 V to 618 V when moving from r = 0.16 m to r = 0.56 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 618 - 693 = -75 V.

Step 2: dr = 0.56 - 0.16 = 0.4 m.

Step 3: dV/dr = -75/0.4 = -187.5 V m^-1.

Step 4: E = -dV/dr = 187.5 N C^-1.

The field points in the direction of decreasing potential.

JEE Main Example 2: A charge of 2 microcoulomb moves through a potential difference of 120 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 2 microcoulomb = 2 x 10^-6 C.

Step 2: ΔV = 120 V.

Step 3: W = (2 x 10^-6)(120) = 2.40e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

JEE Main Example 3: A point charge of +3 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 3 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(3 x 10^-9)/0.16 = 168.75 V.

Step 3: E = kQ/r² = (9 x 10⁹)(3 x 10^-9)/(0.16)² = 1054.688 N C^-1.

Since Q is positive, E is radially outward.

JEE Main Example 4: The electric field between two parallel plates is 750 N/C and plate separation is 0.02 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (750)(0.02) = 15 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

JEE Main Example 5: In a region, V(r) = -35r + 275 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -35.

Step 2: E = -dV/dr = -(-35) = 35 N C^-1.

The positive sign shows the field is along increasing r.

JEE Main Example 6: In a small region, V is constant at 236 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

JEE Main Example 7: The potential is V(x) = 45x² + 5x + 157 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(45)x + 5.

Step 2: At x = 3, dV/dx = 270 + 5 = 275 V/m.

Step 3: E_x = -275 N/C.

The negative sign means field points opposite to increasing x at that point.

JEE Main Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

JEE Main Example 9: The slope of a V-r graph at a point is +55 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +55 V/m.

Step 2: E = -dV/dr = -55 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

JEE Main Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 720 V. Find the work done by electric force on a 2 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 720 - 720 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

JEE Main Example 11: The potential changes from 723 V to 588 V when moving from r = 0.12 m to r = 0.28 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 588 - 723 = -135 V.

Step 2: dr = 0.28 - 0.12 = 0.16 m.

Step 3: dV/dr = -135/0.16 = -843.75 V m^-1.

Step 4: E = -dV/dr = 843.75 N C^-1.

The field points in the direction of decreasing potential.

JEE Main Example 12: A charge of 4 microcoulomb moves through a potential difference of 100 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 4 microcoulomb = 4 x 10^-6 C.

Step 2: ΔV = 100 V.

Step 3: W = (4 x 10^-6)(100) = 4.00e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

JEE Main Example 13: A point charge of +4 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 4 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(4 x 10^-9)/0.16 = 225 V.

Step 3: E = kQ/r² = (9 x 10⁹)(4 x 10^-9)/(0.16)² = 1406.25 N C^-1.

Since Q is positive, E is radially outward.

JEE Main Example 14: The electric field between two parallel plates is 650 N/C and plate separation is 0.05 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (650)(0.05) = 32.5 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

JEE Main Example 15: In a region, V(r) = -30r + 285 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -30.

Step 2: E = -dV/dr = -(-30) = 30 N C^-1.

The positive sign shows the field is along increasing r.

JEE Main Example 16: In a small region, V is constant at 246 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

JEE Main Example 17: The potential is V(x) = 40x² + 9x + 167 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(40)x + 9.

Step 2: At x = 3, dV/dx = 240 + 9 = 249 V/m.

Step 3: E_x = -249 N/C.

The negative sign means field points opposite to increasing x at that point.

JEE Main Example 18: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

JEE Main Example 19: The slope of a V-r graph at a point is +50 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +50 V/m.

Step 2: E = -dV/dr = -50 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

JEE Main Example 20: A charge moves along an equipotential surface from A to B. The potential at both points is 750 V. Find the work done by electric force on a 4 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 750 - 750 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

20 JEE Advanced Solved Examples

JEE Advanced Example 1: The potential changes from 903 V to 768 V when moving from r = 0.12 m to r = 0.44 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 768 - 903 = -135 V.

Step 2: dr = 0.44 - 0.12 = 0.32 m.

Step 3: dV/dr = -135/0.32 = -421.875 V m^-1.

Step 4: E = -dV/dr = 421.875 N C^-1.

The field points in the direction of decreasing potential.

JEE Advanced Example 2: A charge of 8 microcoulomb moves through a potential difference of 100 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 8 microcoulomb = 8 x 10^-6 C.

Step 2: ΔV = 100 V.

Step 3: W = (8 x 10^-6)(100) = 8.00e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

JEE Advanced Example 3: A point charge of +10 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 10 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(10 x 10^-9)/0.16 = 562.5 V.

Step 3: E = kQ/r² = (9 x 10⁹)(10 x 10^-9)/(0.16)² = 3515.625 N C^-1.

Since Q is positive, E is radially outward.

JEE Advanced Example 4: The electric field between two parallel plates is 650 N/C and plate separation is 0.02 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (650)(0.02) = 13 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

JEE Advanced Example 5: In a region, V(r) = -55r + 345 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -55.

Step 2: E = -dV/dr = -(-55) = 55 N C^-1.

The positive sign shows the field is along increasing r.

JEE Advanced Example 6: In a small region, V is constant at 306 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

JEE Advanced Example 7: The potential is V(x) = 65x² + 9x + 227 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(65)x + 9.

Step 2: At x = 3, dV/dx = 390 + 9 = 399 V/m.

Step 3: E_x = -399 N/C.

The negative sign means field points opposite to increasing x at that point.

JEE Advanced Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

JEE Advanced Example 9: The slope of a V-r graph at a point is +75 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +75 V/m.

Step 2: E = -dV/dr = -75 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

JEE Advanced Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 930 V. Find the work done by electric force on a 8 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 930 - 930 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

JEE Advanced Example 11: The potential changes from 933 V to 828 V when moving from r = 0.16 m to r = 0.56 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 828 - 933 = -105 V.

Step 2: dr = 0.56 - 0.16 = 0.4 m.

Step 3: dV/dr = -105/0.4 = -262.5 V m^-1.

Step 4: E = -dV/dr = 262.5 N C^-1.

The field points in the direction of decreasing potential.

JEE Advanced Example 12: A charge of 2 microcoulomb moves through a potential difference of 80 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 2 microcoulomb = 2 x 10^-6 C.

Step 2: ΔV = 80 V.

Step 3: W = (2 x 10^-6)(80) = 1.60e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

JEE Advanced Example 13: A point charge of +11 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 11 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(11 x 10^-9)/0.16 = 618.75 V.

Step 3: E = kQ/r² = (9 x 10⁹)(11 x 10^-9)/(0.16)² = 3867.188 N C^-1.

Since Q is positive, E is radially outward.

JEE Advanced Example 14: The electric field between two parallel plates is 550 N/C and plate separation is 0.05 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (550)(0.05) = 27.5 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

JEE Advanced Example 15: In a region, V(r) = -50r + 355 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -50.

Step 2: E = -dV/dr = -(-50) = 50 N C^-1.

The positive sign shows the field is along increasing r.

JEE Advanced Example 16: In a small region, V is constant at 316 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

JEE Advanced Example 17: The potential is V(x) = 60x² + 7x + 237 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(60)x + 7.

Step 2: At x = 3, dV/dx = 360 + 7 = 367 V/m.

Step 3: E_x = -367 N/C.

The negative sign means field points opposite to increasing x at that point.

JEE Advanced Example 18: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

JEE Advanced Example 19: The slope of a V-r graph at a point is +70 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +70 V/m.

Step 2: E = -dV/dr = -70 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

JEE Advanced Example 20: A charge moves along an equipotential surface from A to B. The potential at both points is 960 V. Find the work done by electric force on a 2 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 960 - 960 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

10 IB Physics Solved Examples

IB Physics Example 1: The potential changes from 1173 V to 1038 V when moving from r = 0.16 m to r = 0.56 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 1038 - 1173 = -135 V.

Step 2: dr = 0.56 - 0.16 = 0.4 m.

Step 3: dV/dr = -135/0.4 = -337.5 V m^-1.

Step 4: E = -dV/dr = 337.5 N C^-1.

The field points in the direction of decreasing potential.

IB Physics Example 2: A charge of 2 microcoulomb moves through a potential difference of 40 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 2 microcoulomb = 2 x 10^-6 C.

Step 2: ΔV = 40 V.

Step 3: W = (2 x 10^-6)(40) = 8.00e-5 J.

The positive answer means external work is required to move a positive charge to higher potential.

IB Physics Example 3: A point charge of +10 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 10 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(10 x 10^-9)/0.16 = 562.5 V.

Step 3: E = kQ/r² = (9 x 10⁹)(10 x 10^-9)/(0.16)² = 3515.625 N C^-1.

Since Q is positive, E is radially outward.

IB Physics Example 4: The electric field between two parallel plates is 350 N/C and plate separation is 0.08 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (350)(0.08) = 28 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

IB Physics Example 5: In a region, V(r) = -65r + 435 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -65.

Step 2: E = -dV/dr = -(-65) = 65 N C^-1.

The positive sign shows the field is along increasing r.

IB Physics Example 6: In a small region, V is constant at 396 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

IB Physics Example 7: The potential is V(x) = 75x² + 9x + 317 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(75)x + 9.

Step 2: At x = 3, dV/dx = 450 + 9 = 459 V/m.

Step 3: E_x = -459 N/C.

The negative sign means field points opposite to increasing x at that point.

IB Physics Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

IB Physics Example 9: The slope of a V-r graph at a point is +30 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +30 V/m.

Step 2: E = -dV/dr = -30 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

IB Physics Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 1200 V. Find the work done by electric force on a 2 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 1200 - 1200 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

10 IGCSE Solved Examples

IGCSE Example 1: The potential changes from 1353 V to 1218 V when moving from r = 0.16 m to r = 0.4 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 1218 - 1353 = -135 V.

Step 2: dr = 0.4 - 0.16 = 0.24 m.

Step 3: dV/dr = -135/0.24 = -562.5 V m^-1.

Step 4: E = -dV/dr = 562.5 N C^-1.

