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Electric Potential

Definition, derivations, point charge potential, dipole potential, properties, solved examples and exam questions.

CBSE Class 12NEETJEE MainJEE AdvancedIB PhysicsIGCSEICSEA-LevelUP Board
DefinitionPoint ChargeDipoleFormula SheetSolved ExamplesQuestion BankCase StudiesRevision
+Q P1 P2 V = W/q = kQ/r Potential is scalar: add algebraically, not vectorially.
Section 1

Definition of Electric Potential

Electric potential at a point is defined as the work done per unit positive test charge in bringing that charge from infinity to the point without acceleration. The phrase without acceleration means the test charge is moved slowly, so that the external agent balances the electric force at every instant.

Potential at a Point

V = W / q

Here W is the work done by an external agent and q is a small positive test charge. Electric potential tells us how much electric potential energy each coulomb of charge would have at that point.

Potential Difference

ΔV = W / q

Potential difference between two points is work done per unit charge in moving the test charge from one point to the other. It is the useful measurable quantity in circuits and electrostatics.

Work Relation

W = qΔV

If charge q moves through potential difference ΔV, the work associated with the movement is the product of charge and potential difference.

IdeaElectric PotentialElectric Field
Physical meaningPotential energy per unit positive test charge.Force per unit positive test charge.
Formula ideaV = W/qE = F/q
NatureScalar quantity; it has sign but no direction.Vector quantity; it has magnitude and direction.
SI unitvolt, where 1 V = 1 J C-1newton per coulomb or volt per metre.
SuperpositionAlgebraic sum of potentials.Vector sum of fields.

Why Potential is Scalar

Work and charge are scalar quantities, so their ratio is also scalar. A positive potential does not mean it points in a positive direction; it means a positive charge has positive potential energy there relative to the chosen zero level.

Conceptual Example

If 6 J of work is required to bring a 2 C positive test charge from infinity to a point, then V = W/q = 6/2 = 3 V. A 5 C charge placed at the same point would have potential energy U = qV = 15 J.

Section 2

Electric Potential Due to a Point Charge

For a point charge Q, electric potential at a distance r is obtained by calculating the work done in bringing a unit positive charge from infinity to that point.

+Q P r V = kQ/r at observation point P
r V negative charge positive charge Magnitude of potential decreases as distance increases

Derivation of V = kQ/r

  1. Electric field due to a point charge at distance x is E = (${k}) Q/x2.
  2. For a small displacement dr, work done by the electric field is related to potential difference by dV = -E dr when displacement is radial.
  3. Therefore V(r) - V(∞) = -∫∞r E dr.
  4. Substitute E = kQ/r2: V(r) - V(∞) = -∫∞r kQ r-2 dr.
  5. Integrating, V(r) - V(∞) = kQ/r.
  6. With the standard convention V(∞) = 0, the potential is V = kQ/r.

Positive Charge

A positive source charge gives positive potential because positive external work is needed to bring a positive test charge closer against repulsion.

Negative Charge

A negative source charge gives negative potential because the electric field attracts a positive test charge, so the external work for slow movement is negative.

Distance Effect

The magnitude |V| = k|Q|/r decreases as distance increases and tends to zero at infinity.

Scalar Addition

If several charges are present, their potentials are added algebraically, taking signs of charges directly.

Section 3

Electric Potential Energy and Potential Difference

Electric potential connects field geometry with energy. If a charge is moved between two points, the potential difference determines how much work is required or released.

Potential Difference

ΔV = VB - VA = Wext/q

For slow movement, external work per unit charge equals change in potential.

Work by Field

Wfield = -qΔV

The electric field does positive work when a positive charge naturally moves toward lower potential.

Potential Energy

U = qV

Potential energy of charge q at potential V is qV.

Potential energy of two point charges

  1. Bring charge q1 first; with no other charge present, no work is required from infinity.
  2. Potential at distance r from q1 is V = kq1/r.
  3. Bring q2 from infinity to that point. External work is W = q2V.
  4. Therefore U = kq1q2/r.
  5. If charges have the same sign, U is positive; if opposite signs, U is negative.
Movement of positive chargePotential changeWork by fieldExternal work for slow motion
High potential to low potentialΔV < 0PositiveNegative
Low potential to high potentialΔV > 0NegativePositive
Along equipotentialΔV = 0ZeroZero
Section 4

Properties of Electric Potential

Electric potential is often easier than electric field because it is scalar, but its signs and reference level must be handled carefully.