The field points in the direction of decreasing potential.

IGCSE Example 2: A charge of 6 microcoulomb moves through a potential difference of 40 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 6 microcoulomb = 6 x 10^-6 C.

Step 2: ΔV = 40 V.

Step 3: W = (6 x 10^-6)(40) = 2.40e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

IGCSE Example 3: A point charge of +7 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 7 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(7 x 10^-9)/0.16 = 393.75 V.

Step 3: E = kQ/r² = (9 x 10⁹)(7 x 10^-9)/(0.16)² = 2460.938 N C^-1.

Since Q is positive, E is radially outward.

IGCSE Example 4: The electric field between two parallel plates is 350 N/C and plate separation is 0.05 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (350)(0.05) = 17.5 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

IGCSE Example 5: In a region, V(r) = -35r + 495 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -35.

Step 2: E = -dV/dr = -(-35) = 35 N C^-1.

The positive sign shows the field is along increasing r.

IGCSE Example 6: In a small region, V is constant at 456 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

IGCSE Example 7: The potential is V(x) = 45x² + 9x + 377 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(45)x + 9.

Step 2: At x = 3, dV/dx = 270 + 9 = 279 V/m.

Step 3: E_x = -279 N/C.

The negative sign means field points opposite to increasing x at that point.

IGCSE Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

IGCSE Example 9: The slope of a V-r graph at a point is +55 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +55 V/m.

Step 2: E = -dV/dr = -55 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

IGCSE Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 1380 V. Find the work done by electric force on a 6 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 1380 - 1380 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

10 ICSE Solved Examples

ICSE Example 1: The potential changes from 1533 V to 1398 V when moving from r = 0.16 m to r = 0.56 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 1398 - 1533 = -135 V.

Step 2: dr = 0.56 - 0.16 = 0.4 m.

Step 3: dV/dr = -135/0.4 = -337.5 V m^-1.

Step 4: E = -dV/dr = 337.5 N C^-1.

The field points in the direction of decreasing potential.

ICSE Example 2: A charge of 2 microcoulomb moves through a potential difference of 40 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 2 microcoulomb = 2 x 10^-6 C.

Step 2: ΔV = 40 V.

Step 3: W = (2 x 10^-6)(40) = 8.00e-5 J.

The positive answer means external work is required to move a positive charge to higher potential.

ICSE Example 3: A point charge of +4 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 4 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(4 x 10^-9)/0.16 = 225 V.

Step 3: E = kQ/r² = (9 x 10⁹)(4 x 10^-9)/(0.16)² = 1406.25 N C^-1.

Since Q is positive, E is radially outward.

ICSE Example 4: The electric field between two parallel plates is 350 N/C and plate separation is 0.02 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (350)(0.02) = 7 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

ICSE Example 5: In a region, V(r) = -60r + 555 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -60.

Step 2: E = -dV/dr = -(-60) = 60 N C^-1.

The positive sign shows the field is along increasing r.

ICSE Example 6: In a small region, V is constant at 516 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

ICSE Example 7: The potential is V(x) = 70x² + 9x + 437 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(70)x + 9.

Step 2: At x = 3, dV/dx = 420 + 9 = 429 V/m.

Step 3: E_x = -429 N/C.

The negative sign means field points opposite to increasing x at that point.

ICSE Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

ICSE Example 9: The slope of a V-r graph at a point is +80 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +80 V/m.

Step 2: E = -dV/dr = -80 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

ICSE Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 1560 V. Find the work done by electric force on a 2 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 1560 - 1560 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

10 UP Board Solved Examples

UP Board Example 1: The potential changes from 1713 V to 1578 V when moving from r = 0.16 m to r = 0.4 m along a straight line. Find the electric field component along r.

Show Answer

Use the graph-slope relation.

Step 1: dV = V₂ - V₁ = 1578 - 1713 = -135 V.

Step 2: dr = 0.4 - 0.16 = 0.24 m.

Step 3: dV/dr = -135/0.24 = -562.5 V m^-1.

Step 4: E = -dV/dr = 562.5 N C^-1.

The field points in the direction of decreasing potential.

UP Board Example 2: A charge of 6 microcoulomb moves through a potential difference of 40 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

Show Answer

For slow motion, external work stored as electric potential energy is W = qΔV.

Step 1: q = 6 microcoulomb = 6 x 10^-6 C.

Step 2: ΔV = 40 V.

Step 3: W = (6 x 10^-6)(40) = 2.40e-4 J.

The positive answer means external work is required to move a positive charge to higher potential.

UP Board Example 3: A point charge of +10 nC is kept in air. Find V and E at a point 16 cm away.

Show Answer

Use point-charge formulas.

Step 1: Q = 10 x 10^-9 C and r = 0.16 m.

Step 2: V = kQ/r = (9 x 10⁹)(10 x 10^-9)/0.16 = 562.5 V.

Step 3: E = kQ/r² = (9 x 10⁹)(10 x 10^-9)/(0.16)² = 3515.625 N C^-1.

Since Q is positive, E is radially outward.

UP Board Example 4: The electric field between two parallel plates is 350 N/C and plate separation is 0.06 m. Find the potential difference between the plates.

Show Answer

For a uniform field, magnitude relation is E = ΔV/d.

Step 1: ΔV = Ed.

Step 2: ΔV = (350)(0.06) = 21 V.

Step 3: Potential decreases in the direction of E; the positive plate is at higher potential.

UP Board Example 5: In a region, V(r) = -30r + 615 volt. Find the electric field along r.

Show Answer

Compare with V = -Er + C.

Step 1: dV/dr = -30.

Step 2: E = -dV/dr = -(-30) = 30 N C^-1.

The positive sign shows the field is along increasing r.

UP Board Example 6: In a small region, V is constant at 576 V. What is the electric field in that region?

Show Answer

Electric field depends on potential gradient, not on the absolute value of potential.

Step 1: V = constant.

Step 2: dV/dr = 0.

Step 3: E = -dV/dr = 0.

A non-zero constant potential still gives zero field because there is no spatial change in V.

UP Board Example 7: The potential is V(x) = 40x² + 9x + 497 volt. Find E_x at x = 3 m.

Show Answer

In one dimension, E_x = -dV/dx.

Step 1: dV/dx = 2(40)x + 9.

Step 2: At x = 3, dV/dx = 240 + 9 = 249 V/m.

Step 3: E_x = -249 N/C.

The negative sign means field points opposite to increasing x at that point.

UP Board Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

Show Answer

Start from V = kQr^-1.

Step 1: dV/dr = kQ(-1)r^-2 = -kQ/r².

Step 2: E = -dV/dr.

Step 3: E = -(-kQ/r²) = kQ/r².

The result is positive in the outward radial direction for Q > 0.

UP Board Example 9: The slope of a V-r graph at a point is +50 V/m. Find the electric field and state its direction relative to increasing r.

Show Answer

Step 1: Slope = dV/dr = +50 V/m.

Step 2: E = -dV/dr = -50 N/C.

Step 3: The negative sign means the field is opposite to the direction of increasing r.

UP Board Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 1740 V. Find the work done by electric force on a 6 microcoulomb charge.

Show Answer

Step 1: ΔV = V_B - V_A = 1740 - 1740 = 0.

Step 2: Change in potential energy ΔU = qΔV = 0.

Step 3: Work done by electric force = -ΔU = 0 J.

No work is done along an equipotential surface.

254 non-repeated practice questions

18. Question Bank

This bank includes MCQs, numerical value questions, assertion-reason questions, match the following, graph interpretation and conceptual reasoning.

52 NEET-level Questions

MCQ

NEET-level Q1: Potential decreases from 1302 V to 1277 V in 0.12 m. What is the magnitude of electric field?

208.333 N/C
104.167 N/C
416.667 N/C
0 N/C

Show Answer

Slope = dV/dr = (1277 - 1302)/0.12 = -208.333 V/m.

E = -slope = 208.333 N/C, so magnitude is 208.333 N/C.

Numerical value

NEET-level Q2: A 9 microcoulomb charge moves through a potential difference of 25 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 9 x 10^-6 C.

ΔU = (9 x 10^-6)(25) = 2.25e-4 J.

Assertion-Reason

NEET-level Q3: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q4: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -22r + C	2. E = 0
C. V = kQ/r	3. E = 22 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -22, so E = 22.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

NEET-level Q5: On a V-x graph, the tangent at x = 2 m has slope -25 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -25 V/m.

E_x = -(-25) = +25 N/C, directed along positive x.

Conceptual reasoning

NEET-level Q6: At a point 20 cm from a +8 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.2 m, E = (9 x 10⁹)(8 x 10^-9)/(0.2)² = 1800 N/C.

The field is non-zero because potential changes with distance.

MCQ

NEET-level Q7: Potential decreases from 1314 V to 1259 V in 0.09 m. What is the magnitude of electric field?

611.111 N/C
305.556 N/C
1222.222 N/C
0 N/C

Show Answer

Slope = dV/dr = (1259 - 1314)/0.09 = -611.111 V/m.

E = -slope = 611.111 N/C, so magnitude is 611.111 N/C.

Numerical value

NEET-level Q8: A 6 microcoulomb charge moves through a potential difference of 55 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 6 x 10^-6 C.

ΔU = (6 x 10^-6)(55) = 3.30e-4 J.

Assertion-Reason

NEET-level Q9: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q10: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -40r + C	2. E = 0
C. V = kQ/r	3. E = 40 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -40, so E = 40.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

NEET-level Q11: On a V-x graph, the tangent at x = 3 m has slope -43 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -43 V/m.

E_x = -(-43) = +43 N/C, directed along positive x.

Conceptual reasoning

NEET-level Q12: At a point 8 cm from a +2 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.08 m, E = (9 x 10⁹)(2 x 10^-9)/(0.08)² = 2812.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

NEET-level Q13: Potential decreases from 1326 V to 1281 V in 0.06 m. What is the magnitude of electric field?

750 N/C
375 N/C
1500 N/C
0 N/C

Show Answer

Slope = dV/dr = (1281 - 1326)/0.06 = -750 V/m.