Scalar quantity

Potential has sign but no direction. Add potentials algebraically.

Positive, negative or zero

Potential may be positive near positive charges, negative near negative charges, and zero at reference points or by cancellation.

Zero at infinity

For isolated charge distributions, infinity is usually chosen as the zero-potential reference.

Superposition

Net potential is the algebraic sum of potentials due to all individual charges.

Field relation

Electric field is the negative gradient of potential: E = -dV/dr or E = -∇V.

Equipotential surfaces

Electric field is perpendicular to equipotential surfaces and no work is done along them.

Conductors

Inside a conductor in electrostatic equilibrium, electric field is zero and potential is constant.

Direction of decrease

Electric field points in the direction in which potential decreases fastest.

Reference dependence

Only potential difference has direct physical meaning; absolute potential depends on chosen zero.

PropertyExam implicationCommon trap
Potential is scalarUse simple algebraic addition.Do not resolve potential into components.
Potential can be zero where field is nonzeroDipole equatorial line is the standard example.Zero potential does not prove zero electric field.
Potential inside conductor is constantEvery point inside and on the surface has same potential.Do not use kQ/r for points inside the conductor.
Equipotential motion needs no workUse W = qΔV = 0.Path length does not matter if endpoints have same potential.
Section 5

Potential Due to Multiple Point Charges

For several point charges, potential is found by adding individual potentials algebraically. The distance used for each charge is the distance from that particular charge to the observation point.

V = (${k}) Σ(qi / ri)

Two Charges

V = k(q1/r1 + q2/r2). Include signs of charges.

Three Charges

Add three terms. If symmetry exists, combine equal-distance terms first.

Triangle

At the centroid or a vertex, calculate separate distances from every charged vertex.

Square

At the centre, each corner is at distance a/√2 for side a.

Why algebraic addition works

  1. Potential energy of a test charge near many source charges is the sum of potential energies due to each source charge.
  2. Dividing total potential energy by the test charge gives total potential.
  3. Because potential is scalar, directions are not added; only signs and magnitudes matter.
  4. The final expression is V = kΣ(qi/ri).
ConfigurationObservation pointKey formula
Two charges on a linePoint between or outsideUse actual distance from each charge.
Equilateral triangleCentre or vertexUse equal distances when symmetry allows.
Square of side aCentreDistance from each corner is a/√2.
Regular polygonCentreEqual distances make scalar addition very fast.
Section 6

Electric Dipole Introduction

An electric dipole consists of two equal and opposite charges +q and -q separated by a small distance 2a. Dipoles are central in electrostatics, molecular physics, polarization, and entrance exam problems.

-q +q dipole moment p separation = 2a
p = q(2a)

Direction of Dipole Moment

Electric dipole moment points from negative charge to positive charge. Its SI unit is coulomb metre, written as C m.

Why It Matters

CBSE focuses on definitions and derivations, NEET emphasizes formula recognition, and JEE often combines dipole potential with approximation, energy, and symmetry.

Section 7

Electric Potential Due to a Dipole at an Axial Point

Let point P lie on the dipole axis at distance r from the centre. The positive charge is nearer, so its contribution has denominator r - a; the negative charge is farther, so its denominator is r + a.

-q +q P centre O r - a distance from -q is r + a
V = k[q/(r-a) - q/(r+a)]
V = k[2qa/(r2 - a2)]
Vaxial = k[p/(r2 - a2)]
For r >> a: Vaxial = kp/r2

Axial potential derivation

  1. Potential due to +q at P is V+ = kq/(r-a).
  2. Potential due to -q at P is V- = -kq/(r+a).
  3. Net potential is V = kq/(r-a) - kq/(r+a).
  4. Combine fractions: V = kq[(r+a-r+a)/(r2-a2)].
  5. Therefore V = k(2qa)/(r2-a2).
  6. Since p = 2qa, Vaxial = kp/(r2-a2).
  7. For a short dipole, r >> a, so r2 - a2 ≈ r2 and Vaxial = kp/r2.
Section 8

Electric Potential Due to a Dipole at an Equatorial Point

At an equatorial point, the observation point is equally distant from +q and -q. The magnitudes of the two potentials are equal, but their signs are opposite.

-q +q P same distance same distance V+ and V- cancel, so V = 0 on the equatorial line
V+ = kq/√(r2 + a2)
V- = -kq/√(r2 + a2)
Vequatorial = 0

Equatorial potential derivation

  1. Distance from P to +q is √(r2 + a2).
  2. Distance from P to -q is the same.
  3. Potential due to positive charge is +kq/√(r2 + a2).
  4. Potential due to negative charge is -kq/√(r2 + a2).
  5. They cancel exactly, so net potential is zero.
Section 9

Electric Potential Due to a Dipole at Any General Point

For a point P at distance r from the centre of a short dipole and making angle θ with the dipole axis, the potential depends on the projection of dipole moment along the position vector.

-q +q P θ r p Short dipole: V = kp cosθ/r2
Exact form: V = k[q/r+ - q/r-]
Short dipole: V = k(p cosθ/r2)
At θ = 0°: V = kp/r2
At θ = 90°: V = 0

General short dipole derivation

  1. Let r+ and r- be distances of point P from +q and -q respectively.
  2. Net potential is V = k(q/r+ - q/r-).
  3. For a short dipole, distance difference is approximately r- - r+ = 2a cosθ.
  4. Also, r+r- ≈ r2.
  5. Thus V = kq(r- - r+)/(r+r-) becomes V = kq(2a cosθ)/r2.
  6. Since p = 2qa, V = kp cosθ/r2.
Section 10

Relation Between Electric Field and Potential

Electric field is related to how rapidly potential changes with position. In one-dimensional radial motion, electric field is the negative rate of change of potential with distance.

One-dimensional relation

E = -dV/dr

The negative sign shows that electric field points toward decreasing potential.

Vector form

E = -∇V

In three dimensions, the gradient points toward increasing potential, so field is opposite to it.

Point charge check

V = kQ/r

Differentiating gives the familiar electric field magnitude.

Recovering electric field from point charge potential

  1. Start with V = kQ/r = kQr-1.
  2. Differentiate: dV/dr = -kQ/r2.
  3. Use E = -dV/dr.
  4. Therefore E = kQ/r2, directed radially away from a positive charge and toward a negative charge.
Section 11

Equipotential Surfaces

An equipotential surface is a surface on which electric potential is the same at every point. Moving a charge along such a surface requires no work because the potential difference is zero.

+Q Point charge Uniform field -q +q Dipole

Definition

All points on an equipotential surface have the same value of V.

No Work

W = qΔV. On an equipotential surface, ΔV = 0, so W = 0.

Perpendicular Field

Electric field cannot have a tangential component along an equipotential surface; otherwise work would be done.

Shapes

Point charge surfaces are concentric spheres; uniform field surfaces are parallel planes; dipole surfaces are more complex.

Section 12

Conductor and Electric Potential

In electrostatic equilibrium, the electric field inside a conductor is zero. Therefore the potential does not change inside the conductor, and the whole conductor is an equipotential body.

R V r inside: constant outside: kQ/r Potential of a charged conducting sphere
Vsurface = kQ/R
Outside: V = kQ/r
Inside: V = constant = kQ/R

Potential of a charged conducting sphere

  1. For a charged conducting sphere of radius R, all charge resides on the outer surface.
  2. For an outside point at distance r ≥ R, the sphere behaves like a point charge at the centre, so V = kQ/r.
  3. At the surface, put r = R, giving V = kQ/R.
  4. Inside the conductor, E = 0, so dV/dr = 0.
  5. Hence potential inside is constant and equal to the surface value kQ/R.
RegionElectric fieldPotential
Inside conductorZeroConstant = kQ/R
On surfaceJust outside: kQ/R2kQ/R
Outside conductorkQ/r2kQ/r
Section 13

Charged Liquid Droplet Problems

Charged droplet problems are common in NEET and JEE because they combine conservation of charge, conservation of volume, and the potential of a conducting sphere.

Final radius

R = n1/3r

Volume is conserved: n(4/3)πr3 = (4/3)πR3.

Final charge

Q = nq

Charges add directly when droplets merge.

Final potential

V' = n2/3V

Use V = kq/r, Q = nq, and R = n1/3r.

Droplet potential formula

  1. Potential of each small droplet is V = kq/r.
  2. When n droplets combine, total charge is Q = nq.
  3. Final radius is R = n1/3r because total volume is conserved.
  4. Potential of big droplet is V' = kQ/R = k(nq)/(n1/3r).
  5. Therefore V' = n2/3(kq/r) = n2/3V.
Droplet Example 1

8 identical droplets

Question: Eight identical droplets each have potential 10 V. Find the potential of the larger drop.

Solution:
  1. n = 8.
  2. V' = n2/3V.
  3. 82/3 = (2)2 = 4.
  4. V' = 4 x 10 = 40 V.