E = -slope = 750 N/C, so magnitude is 750 N/C.

Numerical value

NEET-level Q14: A 3 microcoulomb charge moves through a potential difference of 25 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 3 x 10^-6 C.

ΔU = (3 x 10^-6)(25) = 7.50e-5 J.

Assertion-Reason

NEET-level Q15: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q16: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -13r + C	2. E = 0
C. V = kQ/r	3. E = 13 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -13, so E = 13.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

NEET-level Q17: On a V-x graph, the tangent at x = 4 m has slope -16 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -16 V/m.

E_x = -(-16) = +16 N/C, directed along positive x.

Conceptual reasoning

NEET-level Q18: At a point 20 cm from a +8 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.2 m, E = (9 x 10⁹)(8 x 10^-9)/(0.2)² = 1800 N/C.

The field is non-zero because potential changes with distance.

MCQ

NEET-level Q19: Potential decreases from 1338 V to 1303 V in 0.12 m. What is the magnitude of electric field?

291.667 N/C
145.833 N/C
583.333 N/C
0 N/C

Show Answer

Slope = dV/dr = (1303 - 1338)/0.12 = -291.667 V/m.

E = -slope = 291.667 N/C, so magnitude is 291.667 N/C.

Numerical value

NEET-level Q20: A 9 microcoulomb charge moves through a potential difference of 55 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 9 x 10^-6 C.

ΔU = (9 x 10^-6)(55) = 4.95e-4 J.

Assertion-Reason

NEET-level Q21: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q22: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -31r + C	2. E = 0
C. V = kQ/r	3. E = 31 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -31, so E = 31.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

NEET-level Q23: On a V-x graph, the tangent at x = 5 m has slope -34 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -34 V/m.

E_x = -(-34) = +34 N/C, directed along positive x.

Conceptual reasoning

NEET-level Q24: At a point 8 cm from a +2 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.08 m, E = (9 x 10⁹)(2 x 10^-9)/(0.08)² = 2812.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

NEET-level Q25: Potential decreases from 1350 V to 1325 V in 0.09 m. What is the magnitude of electric field?

277.778 N/C
138.889 N/C
555.556 N/C
0 N/C

Show Answer

Slope = dV/dr = (1325 - 1350)/0.09 = -277.778 V/m.

E = -slope = 277.778 N/C, so magnitude is 277.778 N/C.

Numerical value

NEET-level Q26: A 6 microcoulomb charge moves through a potential difference of 25 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 6 x 10^-6 C.

ΔU = (6 x 10^-6)(25) = 1.50e-4 J.

Assertion-Reason

NEET-level Q27: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q28: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -49r + C	2. E = 0
C. V = kQ/r	3. E = 49 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -49, so E = 49.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

NEET-level Q29: On a V-x graph, the tangent at x = 6 m has slope -52 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -52 V/m.

E_x = -(-52) = +52 N/C, directed along positive x.

Conceptual reasoning

NEET-level Q30: At a point 20 cm from a +8 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.2 m, E = (9 x 10⁹)(8 x 10^-9)/(0.2)² = 1800 N/C.

The field is non-zero because potential changes with distance.

MCQ

NEET-level Q31: Potential decreases from 1362 V to 1307 V in 0.06 m. What is the magnitude of electric field?

916.667 N/C
458.333 N/C
1833.333 N/C
0 N/C

Show Answer

Slope = dV/dr = (1307 - 1362)/0.06 = -916.667 V/m.

E = -slope = 916.667 N/C, so magnitude is 916.667 N/C.

Numerical value

NEET-level Q32: A 3 microcoulomb charge moves through a potential difference of 55 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 3 x 10^-6 C.

ΔU = (3 x 10^-6)(55) = 1.65e-4 J.

Assertion-Reason

NEET-level Q33: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q34: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -22r + C	2. E = 0
C. V = kQ/r	3. E = 22 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -22, so E = 22.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

NEET-level Q35: On a V-x graph, the tangent at x = 2 m has slope -25 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -25 V/m.

E_x = -(-25) = +25 N/C, directed along positive x.

Conceptual reasoning

NEET-level Q36: At a point 8 cm from a +2 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.08 m, E = (9 x 10⁹)(2 x 10^-9)/(0.08)² = 2812.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

NEET-level Q37: Potential decreases from 1374 V to 1329 V in 0.12 m. What is the magnitude of electric field?

375 N/C
187.5 N/C
750 N/C
0 N/C

Show Answer

Slope = dV/dr = (1329 - 1374)/0.12 = -375 V/m.

E = -slope = 375 N/C, so magnitude is 375 N/C.

Numerical value

NEET-level Q38: A 9 microcoulomb charge moves through a potential difference of 25 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 9 x 10^-6 C.

ΔU = (9 x 10^-6)(25) = 2.25e-4 J.

Assertion-Reason

NEET-level Q39: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q40: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -40r + C	2. E = 0
C. V = kQ/r	3. E = 40 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -40, so E = 40.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

NEET-level Q41: On a V-x graph, the tangent at x = 3 m has slope -43 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -43 V/m.

E_x = -(-43) = +43 N/C, directed along positive x.

Conceptual reasoning

NEET-level Q42: At a point 20 cm from a +8 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.2 m, E = (9 x 10⁹)(8 x 10^-9)/(0.2)² = 1800 N/C.

The field is non-zero because potential changes with distance.

MCQ

NEET-level Q43: Potential decreases from 1386 V to 1351 V in 0.09 m. What is the magnitude of electric field?

388.889 N/C
194.444 N/C
777.778 N/C
0 N/C

Show Answer

Slope = dV/dr = (1351 - 1386)/0.09 = -388.889 V/m.

E = -slope = 388.889 N/C, so magnitude is 388.889 N/C.

Numerical value

NEET-level Q44: A 6 microcoulomb charge moves through a potential difference of 55 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 6 x 10^-6 C.

ΔU = (6 x 10^-6)(55) = 3.30e-4 J.

Assertion-Reason

NEET-level Q45: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q46: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -13r + C	2. E = 0
C. V = kQ/r	3. E = 13 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -13, so E = 13.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

NEET-level Q47: On a V-x graph, the tangent at x = 4 m has slope -16 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -16 V/m.

E_x = -(-16) = +16 N/C, directed along positive x.

Conceptual reasoning

NEET-level Q48: At a point 8 cm from a +2 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.08 m, E = (9 x 10⁹)(2 x 10^-9)/(0.08)² = 2812.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

NEET-level Q49: Potential decreases from 1398 V to 1373 V in 0.06 m. What is the magnitude of electric field?

416.667 N/C
208.333 N/C
833.333 N/C
0 N/C

Show Answer

Slope = dV/dr = (1373 - 1398)/0.06 = -416.667 V/m.

E = -slope = 416.667 N/C, so magnitude is 416.667 N/C.

Numerical value

NEET-level Q50: A 3 microcoulomb charge moves through a potential difference of 25 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 3 x 10^-6 C.

ΔU = (3 x 10^-6)(25) = 7.50e-5 J.

Assertion-Reason

NEET-level Q51: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

NEET-level Q52: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -31r + C	2. E = 0
C. V = kQ/r	3. E = 31 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -31, so E = 31.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

52 JEE Main Questions

MCQ

JEE Main Q1: Potential decreases from 1622 V to 1597 V in 0.1 m. What is the magnitude of electric field?

250 N/C
125 N/C
500 N/C
0 N/C

Show Answer

Slope = dV/dr = (1597 - 1622)/0.1 = -250 V/m.

E = -slope = 250 N/C, so magnitude is 250 N/C.

Numerical value

JEE Main Q2: A 7 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 7 x 10^-6 C.

ΔU = (7 x 10^-6)(45) = 3.15e-4 J.

Assertion-Reason

JEE Main Q3: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q4: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -52r + C	2. E = 0
C. V = kQ/r	3. E = 52 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -52, so E = 52.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Main Q5: On a V-x graph, the tangent at x = 2 m has slope -10 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -10 V/m.

E_x = -(-10) = +10 N/C, directed along positive x.

Conceptual reasoning

JEE Main Q6: At a point 28 cm from a +12 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.28 m, E = (9 x 10⁹)(12 x 10^-9)/(0.28)² = 1377.551 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Main Q7: Potential decreases from 1634 V to 1579 V in 0.07 m. What is the magnitude of electric field?

785.714 N/C
392.857 N/C
1571.429 N/C
0 N/C

Show Answer

Slope = dV/dr = (1579 - 1634)/0.07 = -785.714 V/m.

E = -slope = 785.714 N/C, so magnitude is 785.714 N/C.

Numerical value

JEE Main Q8: A 4 microcoulomb charge moves through a potential difference of 15 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 4 x 10^-6 C.

ΔU = (4 x 10^-6)(15) = 6.00e-5 J.

Assertion-Reason

JEE Main Q9: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q10: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -25r + C	2. E = 0
C. V = kQ/r	3. E = 25 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -25, so E = 25.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Main Q11: On a V-x graph, the tangent at x = 3 m has slope -28 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -28 V/m.

E_x = -(-28) = +28 N/C, directed along positive x.

Conceptual reasoning

JEE Main Q12: At a point 16 cm from a +6 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.16 m, E = (9 x 10⁹)(6 x 10^-9)/(0.16)² = 2109.375 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Main Q13: Potential decreases from 1646 V to 1601 V in 0.13 m. What is the magnitude of electric field?

346.154 N/C
173.077 N/C
692.308 N/C
0 N/C

Show Answer

Slope = dV/dr = (1601 - 1646)/0.13 = -346.154 V/m.

E = -slope = 346.154 N/C, so magnitude is 346.154 N/C.

Numerical value

JEE Main Q14: A 1 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 1 x 10^-6 C.

ΔU = (1 x 10^-6)(45) = 4.50e-5 J.

Assertion-Reason

JEE Main Q15: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q16: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -43r + C	2. E = 0
C. V = kQ/r	3. E = 43 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -43, so E = 43.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Main Q17: On a V-x graph, the tangent at x = 4 m has slope -46 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -46 V/m.

E_x = -(-46) = +46 N/C, directed along positive x.

Conceptual reasoning

JEE Main Q18: At a point 28 cm from a +12 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.28 m, E = (9 x 10⁹)(12 x 10^-9)/(0.28)² = 1377.551 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Main Q19: Potential decreases from 1658 V to 1623 V in 0.1 m. What is the magnitude of electric field?

350 N/C
175 N/C
700 N/C
0 N/C

Show Answer

Slope = dV/dr = (1623 - 1658)/0.1 = -350 V/m.

E = -slope = 350 N/C, so magnitude is 350 N/C.

Numerical value

JEE Main Q20: A 7 microcoulomb charge moves through a potential difference of 15 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 7 x 10^-6 C.

ΔU = (7 x 10^-6)(15) = 1.05e-4 J.

Assertion-Reason

JEE Main Q21: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q22: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -16r + C	2. E = 0
C. V = kQ/r	3. E = 16 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -16, so E = 16.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Main Q23: On a V-x graph, the tangent at x = 5 m has slope -19 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -19 V/m.

E_x = -(-19) = +19 N/C, directed along positive x.

Conceptual reasoning

JEE Main Q24: At a point 16 cm from a +6 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.16 m, E = (9 x 10⁹)(6 x 10^-9)/(0.16)² = 2109.375 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Main Q25: Potential decreases from 1670 V to 1645 V in 0.07 m. What is the magnitude of electric field?

357.143 N/C
178.571 N/C
714.286 N/C
0 N/C

Show Answer

Slope = dV/dr = (1645 - 1670)/0.07 = -357.143 V/m.

E = -slope = 357.143 N/C, so magnitude is 357.143 N/C.

Numerical value

JEE Main Q26: A 4 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 4 x 10^-6 C.

ΔU = (4 x 10^-6)(45) = 1.80e-4 J.

Assertion-Reason

JEE Main Q27: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q28: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -34r + C	2. E = 0
C. V = kQ/r	3. E = 34 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -34, so E = 34.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Main Q29: On a V-x graph, the tangent at x = 6 m has slope -37 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -37 V/m.

E_x = -(-37) = +37 N/C, directed along positive x.

Conceptual reasoning

JEE Main Q30: At a point 28 cm from a +12 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.28 m, E = (9 x 10⁹)(12 x 10^-9)/(0.28)² = 1377.551 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Main Q31: Potential decreases from 1682 V to 1627 V in 0.13 m. What is the magnitude of electric field?

423.077 N/C
211.538 N/C
846.154 N/C
0 N/C

Show Answer

Slope = dV/dr = (1627 - 1682)/0.13 = -423.077 V/m.

E = -slope = 423.077 N/C, so magnitude is 423.077 N/C.

Numerical value

JEE Main Q32: A 1 microcoulomb charge moves through a potential difference of 15 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 1 x 10^-6 C.

ΔU = (1 x 10^-6)(15) = 1.50e-5 J.

Assertion-Reason

JEE Main Q33: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q34: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -52r + C	2. E = 0
C. V = kQ/r	3. E = 52 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -52, so E = 52.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Main Q35: On a V-x graph, the tangent at x = 2 m has slope -10 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -10 V/m.

E_x = -(-10) = +10 N/C, directed along positive x.

Conceptual reasoning

JEE Main Q36: At a point 16 cm from a +6 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.16 m, E = (9 x 10⁹)(6 x 10^-9)/(0.16)² = 2109.375 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Main Q37: Potential decreases from 1694 V to 1649 V in 0.1 m. What is the magnitude of electric field?

450 N/C
225 N/C
900 N/C
0 N/C

Show Answer

Slope = dV/dr = (1649 - 1694)/0.1 = -450 V/m.

E = -slope = 450 N/C, so magnitude is 450 N/C.

Numerical value

JEE Main Q38: A 7 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 7 x 10^-6 C.

ΔU = (7 x 10^-6)(45) = 3.15e-4 J.

Assertion-Reason

JEE Main Q39: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q40: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -25r + C	2. E = 0
C. V = kQ/r	3. E = 25 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -25, so E = 25.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Main Q41: On a V-x graph, the tangent at x = 3 m has slope -28 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -28 V/m.

E_x = -(-28) = +28 N/C, directed along positive x.

Conceptual reasoning

JEE Main Q42: At a point 28 cm from a +12 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.28 m, E = (9 x 10⁹)(12 x 10^-9)/(0.28)² = 1377.551 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Main Q43: Potential decreases from 1706 V to 1671 V in 0.07 m. What is the magnitude of electric field?

500 N/C
250 N/C
1000 N/C
0 N/C

Show Answer

Slope = dV/dr = (1671 - 1706)/0.07 = -500 V/m.

E = -slope = 500 N/C, so magnitude is 500 N/C.

Numerical value

JEE Main Q44: A 4 microcoulomb charge moves through a potential difference of 15 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 4 x 10^-6 C.

ΔU = (4 x 10^-6)(15) = 6.00e-5 J.

Assertion-Reason

JEE Main Q45: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q46: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -43r + C	2. E = 0
C. V = kQ/r	3. E = 43 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -43, so E = 43.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Main Q47: On a V-x graph, the tangent at x = 4 m has slope -46 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -46 V/m.

E_x = -(-46) = +46 N/C, directed along positive x.

Conceptual reasoning

JEE Main Q48: At a point 16 cm from a +6 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.16 m, E = (9 x 10⁹)(6 x 10^-9)/(0.16)² = 2109.375 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Main Q49: Potential decreases from 1718 V to 1693 V in 0.13 m. What is the magnitude of electric field?

192.308 N/C
96.154 N/C
384.615 N/C
0 N/C

Show Answer

Slope = dV/dr = (1693 - 1718)/0.13 = -192.308 V/m.

E = -slope = 192.308 N/C, so magnitude is 192.308 N/C.

Numerical value

JEE Main Q50: A 1 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 1 x 10^-6 C.

ΔU = (1 x 10^-6)(45) = 4.50e-5 J.

Assertion-Reason

JEE Main Q51: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Main Q52: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -16r + C	2. E = 0
C. V = kQ/r	3. E = 16 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -16, so E = 16.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

42 JEE Advanced Questions

MCQ

JEE Advanced Q1: Potential decreases from 1942 V to 1917 V in 0.08 m. What is the magnitude of electric field?

312.5 N/C
156.25 N/C
625 N/C
0 N/C

Show Answer

Slope = dV/dr = (1917 - 1942)/0.08 = -312.5 V/m.

E = -slope = 312.5 N/C, so magnitude is 312.5 N/C.

Numerical value

JEE Advanced Q2: A 5 microcoulomb charge moves through a potential difference of 65 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 5 x 10^-6 C.

ΔU = (5 x 10^-6)(65) = 3.25e-4 J.

Assertion-Reason

JEE Advanced Q3: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Advanced Q4: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -37r + C	2. E = 0
C. V = kQ/r	3. E = 37 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -37, so E = 37.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Advanced Q5: On a V-x graph, the tangent at x = 2 m has slope -40 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -40 V/m.

E_x = -(-40) = +40 N/C, directed along positive x.

Conceptual reasoning

JEE Advanced Q6: At a point 12 cm from a +4 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.12 m, E = (9 x 10⁹)(4 x 10^-9)/(0.12)² = 2500 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Advanced Q7: Potential decreases from 1954 V to 1899 V in 0.05 m. What is the magnitude of electric field?

1100 N/C
550 N/C
2200 N/C
0 N/C

Show Answer

Slope = dV/dr = (1899 - 1954)/0.05 = -1100 V/m.

E = -slope = 1100 N/C, so magnitude is 1100 N/C.

Numerical value

JEE Advanced Q8: A 2 microcoulomb charge moves through a potential difference of 35 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 2 x 10^-6 C.

ΔU = (2 x 10^-6)(35) = 7.00e-5 J.

Assertion-Reason

JEE Advanced Q9: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Advanced Q10: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -10r + C	2. E = 0
C. V = kQ/r	3. E = 10 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -10, so E = 10.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Advanced Q11: On a V-x graph, the tangent at x = 3 m has slope -13 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -13 V/m.

E_x = -(-13) = +13 N/C, directed along positive x.

Conceptual reasoning

JEE Advanced Q12: At a point 24 cm from a +10 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.24 m, E = (9 x 10⁹)(10 x 10^-9)/(0.24)² = 1562.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Advanced Q13: Potential decreases from 1966 V to 1921 V in 0.11 m. What is the magnitude of electric field?

409.091 N/C
204.545 N/C
818.182 N/C
0 N/C

Show Answer

Slope = dV/dr = (1921 - 1966)/0.11 = -409.091 V/m.

E = -slope = 409.091 N/C, so magnitude is 409.091 N/C.

Numerical value

JEE Advanced Q14: A 8 microcoulomb charge moves through a potential difference of 65 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 8 x 10^-6 C.

ΔU = (8 x 10^-6)(65) = 5.20e-4 J.

Assertion-Reason

JEE Advanced Q15: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Advanced Q16: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -28r + C	2. E = 0
C. V = kQ/r	3. E = 28 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -28, so E = 28.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Advanced Q17: On a V-x graph, the tangent at x = 4 m has slope -31 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -31 V/m.

E_x = -(-31) = +31 N/C, directed along positive x.

Conceptual reasoning

JEE Advanced Q18: At a point 12 cm from a +4 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.12 m, E = (9 x 10⁹)(4 x 10^-9)/(0.12)² = 2500 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Advanced Q19: Potential decreases from 1978 V to 1943 V in 0.08 m. What is the magnitude of electric field?

437.5 N/C
218.75 N/C
875 N/C
0 N/C

Show Answer

Slope = dV/dr = (1943 - 1978)/0.08 = -437.5 V/m.

E = -slope = 437.5 N/C, so magnitude is 437.5 N/C.

Numerical value

JEE Advanced Q20: A 5 microcoulomb charge moves through a potential difference of 35 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 5 x 10^-6 C.

ΔU = (5 x 10^-6)(35) = 1.75e-4 J.