Answer: 40 V

Droplet Example 2

27 droplets

Question: Twenty-seven droplets each of radius r and charge q combine. Express final potential in terms of small droplet potential V.

Solution:
  1. n = 27.
  2. V' = n2/3V.
  3. 272/3 = 9.
  4. V' = 9V.

Answer: 9V

Droplet Example 3

Radius relation

Question: If 64 identical charged droplets combine, what is the radius of the new drop?

Solution:
  1. Volume is conserved.
  2. R = n1/3r.
  3. 641/3 = 4.
  4. R = 4r.

Answer: 4r

Droplet Example 4

Potential multiplier

Question: How many identical droplets must combine to make final potential 16 times the original?

Solution:
  1. Use V' = n2/3V.
  2. n2/3 = 16.
  3. Raise both sides to power 3/2: n = 163/2.
  4. n = 64.

Answer: 64 droplets

Droplet Example 5

Charge and radius

Question: 125 droplets each of charge 2 nC and radius 1 mm combine. Find final charge and radius.

Solution:
  1. Total charge Q = nq = 125 x 2 nC = 250 nC.
  2. Final radius R = n1/3r = 5 x 1 mm.
  3. R = 5 mm.
  4. Both charge and volume are conserved.

Answer: Q = 250 nC, R = 5 mm

Droplet Example 6

Potential from data

Question: Each small droplet has potential 3 V. If 216 droplets combine, find new potential.

Solution:
  1. n = 216 = 63.
  2. n2/3 = 62 = 36.
  3. V' = 36 x 3 V.
  4. V' = 108 V.

Answer: 108 V

Droplet Example 7

Find original potential

Question: A large drop formed from 27 equal droplets has potential 180 V. What was each small droplet's potential?

Solution:
  1. For n = 27, multiplier is 9.
  2. V' = 9V.
  3. V = V'/9.
  4. V = 180/9 = 20 V.

Answer: 20 V

Droplet Example 8

Unequal interpretation warning

Question: A problem states identical droplets. Which quantities become n times, n1/3 times, and n2/3 times?

Solution:
  1. Charge becomes Q = nq.
  2. Radius becomes R = n1/3r.
  3. Potential becomes V' = n2/3V.
  4. These results come from charge conservation and volume conservation.

Answer: Charge n times, radius n1/3 times, potential n2/3 times

Droplet Example 9

Potential ratio

Question: If 1000 identical droplets combine, what is V'/V?

Solution:
  1. n = 1000 = 103.
  2. V'/V = n2/3.
  3. V'/V = 102.
  4. V'/V = 100.

Answer: 100

Droplet Example 10

NEET-style droplet

Question: A charged droplet at potential V splits into 8 identical droplets. Find potential of each small droplet.

Solution:
  1. Reverse the combination formula.
  2. If 8 small droplets combine to make the original large droplet, Vlarge = 82/3Vsmall.
  3. V = 4Vsmall.
  4. Vsmall = V/4.

Answer: V/4

Section 14

Important Formula Sheet

Use this formula sheet for quick revision before CBSE board exams, NEET, JEE Main, JEE Advanced, IB Physics, IGCSE, ICSE, A-Level, and UP Board exams.

DefinitionV = W/q
Potential differenceΔV = W/q
Work relationW = qΔV
Point chargeV = kQ/r
Potential energyU = qV
Two charges energyU = kq1q2/r
Multiple chargesVtotal = kΣ(qi/ri)
Dipole axialV = k[p/(r2-a2)]
Short dipole axialV = kp/r2
Dipole equatorialV = 0
General dipoleV = k(p cosθ/r2)
Field relationE = -dV/dr
Droplet radiusR = n1/3r
Droplet potentialV' = n2/3V
Conducting sphereInside V = kQ/R; outside V = kQ/r
Section 15

Solved Examples

These solved examples are designed to avoid repetition. Each example tests a different idea: definition, sign, superposition, dipole potential, conducting sphere, equipotential surface, work, energy, and charged droplets.

CBSE-Level Solved Examples

CBSE-Level Solved Examples Example 1

Basic definition

Question: A 3 C positive test charge is brought from infinity to a point. If 24 J work is done by the external agent, find the potential.

Solution:
  1. Use V = W/q.
  2. Substitute W = 24 J and q = 3 C.
  3. V = 24/3.
  4. V = 8 V.

Answer: 8 V

CBSE-Level Solved Examples Example 2

Potential difference

Question: A charge of 5 C moves between two points with potential difference 12 V. Find the work done by the external agent for slow movement.