Assertion-Reason

JEE Advanced Q21: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Advanced Q22: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -46r + C	2. E = 0
C. V = kQ/r	3. E = 46 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -46, so E = 46.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Advanced Q23: On a V-x graph, the tangent at x = 5 m has slope -49 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -49 V/m.

E_x = -(-49) = +49 N/C, directed along positive x.

Conceptual reasoning

JEE Advanced Q24: At a point 24 cm from a +10 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.24 m, E = (9 x 10⁹)(10 x 10^-9)/(0.24)² = 1562.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Advanced Q25: Potential decreases from 1990 V to 1965 V in 0.05 m. What is the magnitude of electric field?

500 N/C
250 N/C
1000 N/C
0 N/C

Show Answer

Slope = dV/dr = (1965 - 1990)/0.05 = -500 V/m.

E = -slope = 500 N/C, so magnitude is 500 N/C.

Numerical value

JEE Advanced Q26: A 2 microcoulomb charge moves through a potential difference of 65 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 2 x 10^-6 C.

ΔU = (2 x 10^-6)(65) = 1.30e-4 J.

Assertion-Reason

JEE Advanced Q27: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Advanced Q28: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -19r + C	2. E = 0
C. V = kQ/r	3. E = 19 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -19, so E = 19.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Advanced Q29: On a V-x graph, the tangent at x = 6 m has slope -22 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -22 V/m.

E_x = -(-22) = +22 N/C, directed along positive x.

Conceptual reasoning

JEE Advanced Q30: At a point 12 cm from a +4 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.12 m, E = (9 x 10⁹)(4 x 10^-9)/(0.12)² = 2500 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Advanced Q31: Potential decreases from 2002 V to 1947 V in 0.11 m. What is the magnitude of electric field?

500 N/C
250 N/C
1000 N/C
0 N/C

Show Answer

Slope = dV/dr = (1947 - 2002)/0.11 = -500 V/m.

E = -slope = 500 N/C, so magnitude is 500 N/C.

Numerical value

JEE Advanced Q32: A 8 microcoulomb charge moves through a potential difference of 35 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 8 x 10^-6 C.

ΔU = (8 x 10^-6)(35) = 2.80e-4 J.

Assertion-Reason

JEE Advanced Q33: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Advanced Q34: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -37r + C	2. E = 0
C. V = kQ/r	3. E = 37 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -37, so E = 37.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Advanced Q35: On a V-x graph, the tangent at x = 2 m has slope -40 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -40 V/m.

E_x = -(-40) = +40 N/C, directed along positive x.

Conceptual reasoning

JEE Advanced Q36: At a point 24 cm from a +10 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.24 m, E = (9 x 10⁹)(10 x 10^-9)/(0.24)² = 1562.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

JEE Advanced Q37: Potential decreases from 2014 V to 1969 V in 0.08 m. What is the magnitude of electric field?

562.5 N/C
281.25 N/C
1125 N/C
0 N/C

Show Answer

Slope = dV/dr = (1969 - 2014)/0.08 = -562.5 V/m.

E = -slope = 562.5 N/C, so magnitude is 562.5 N/C.

Numerical value

JEE Advanced Q38: A 5 microcoulomb charge moves through a potential difference of 65 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 5 x 10^-6 C.

ΔU = (5 x 10^-6)(65) = 3.25e-4 J.

Assertion-Reason

JEE Advanced Q39: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

JEE Advanced Q40: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -10r + C	2. E = 0
C. V = kQ/r	3. E = 10 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -10, so E = 10.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

JEE Advanced Q41: On a V-x graph, the tangent at x = 3 m has slope -13 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -13 V/m.

E_x = -(-13) = +13 N/C, directed along positive x.

Conceptual reasoning

JEE Advanced Q42: At a point 12 cm from a +4 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.12 m, E = (9 x 10⁹)(4 x 10^-9)/(0.12)² = 2500 N/C.

The field is non-zero because potential changes with distance.

32 IB Physics Questions

MCQ

IB Physics Q1: Potential decreases from 2262 V to 2237 V in 0.06 m. What is the magnitude of electric field?

416.667 N/C
208.333 N/C
833.333 N/C
0 N/C

Show Answer

Slope = dV/dr = (2237 - 2262)/0.06 = -416.667 V/m.

E = -slope = 416.667 N/C, so magnitude is 416.667 N/C.

Numerical value

IB Physics Q2: A 3 microcoulomb charge moves through a potential difference of 25 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 3 x 10^-6 C.

ΔU = (3 x 10^-6)(25) = 7.50e-5 J.

Assertion-Reason

IB Physics Q3: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IB Physics Q4: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -22r + C	2. E = 0
C. V = kQ/r	3. E = 22 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -22, so E = 22.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IB Physics Q5: On a V-x graph, the tangent at x = 2 m has slope -25 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -25 V/m.

E_x = -(-25) = +25 N/C, directed along positive x.

Conceptual reasoning

IB Physics Q6: At a point 20 cm from a +8 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.2 m, E = (9 x 10⁹)(8 x 10^-9)/(0.2)² = 1800 N/C.

The field is non-zero because potential changes with distance.

MCQ

IB Physics Q7: Potential decreases from 2274 V to 2219 V in 0.12 m. What is the magnitude of electric field?

458.333 N/C
229.167 N/C
916.667 N/C
0 N/C

Show Answer

Slope = dV/dr = (2219 - 2274)/0.12 = -458.333 V/m.

E = -slope = 458.333 N/C, so magnitude is 458.333 N/C.

Numerical value

IB Physics Q8: A 9 microcoulomb charge moves through a potential difference of 55 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 9 x 10^-6 C.

ΔU = (9 x 10^-6)(55) = 4.95e-4 J.

Assertion-Reason

IB Physics Q9: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IB Physics Q10: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -40r + C	2. E = 0
C. V = kQ/r	3. E = 40 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -40, so E = 40.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IB Physics Q11: On a V-x graph, the tangent at x = 3 m has slope -43 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -43 V/m.

E_x = -(-43) = +43 N/C, directed along positive x.

Conceptual reasoning

IB Physics Q12: At a point 8 cm from a +2 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.08 m, E = (9 x 10⁹)(2 x 10^-9)/(0.08)² = 2812.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

IB Physics Q13: Potential decreases from 2286 V to 2241 V in 0.09 m. What is the magnitude of electric field?

500 N/C
250 N/C
1000 N/C
0 N/C

Show Answer

Slope = dV/dr = (2241 - 2286)/0.09 = -500 V/m.

E = -slope = 500 N/C, so magnitude is 500 N/C.

Numerical value

IB Physics Q14: A 6 microcoulomb charge moves through a potential difference of 25 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 6 x 10^-6 C.

ΔU = (6 x 10^-6)(25) = 1.50e-4 J.

Assertion-Reason

IB Physics Q15: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IB Physics Q16: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -13r + C	2. E = 0
C. V = kQ/r	3. E = 13 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -13, so E = 13.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IB Physics Q17: On a V-x graph, the tangent at x = 4 m has slope -16 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -16 V/m.

E_x = -(-16) = +16 N/C, directed along positive x.

Conceptual reasoning

IB Physics Q18: At a point 20 cm from a +8 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.2 m, E = (9 x 10⁹)(8 x 10^-9)/(0.2)² = 1800 N/C.

The field is non-zero because potential changes with distance.

MCQ

IB Physics Q19: Potential decreases from 2298 V to 2263 V in 0.06 m. What is the magnitude of electric field?

583.333 N/C
291.667 N/C
1166.667 N/C
0 N/C

Show Answer

Slope = dV/dr = (2263 - 2298)/0.06 = -583.333 V/m.

E = -slope = 583.333 N/C, so magnitude is 583.333 N/C.

Numerical value

IB Physics Q20: A 3 microcoulomb charge moves through a potential difference of 55 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 3 x 10^-6 C.

ΔU = (3 x 10^-6)(55) = 1.65e-4 J.

Assertion-Reason

IB Physics Q21: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IB Physics Q22: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -31r + C	2. E = 0
C. V = kQ/r	3. E = 31 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -31, so E = 31.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IB Physics Q23: On a V-x graph, the tangent at x = 5 m has slope -34 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -34 V/m.

E_x = -(-34) = +34 N/C, directed along positive x.

Conceptual reasoning

IB Physics Q24: At a point 8 cm from a +2 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.08 m, E = (9 x 10⁹)(2 x 10^-9)/(0.08)² = 2812.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

IB Physics Q25: Potential decreases from 2310 V to 2285 V in 0.12 m. What is the magnitude of electric field?

208.333 N/C
104.167 N/C
416.667 N/C
0 N/C

Show Answer

Slope = dV/dr = (2285 - 2310)/0.12 = -208.333 V/m.

E = -slope = 208.333 N/C, so magnitude is 208.333 N/C.

Numerical value

IB Physics Q26: A 9 microcoulomb charge moves through a potential difference of 25 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 9 x 10^-6 C.

ΔU = (9 x 10^-6)(25) = 2.25e-4 J.

Assertion-Reason

IB Physics Q27: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IB Physics Q28: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -49r + C	2. E = 0
C. V = kQ/r	3. E = 49 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -49, so E = 49.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IB Physics Q29: On a V-x graph, the tangent at x = 6 m has slope -52 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -52 V/m.

E_x = -(-52) = +52 N/C, directed along positive x.

Conceptual reasoning

IB Physics Q30: At a point 20 cm from a +8 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.2 m, E = (9 x 10⁹)(8 x 10^-9)/(0.2)² = 1800 N/C.

The field is non-zero because potential changes with distance.

MCQ

IB Physics Q31: Potential decreases from 2322 V to 2267 V in 0.09 m. What is the magnitude of electric field?

611.111 N/C
305.556 N/C
1222.222 N/C
0 N/C

Show Answer

Slope = dV/dr = (2267 - 2322)/0.09 = -611.111 V/m.

E = -slope = 611.111 N/C, so magnitude is 611.111 N/C.

Numerical value

IB Physics Q32: A 6 microcoulomb charge moves through a potential difference of 55 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 6 x 10^-6 C.

ΔU = (6 x 10^-6)(55) = 3.30e-4 J.

32 IGCSE Questions

MCQ

IGCSE Q1: Potential decreases from 2542 V to 2497 V in 0.11 m. What is the magnitude of electric field?

409.091 N/C
204.545 N/C
818.182 N/C
0 N/C

Show Answer

Slope = dV/dr = (2497 - 2542)/0.11 = -409.091 V/m.

E = -slope = 409.091 N/C, so magnitude is 409.091 N/C.

Numerical value

IGCSE Q2: A 8 microcoulomb charge moves through a potential difference of 65 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 8 x 10^-6 C.

ΔU = (8 x 10^-6)(65) = 5.20e-4 J.

Assertion-Reason

IGCSE Q3: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IGCSE Q4: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -37r + C	2. E = 0
C. V = kQ/r	3. E = 37 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -37, so E = 37.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IGCSE Q5: On a V-x graph, the tangent at x = 2 m has slope -40 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -40 V/m.

E_x = -(-40) = +40 N/C, directed along positive x.

Conceptual reasoning

IGCSE Q6: At a point 12 cm from a +4 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.12 m, E = (9 x 10⁹)(4 x 10^-9)/(0.12)² = 2500 N/C.

The field is non-zero because potential changes with distance.

MCQ

IGCSE Q7: Potential decreases from 2554 V to 2519 V in 0.08 m. What is the magnitude of electric field?

437.5 N/C
218.75 N/C
875 N/C
0 N/C

Show Answer

Slope = dV/dr = (2519 - 2554)/0.08 = -437.5 V/m.

E = -slope = 437.5 N/C, so magnitude is 437.5 N/C.

Numerical value

IGCSE Q8: A 5 microcoulomb charge moves through a potential difference of 35 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 5 x 10^-6 C.

ΔU = (5 x 10^-6)(35) = 1.75e-4 J.

Assertion-Reason

IGCSE Q9: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IGCSE Q10: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -10r + C	2. E = 0
C. V = kQ/r	3. E = 10 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -10, so E = 10.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IGCSE Q11: On a V-x graph, the tangent at x = 3 m has slope -13 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -13 V/m.

E_x = -(-13) = +13 N/C, directed along positive x.

Conceptual reasoning

IGCSE Q12: At a point 24 cm from a +10 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.24 m, E = (9 x 10⁹)(10 x 10^-9)/(0.24)² = 1562.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

IGCSE Q13: Potential decreases from 2566 V to 2541 V in 0.05 m. What is the magnitude of electric field?

500 N/C
250 N/C
1000 N/C
0 N/C

Show Answer

Slope = dV/dr = (2541 - 2566)/0.05 = -500 V/m.

E = -slope = 500 N/C, so magnitude is 500 N/C.

Numerical value

IGCSE Q14: A 2 microcoulomb charge moves through a potential difference of 65 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 2 x 10^-6 C.

ΔU = (2 x 10^-6)(65) = 1.30e-4 J.

Assertion-Reason

IGCSE Q15: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IGCSE Q16: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -28r + C	2. E = 0
C. V = kQ/r	3. E = 28 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -28, so E = 28.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IGCSE Q17: On a V-x graph, the tangent at x = 4 m has slope -31 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -31 V/m.

E_x = -(-31) = +31 N/C, directed along positive x.

Conceptual reasoning

IGCSE Q18: At a point 12 cm from a +4 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.12 m, E = (9 x 10⁹)(4 x 10^-9)/(0.12)² = 2500 N/C.

The field is non-zero because potential changes with distance.

MCQ

IGCSE Q19: Potential decreases from 2578 V to 2523 V in 0.11 m. What is the magnitude of electric field?

500 N/C
250 N/C
1000 N/C
0 N/C

Show Answer

Slope = dV/dr = (2523 - 2578)/0.11 = -500 V/m.

E = -slope = 500 N/C, so magnitude is 500 N/C.

Numerical value

IGCSE Q20: A 8 microcoulomb charge moves through a potential difference of 35 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 8 x 10^-6 C.

ΔU = (8 x 10^-6)(35) = 2.80e-4 J.

Assertion-Reason

IGCSE Q21: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IGCSE Q22: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -46r + C	2. E = 0
C. V = kQ/r	3. E = 46 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -46, so E = 46.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IGCSE Q23: On a V-x graph, the tangent at x = 5 m has slope -49 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -49 V/m.

E_x = -(-49) = +49 N/C, directed along positive x.

Conceptual reasoning

IGCSE Q24: At a point 24 cm from a +10 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.24 m, E = (9 x 10⁹)(10 x 10^-9)/(0.24)² = 1562.5 N/C.

The field is non-zero because potential changes with distance.

MCQ

IGCSE Q25: Potential decreases from 2590 V to 2545 V in 0.08 m. What is the magnitude of electric field?

562.5 N/C
281.25 N/C
1125 N/C
0 N/C

Show Answer

Slope = dV/dr = (2545 - 2590)/0.08 = -562.5 V/m.

E = -slope = 562.5 N/C, so magnitude is 562.5 N/C.

Numerical value

IGCSE Q26: A 5 microcoulomb charge moves through a potential difference of 65 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 5 x 10^-6 C.

ΔU = (5 x 10^-6)(65) = 3.25e-4 J.

Assertion-Reason

IGCSE Q27: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

IGCSE Q28: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -19r + C	2. E = 0
C. V = kQ/r	3. E = 19 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -19, so E = 19.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

IGCSE Q29: On a V-x graph, the tangent at x = 6 m has slope -22 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -22 V/m.

E_x = -(-22) = +22 N/C, directed along positive x.

Conceptual reasoning

IGCSE Q30: At a point 12 cm from a +4 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.12 m, E = (9 x 10⁹)(4 x 10^-9)/(0.12)² = 2500 N/C.

The field is non-zero because potential changes with distance.

MCQ

IGCSE Q31: Potential decreases from 2602 V to 2567 V in 0.05 m. What is the magnitude of electric field?

700 N/C
350 N/C
1400 N/C
0 N/C

Show Answer

Slope = dV/dr = (2567 - 2602)/0.05 = -700 V/m.

E = -slope = 700 N/C, so magnitude is 700 N/C.

Numerical value

IGCSE Q32: A 2 microcoulomb charge moves through a potential difference of 35 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 2 x 10^-6 C.

ΔU = (2 x 10^-6)(35) = 7.00e-5 J.

22 ICSE Questions

MCQ

ICSE Q1: Potential decreases from 2822 V to 2797 V in 0.07 m. What is the magnitude of electric field?

357.143 N/C
178.571 N/C
714.286 N/C
0 N/C

Show Answer

Slope = dV/dr = (2797 - 2822)/0.07 = -357.143 V/m.

E = -slope = 357.143 N/C, so magnitude is 357.143 N/C.

Numerical value

ICSE Q2: A 4 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 4 x 10^-6 C.

ΔU = (4 x 10^-6)(45) = 1.80e-4 J.

Assertion-Reason

ICSE Q3: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

ICSE Q4: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -52r + C	2. E = 0
C. V = kQ/r	3. E = 52 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -52, so E = 52.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

ICSE Q5: On a V-x graph, the tangent at x = 2 m has slope -10 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -10 V/m.

E_x = -(-10) = +10 N/C, directed along positive x.

Conceptual reasoning

ICSE Q6: At a point 28 cm from a +12 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.28 m, E = (9 x 10⁹)(12 x 10^-9)/(0.28)² = 1377.551 N/C.

The field is non-zero because potential changes with distance.

MCQ

ICSE Q7: Potential decreases from 2834 V to 2779 V in 0.13 m. What is the magnitude of electric field?

423.077 N/C
211.538 N/C
846.154 N/C
0 N/C

Show Answer

Slope = dV/dr = (2779 - 2834)/0.13 = -423.077 V/m.

E = -slope = 423.077 N/C, so magnitude is 423.077 N/C.

Numerical value

ICSE Q8: A 1 microcoulomb charge moves through a potential difference of 15 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 1 x 10^-6 C.

ΔU = (1 x 10^-6)(15) = 1.50e-5 J.

Assertion-Reason

ICSE Q9: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

ICSE Q10: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -25r + C	2. E = 0
C. V = kQ/r	3. E = 25 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -25, so E = 25.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

ICSE Q11: On a V-x graph, the tangent at x = 3 m has slope -28 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -28 V/m.

E_x = -(-28) = +28 N/C, directed along positive x.

Conceptual reasoning

ICSE Q12: At a point 16 cm from a +6 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.16 m, E = (9 x 10⁹)(6 x 10^-9)/(0.16)² = 2109.375 N/C.

The field is non-zero because potential changes with distance.

MCQ

ICSE Q13: Potential decreases from 2846 V to 2801 V in 0.1 m. What is the magnitude of electric field?

450 N/C
225 N/C
900 N/C
0 N/C

Show Answer

Slope = dV/dr = (2801 - 2846)/0.1 = -450 V/m.

E = -slope = 450 N/C, so magnitude is 450 N/C.

Numerical value

ICSE Q14: A 7 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 7 x 10^-6 C.

ΔU = (7 x 10^-6)(45) = 3.15e-4 J.

Assertion-Reason

ICSE Q15: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

ICSE Q16: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -43r + C	2. E = 0
C. V = kQ/r	3. E = 43 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -43, so E = 43.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

ICSE Q17: On a V-x graph, the tangent at x = 4 m has slope -46 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -46 V/m.

E_x = -(-46) = +46 N/C, directed along positive x.

Conceptual reasoning

ICSE Q18: At a point 28 cm from a +12 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.28 m, E = (9 x 10⁹)(12 x 10^-9)/(0.28)² = 1377.551 N/C.

The field is non-zero because potential changes with distance.

MCQ

ICSE Q19: Potential decreases from 2858 V to 2823 V in 0.07 m. What is the magnitude of electric field?

500 N/C
250 N/C
1000 N/C
0 N/C

Show Answer

Slope = dV/dr = (2823 - 2858)/0.07 = -500 V/m.

E = -slope = 500 N/C, so magnitude is 500 N/C.