Solution:
  1. Use W = qΔV.
  2. Substitute q = 5 C and ΔV = 12 V.
  3. W = 60 J.
  4. Positive sign means external work increases potential energy.

Answer: 60 J

CBSE-Level Solved Examples Example 3

Potential energy

Question: A 2 C charge is placed at a point where potential is 15 V. Find its potential energy.

Solution:
  1. Use U = qV.
  2. U = 2 x 15.
  3. U = 30 J.
  4. Potential energy depends on the charge placed there.

Answer: 30 J

CBSE-Level Solved Examples Example 4

Unit conversion

Question: Show that 1 volt is 1 joule per coulomb.

Solution:
  1. From V = W/q.
  2. If W = 1 J and q = 1 C, V = 1 J/C.
  3. Therefore 1 V = 1 J C-1.
  4. This is the SI definition of volt.

Answer: 1 V = 1 J C-1

CBSE-Level Solved Examples Example 5

Scalar nature

Question: Two charges produce potentials +12 V and -5 V at the same point. Find net potential.

Solution:
  1. Potential is scalar.
  2. Add algebraically: V = 12 + (-5).
  3. V = 7 V.
  4. No vector direction is required.

Answer: 7 V

CBSE-Level Solved Examples Example 6

Zero work

Question: A 4 C charge moves on an equipotential surface. Find work done.

Solution:
  1. On equipotential surface, ΔV = 0.
  2. W = qΔV.
  3. W = 4 x 0.
  4. No work is done.

Answer: 0 J

CBSE-Level Solved Examples Example 7

Sign of potential

Question: What is the sign of potential near an isolated negative charge?

Solution:
  1. Potential due to a point charge is V = kQ/r.
  2. For negative Q, numerator is negative.
  3. Distance r is positive.
  4. Therefore potential is negative.

Answer: Negative

CBSE-Level Solved Examples Example 8

Point charge formula

Question: Find potential at 0.3 m from a 2 nC charge. Take k = 9 x 109 SI.

Solution:
  1. Use V = kQ/r.
  2. Q = 2 x 10-9 C and r = 0.3 m.
  3. V = (9 x 109 x 2 x 10-9)/0.3.
  4. V = 60 V.

Answer: 60 V

CBSE-Level Solved Examples Example 9

Conductor inside

Question: A charged conducting sphere has surface potential 200 V. What is potential at its centre?

Solution:
  1. Inside a conductor in electrostatic equilibrium, potential is constant.
  2. The constant value equals surface potential.
  3. Therefore potential at centre is 200 V.
  4. Electric field at centre is zero, but potential is not necessarily zero.

Answer: 200 V

CBSE-Level Solved Examples Example 10

Potential difference direction

Question: A positive charge moves from 80 V to 20 V. Is work done by electric field positive or negative?

Solution:
  1. ΔV = 20 - 80 = -60 V.
  2. Work by field = -qΔV.
  3. For positive q, this is positive.
  4. The field does positive work when positive charge moves to lower potential.

Answer: Positive

CBSE-Level Solved Examples Example 11

Multiple charges

Question: At point P, potentials due to three charges are 20 V, -8 V and 3 V. Find net potential.

Solution:
  1. Use superposition.
  2. V = 20 - 8 + 3.
  3. V = 15 V.
  4. Potential is algebraic.

Answer: 15 V

CBSE-Level Solved Examples Example 12

Dipole equatorial

Question: What is the potential at an equatorial point of an electric dipole?

Solution:
  1. Distances from +q and -q are equal.
  2. Potentials have equal magnitudes and opposite signs.
  3. They cancel.
  4. Net potential is zero.

Answer: 0

CBSE-Level Solved Examples Example 13

Axial short dipole

Question: A short dipole has p = 4 x 10-9 C m. Find axial potential at r = 2 m using k = 9 x 109.

Solution:
  1. Use V = kp/r2.
  2. V = (9 x 109)(4 x 10-9)/4.
  3. V = 36/4.
  4. V = 9 V.

Answer: 9 V

CBSE-Level Solved Examples Example 14

Field from potential

Question: If V = 40/r, find electric field magnitude at r = 2 m.

Solution:
  1. E = -dV/dr.
  2. dV/dr = -40/r2.
  3. E = 40/r2.
  4. At r = 2 m, E = 10 N/C.

Answer: 10 N/C

CBSE-Level Solved Examples Example 15

Droplet relation

Question: If 8 identical droplets combine, by what factor does potential increase?