Numerical value

ICSE Q20: A 4 microcoulomb charge moves through a potential difference of 15 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 4 x 10^-6 C.

ΔU = (4 x 10^-6)(15) = 6.00e-5 J.

Assertion-Reason

ICSE Q21: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

ICSE Q22: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -16r + C	2. E = 0
C. V = kQ/r	3. E = 16 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -16, so E = 16.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

22 UP Board Questions

MCQ

UP Board Q1: Potential decreases from 3062 V to 3037 V in 0.1 m. What is the magnitude of electric field?

250 N/C
125 N/C
500 N/C
0 N/C

Show Answer

Slope = dV/dr = (3037 - 3062)/0.1 = -250 V/m.

E = -slope = 250 N/C, so magnitude is 250 N/C.

Numerical value

UP Board Q2: A 7 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 7 x 10^-6 C.

ΔU = (7 x 10^-6)(45) = 3.15e-4 J.

Assertion-Reason

UP Board Q3: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

UP Board Q4: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -52r + C	2. E = 0
C. V = kQ/r	3. E = 52 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -52, so E = 52.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

UP Board Q5: On a V-x graph, the tangent at x = 2 m has slope -10 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -10 V/m.

E_x = -(-10) = +10 N/C, directed along positive x.

Conceptual reasoning

UP Board Q6: At a point 28 cm from a +12 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.28 m, E = (9 x 10⁹)(12 x 10^-9)/(0.28)² = 1377.551 N/C.

The field is non-zero because potential changes with distance.

MCQ

UP Board Q7: Potential decreases from 3074 V to 3019 V in 0.07 m. What is the magnitude of electric field?

785.714 N/C
392.857 N/C
1571.429 N/C
0 N/C

Show Answer

Slope = dV/dr = (3019 - 3074)/0.07 = -785.714 V/m.

E = -slope = 785.714 N/C, so magnitude is 785.714 N/C.

Numerical value

UP Board Q8: A 4 microcoulomb charge moves through a potential difference of 15 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 4 x 10^-6 C.

ΔU = (4 x 10^-6)(15) = 6.00e-5 J.

Assertion-Reason

UP Board Q9: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

UP Board Q10: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -25r + C	2. E = 0
C. V = kQ/r	3. E = 25 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -25, so E = 25.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

UP Board Q11: On a V-x graph, the tangent at x = 3 m has slope -28 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -28 V/m.

E_x = -(-28) = +28 N/C, directed along positive x.

Conceptual reasoning

UP Board Q12: At a point 16 cm from a +6 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.16 m, E = (9 x 10⁹)(6 x 10^-9)/(0.16)² = 2109.375 N/C.

The field is non-zero because potential changes with distance.

MCQ

UP Board Q13: Potential decreases from 3086 V to 3041 V in 0.13 m. What is the magnitude of electric field?

346.154 N/C
173.077 N/C
692.308 N/C
0 N/C

Show Answer

Slope = dV/dr = (3041 - 3086)/0.13 = -346.154 V/m.

E = -slope = 346.154 N/C, so magnitude is 346.154 N/C.

Numerical value

UP Board Q14: A 1 microcoulomb charge moves through a potential difference of 45 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 1 x 10^-6 C.

ΔU = (1 x 10^-6)(45) = 4.50e-5 J.

Assertion-Reason

UP Board Q15: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

UP Board Q16: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -43r + C	2. E = 0
C. V = kQ/r	3. E = 43 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -43, so E = 43.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

Graph interpretation

UP Board Q17: On a V-x graph, the tangent at x = 4 m has slope -46 V/m. State the electric field at that point.

Show Answer

E_x = -dV/dx.

Here dV/dx = -46 V/m.

E_x = -(-46) = +46 N/C, directed along positive x.

Conceptual reasoning

UP Board Q18: At a point 28 cm from a +12 nC charge, potential is positive. Does that mean electric field must be zero? Explain.

Show Answer

No. Positive potential does not mean zero field.

For a point charge, V = kQ/r and E = kQ/r².

At r = 0.28 m, E = (9 x 10⁹)(12 x 10^-9)/(0.28)² = 1377.551 N/C.

The field is non-zero because potential changes with distance.

MCQ

UP Board Q19: Potential decreases from 3098 V to 3063 V in 0.1 m. What is the magnitude of electric field?

350 N/C
175 N/C
700 N/C
0 N/C

Show Answer

Slope = dV/dr = (3063 - 3098)/0.1 = -350 V/m.

E = -slope = 350 N/C, so magnitude is 350 N/C.

Numerical value

UP Board Q20: A 7 microcoulomb charge moves through a potential difference of 15 V. Find the change in electric potential energy in joule.

Show Answer

ΔU = qΔV.

q = 7 x 10^-6 C.

ΔU = (7 x 10^-6)(15) = 1.05e-4 J.

Assertion-Reason

UP Board Q21: Assertion: If potential is constant throughout a region, electric field in that region is zero. Reason: Electric field is negative gradient of potential.

Both assertion and reason are true, and reason explains assertion.
Both are true, but reason does not explain assertion.
Assertion is true but reason is false.
Assertion is false but reason is true.

Show Answer

Correct option: A.

If V is constant, dV/dr = 0. Since E = -dV/dr, E = 0. The reason directly explains the assertion.

Match the following

UP Board Q22: Match the potential condition with the field conclusion.

A. V = constant	1. E = kQ/r²
B. V = -16r + C	2. E = 0
C. V = kQ/r	3. E = 16 N/C
D. Positive slope graph	4. E is negative along r

Show Answer

A-2 because constant potential has zero slope.

B-3 because dV/dr = -16, so E = 16.

C-1 because differentiating kQ/r gives E = kQ/r².

D-4 because E = -slope.

10 passages with 5 questions each

19. CBSE Case Studies

Case Study 1: Potential Graph

A student plots electric potential V against distance x in a laboratory simulation. The graph is a straight line falling from 120 V at x = 0 m to 20 V at x = 0.5 m. The electric field is uniform in this region because the slope of V-x is constant.

1.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

1.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

1.3 If potential decreases by 37 V in 0.10 m, what is the field magnitude?

37 N/C
370 N/C
3.7 N/C
0 N/C

Show Answer

dV/dr = -37/0.10 = -370 V/m. Therefore E = -dV/dr = +370 N/C, so the magnitude is 370 N/C.

1.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

1.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 2: Potential vs Field

In a one-dimensional region, potential changes rapidly near x = 0.1 m but slowly near x = 0.8 m. The teacher asks students to connect steepness of the potential curve with electric field strength.

2.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

2.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

2.3 If potential decreases by 44 V in 0.10 m, what is the field magnitude?

44 N/C
440 N/C
4.4 N/C
0 N/C

Show Answer

dV/dr = -44/0.10 = -440 V/m. Therefore E = -dV/dr = +440 N/C, so the magnitude is 440 N/C.

2.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

2.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 3: Equipotential Surfaces

A charge moves from one point to another on the same equipotential surface around a charged conductor. The path is curved, but every point on it has the same potential.

3.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

3.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

3.3 If potential decreases by 51 V in 0.10 m, what is the field magnitude?

51 N/C
510 N/C
5.1 N/C
0 N/C

Show Answer

dV/dr = -51/0.10 = -510 V/m. Therefore E = -dV/dr = +510 N/C, so the magnitude is 510 N/C.

3.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

3.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 4: Conductors

A hollow metal sphere is kept in electrostatic equilibrium. The potential is the same everywhere inside the metal and throughout the cavity when no charge is placed in the cavity.

4.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

4.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

4.3 If potential decreases by 58 V in 0.10 m, what is the field magnitude?

58 N/C
580 N/C
5.8 N/C
0 N/C

Show Answer

dV/dr = -58/0.10 = -580 V/m. Therefore E = -dV/dr = +580 N/C, so the magnitude is 580 N/C.

4.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

4.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 5: Point Charge

The potential due to a positive point charge is measured at different distances. Students observe that potential falls as 1/r while electric field falls faster as 1/r squared.

5.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

5.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

5.3 If potential decreases by 65 V in 0.10 m, what is the field magnitude?

65 N/C
650 N/C
6.5 N/C
0 N/C

Show Answer

dV/dr = -65/0.10 = -650 V/m. Therefore E = -dV/dr = +650 N/C, so the magnitude is 650 N/C.

5.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

5.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 6: Parallel Plates

Two large parallel plates are connected to a battery. The field between the plates is almost uniform, and potential changes linearly from the positive plate to the negative plate.

6.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

6.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

6.3 If potential decreases by 72 V in 0.10 m, what is the field magnitude?

72 N/C
720 N/C
7.2 N/C
0 N/C

Show Answer

dV/dr = -72/0.10 = -720 V/m. Therefore E = -dV/dr = +720 N/C, so the magnitude is 720 N/C.

6.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

6.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 7: Negative Charge

A negative point charge creates negative potential around it. As distance increases, potential rises toward zero, but the electric field points toward the charge.

7.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

7.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

7.3 If potential decreases by 79 V in 0.10 m, what is the field magnitude?

79 N/C
790 N/C
7.9 N/C
0 N/C

Show Answer

dV/dr = -79/0.10 = -790 V/m. Therefore E = -dV/dr = +790 N/C, so the magnitude is 790 N/C.

7.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

7.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 8: Graph Slope

A V-x graph has three regions: a falling straight line, a flat segment, and a rising straight line. Students must identify direction and magnitude of electric field from slope.

8.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

8.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

8.3 If potential decreases by 86 V in 0.10 m, what is the field magnitude?

86 N/C
860 N/C
8.6 N/C
0 N/C

Show Answer

dV/dr = -86/0.10 = -860 V/m. Therefore E = -dV/dr = +860 N/C, so the magnitude is 860 N/C.

8.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

8.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 9: Gradient in Two Dimensions

In an advanced problem, potential depends on x and y. The field component in each direction is found by taking the negative partial derivative of potential with respect to that coordinate.

9.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

9.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

9.3 If potential decreases by 93 V in 0.10 m, what is the field magnitude?

93 N/C
930 N/C
9.3 N/C
0 N/C

Show Answer

dV/dr = -93/0.10 = -930 V/m. Therefore E = -dV/dr = +930 N/C, so the magnitude is 930 N/C.

9.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

9.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Case Study 10: Zero Potential Point

Between unlike charges, there may be a point where potential is zero. Students often think field must also be zero there, but field depends on gradient, not absolute value.

10.1 What quantity gives electric field from a potential-distance graph in this case study?

Area under the graph
Negative of the slope
Product of V and r
Square of the intercept

Show Answer

Correct option: B. Electric field along the chosen coordinate is E = -dV/dr, so it is the negative of the graph slope.

10.2 If the potential is constant across a portion of the region, what is the field there?

Maximum
Minimum but non-zero
Zero
Infinite

Show Answer

Correct option: C. Constant potential means zero gradient. Therefore E = 0 in that portion.

10.3 If potential decreases by 100 V in 0.10 m, what is the field magnitude?

100 N/C
1000 N/C
10 N/C
0 N/C

Show Answer

dV/dr = -100/0.10 = -1000 V/m. Therefore E = -dV/dr = +1000 N/C, so the magnitude is 1000 N/C.

10.4 Which statement is the safest conceptual conclusion?

Field depends only on the value of potential.
Field depends on how potential changes with position.
Zero potential always means zero field.
Positive potential always means field points left.

Show Answer

Correct option: B. Electric field is controlled by spatial rate of change of potential, not merely by the numerical value of potential.

10.5 Why is the negative sign important in this passage?

It changes scalar potential into energy.
It shows field points toward decreasing potential.
It removes the unit of distance.
It applies only to negative charges.

Show Answer

Correct option: B. The minus sign tells us that electric field points in the direction in which potential decreases fastest.

Error prevention

20. Common Student Mistakes

Wrong sign of slope

Students often write E = dV/dr. Always remember E is the negative slope of potential.

Ignoring negative sign

The negative sign gives direction. Dropping it changes the physical meaning of the result.

Confusing V and E

Potential is scalar value; electric field is vector slope information. They are related but not identical.

Wrong graph interpretation

The height of a V graph is potential; the slope of the V graph gives field.

Assuming zero potential means zero field

Zero potential can occur by choice of reference or cancellation. Field may still be non-zero.

Wrong differentiation

For V = kQ/r, write V = kQr^-1, then dV/dr = -kQ/r^2. Missing the negative sign ruins the direction.

Board and entrance focus

21. Exam Strategy

CBSE

Write definitions, derivations and sign explanation clearly. Practice case studies and graph-slope reasoning.

NEET

Memorize E = -dV/dr, point-charge results and capacitor relation E = V/d. Solve quickly using slope and sign.

JEE Main

Focus on numerical graphs, point charges, conductors and relation between potential energy and potential.

JEE Advanced

Master calculus, gradients, multi-dimensional potential functions and conceptual traps like V = 0 but E not zero.

IB Physics

Explain physical meaning in words, connect fields to energy, and use correct vector language.

IGCSE

Use clear definitions, field-line diagrams and simple graph interpretation.

ICSE

Be precise with units, potential difference and work done. Draw neat electric field diagrams.

UP Board

Prepare derivations step by step and revise formulas with units and meanings.

One-page revision sheet

22. Final Quick Revision

Important formulas

E = F/q
V = W/q
W = qΔV
E = -dV/dr
E = -∇V
V = kQ/r
E = kQ/r²
V = -Er + C

Graph summary

V-r slope = dV/dr
E = -slope
Flat V graph means E = 0
Steeper V graph means stronger E
Constant E means linear V

Concept summary

Potential is scalar.
Electric field is vector.
Field points to decreasing V.
Equipotentials are perpendicular to field.
Inside a conductor, E = 0 and V is constant.

Exam traps

Do not forget the negative sign.
Do not confuse zero V with zero E.
Differentiate 1/r correctly.
Use SI units before calculating.
Read graph axes carefully.

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Relationship Between Electric Field and Electric Potential

1. Introduction

What is electric field?

What is electric potential?

Why are they related?

Why students find it difficult

2. Revision of Electric Potential

Physical meaning

Unit of potential

3. Revision of Electric Field

Electric field intensity

Direction

Field lines

4. Complete Derivation of E = -dV/dr

5. Physical Meaning of the Negative Sign

Why does the sign appear?

Three statements to remember

6. Conceptual Interpretation

Case 1: V decreases rapidly

Case 2: V changes slowly

Case 3: V is constant

7. Constant Electric Field

Parallel plate example

8. Constant Potential

Conductors in electrostatic equilibrium

9. Point Charge Analysis

Potential variation

Field variation

10. Graph Analysis

11. Electric Field from Potential Graph

Positive slope

Negative slope

Zero slope

12. Equipotential Surfaces

Definition

No work done

Relation with field

13. Multi-Dimensional Form

14. Common Confusions

Is potential zero when electric field is zero?

Is electric field zero when potential is constant?

Can potential be zero but electric field non-zero?

Can electric field be non-zero while potential is positive?

Can electric field exist on equipotential surfaces?

Does larger potential always mean larger field?

15. Comparison Table: Electric Field vs Electric Potential

16. Important Formulas

17. Solved Examples

20 CBSE Solved Examples

CBSE Example 1: The potential changes from 273 V to 138 V when moving from r = 0.16 m to r = 0.4 m along a straight line. Find the electric field component along r.

CBSE Example 2: A charge of 6 microcoulomb moves through a potential difference of 40 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

CBSE Example 3: A point charge of +7 nC is kept in air. Find V and E at a point 16 cm away.

CBSE Example 4: The electric field between two parallel plates is 350 N/C and plate separation is 0.02 m. Find the potential difference between the plates.

CBSE Example 5: In a region, V(r) = -50r + 135 volt. Find the electric field along r.

CBSE Example 6: In a small region, V is constant at 96 V. What is the electric field in that region?

CBSE Example 7: The potential is V(x) = 60x2 + 9x + 17 volt. Find Ex at x = 3 m.

CBSE Example 8: If V = kQ/r, prove by differentiation that E = kQ/r2 for a positive point charge.

CBSE Example 9: The slope of a V-r graph at a point is +70 V/m. Find the electric field and state its direction relative to increasing r.

CBSE Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 300 V. Find the work done by electric force on a 6 microcoulomb charge.

CBSE Example 11: The potential changes from 303 V to 198 V when moving from r = 0.12 m to r = 0.44 m along a straight line. Find the electric field component along r.

CBSE Example 12: A charge of 8 microcoulomb moves through a potential difference of 140 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

CBSE Example 13: A point charge of +8 nC is kept in air. Find V and E at a point 16 cm away.

CBSE Example 14: The electric field between two parallel plates is 250 N/C and plate separation is 0.05 m. Find the potential difference between the plates.

CBSE Example 15: In a region, V(r) = -45r + 145 volt. Find the electric field along r.

CBSE Example 16: In a small region, V is constant at 106 V. What is the electric field in that region?

CBSE Example 17: The potential is V(x) = 55x2 + 7x + 27 volt. Find Ex at x = 3 m.

CBSE Example 18: If V = kQ/r, prove by differentiation that E = kQ/r2 for a positive point charge.

CBSE Example 19: The slope of a V-r graph at a point is +65 V/m. Find the electric field and state its direction relative to increasing r.

CBSE Example 20: A charge moves along an equipotential surface from A to B. The potential at both points is 330 V. Find the work done by electric force on a 8 microcoulomb charge.

20 NEET Solved Examples

NEET Example 1: The potential changes from 483 V to 378 V when moving from r = 0.12 m to r = 0.28 m along a straight line. Find the electric field component along r.

NEET Example 2: A charge of 4 microcoulomb moves through a potential difference of 140 V. Calculate the work done by the external agent in slowly moving it to the higher potential point.

NEET Example 3: A point charge of +5 nC is kept in air. Find V and E at a point 16 cm away.

NEET Example 4: The electric field between two parallel plates is 250 N/C and plate separation is 0.02 m. Find the potential difference between the plates.

NEET Example 5: In a region, V(r) = -70r + 205 volt. Find the electric field along r.

NEET Example 6: In a small region, V is constant at 166 V. What is the electric field in that region?

NEET Example 7: The potential is V(x) = 80x2 + 7x + 87 volt. Find Ex at x = 3 m.

NEET Example 8: If V = kQ/r, prove by differentiation that E = kQ/r2 for a positive point charge.

NEET Example 9: The slope of a V-r graph at a point is +35 V/m. Find the electric field and state its direction relative to increasing r.

NEET Example 10: A charge moves along an equipotential surface from A to B. The potential at both points is 510 V. Find the work done by electric force on a 4 microcoulomb charge.

CBSE Example 7: The potential is V(x) = 60x² + 9x + 17 volt. Find E_x at x = 3 m.

CBSE Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

CBSE Example 17: The potential is V(x) = 55x² + 7x + 27 volt. Find E_x at x = 3 m.

CBSE Example 18: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

NEET Example 7: The potential is V(x) = 80x² + 7x + 87 volt. Find E_x at x = 3 m.

NEET Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

NEET Example 17: The potential is V(x) = 75x² + 5x + 97 volt. Find E_x at x = 3 m.

NEET Example 18: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

JEE Main Example 7: The potential is V(x) = 45x² + 5x + 157 volt. Find E_x at x = 3 m.

JEE Main Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

JEE Main Example 17: The potential is V(x) = 40x² + 9x + 167 volt. Find E_x at x = 3 m.

JEE Main Example 18: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

JEE Advanced Example 7: The potential is V(x) = 65x² + 9x + 227 volt. Find E_x at x = 3 m.

JEE Advanced Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

JEE Advanced Example 17: The potential is V(x) = 60x² + 7x + 237 volt. Find E_x at x = 3 m.

JEE Advanced Example 18: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.

IB Physics Example 7: The potential is V(x) = 75x² + 9x + 317 volt. Find E_x at x = 3 m.

IB Physics Example 8: If V = kQ/r, prove by differentiation that E = kQ/r² for a positive point charge.