Solution:
  1. Use V' = n2/3V.
  2. n = 8.
  3. 82/3 = 4.
  4. Potential becomes 4 times.

Answer: 4 times

NEET-Level Solved Examples

NEET-Level Solved Examples Example 1

Fast point potential

Question: A 6 nC charge creates potential at 0.2 m. Find V.

Solution:
  1. V = kQ/r.
  2. V = 9 x 109 x 6 x 10-9 / 0.2.
  3. Numerator = 54.
  4. V = 270 V.

Answer: 270 V

NEET-Level Solved Examples Example 2

Negative charge potential

Question: Find potential 0.5 m from a -4 nC charge.

Solution:
  1. Use V = kQ/r.
  2. V = 9 x 109 x (-4 x 10-9)/0.5.
  3. V = -36/0.5.
  4. V = -72 V.

Answer: -72 V

NEET-Level Solved Examples Example 3

Work in moving charge

Question: A 2 microcoulomb charge moves through a potential difference of 300 V. Find work.

Solution:
  1. Use W = qΔV.
  2. q = 2 x 10-6 C.
  3. W = 2 x 10-6 x 300.
  4. W = 6 x 10-4 J.

Answer: 6 x 10-4 J

NEET-Level Solved Examples Example 4

Two equal charges

Question: Two +2 nC charges are each 0.3 m from point P. Find potential at P.

Solution:
  1. Potential from one charge = kq/r.
  2. For two equal charges, V = 2kq/r.
  3. V = 2 x 9 x 109 x 2 x 10-9/0.3.
  4. V = 120 V.

Answer: 120 V

NEET-Level Solved Examples Example 5

Opposite equal charges

Question: +q and -q are at equal distance from P. Find net potential.

Solution:
  1. Potential due to +q is +kq/r.
  2. Potential due to -q is -kq/r.
  3. The algebraic sum is zero.
  4. This does not necessarily mean electric field is zero.

Answer: 0

NEET-Level Solved Examples Example 6

Dipole axial recognition

Question: For short dipole, if distance from centre is doubled on axial line, how does potential change?

Solution:
  1. Vaxial = kp/r2.
  2. If r becomes 2r, V becomes kp/(4r2).
  3. Potential becomes one-fourth.
  4. This inverse-square dependence is for short dipole potential, not point charge potential.

Answer: One-fourth

NEET-Level Solved Examples Example 7

Dipole general angle

Question: For a short dipole, potential at angle 60 degrees is what fraction of axial potential at same r?

Solution:
  1. V = kp cosθ/r2.
  2. Axial potential = kp/r2.
  3. Fraction = cos 60 degrees.
  4. Fraction = 1/2.

Answer: 1/2

NEET-Level Solved Examples Example 8

Equatorial field concept

Question: Potential at equatorial point of dipole is zero. Is electric field also zero?

Solution:
  1. Potential is scalar and cancels.
  2. Electric field is vector.
  3. Components along the dipole axis add.
  4. Therefore field is not zero.

Answer: No, electric field is not zero

NEET-Level Solved Examples Example 9

Droplet factor

Question: 27 droplets, each at potential 5 V, merge. Find final potential.

Solution:
  1. V' = n2/3V.
  2. 272/3 = 9.
  3. V' = 9 x 5.
  4. V' = 45 V.

Answer: 45 V

NEET-Level Solved Examples Example 10

Conducting sphere

Question: A conducting sphere has Q = 2 microcoulomb and R = 0.1 m. Find surface potential.

Solution:
  1. Use V = kQ/R.
  2. V = 9 x 109 x 2 x 10-6/0.1.
  3. V = 18 x 103/0.1.
  4. V = 1.8 x 105 V.

Answer: 1.8 x 105 V

NEET-Level Solved Examples Example 11

Potential energy sign

Question: Two opposite charges are separated by r. What is sign of potential energy?

Solution:
  1. U = kq1q2/r.
  2. Opposite charges have product negative.
  3. r is positive.
  4. Therefore U is negative.

Answer: Negative

NEET-Level Solved Examples Example 12

Zero potential point

Question: A point has V = 0 due to two charges. What can be concluded about potential energy of a charge placed there?

Solution:
  1. U = qV.
  2. If V = 0, U = 0 for any finite charge q.
  3. This statement concerns potential energy only.
  4. It does not imply electric field is zero.

Answer: Potential energy is zero

NEET-Level Solved Examples Example 13

Equipotential work

Question: A 10 microcoulomb charge moves along an equipotential line for 5 cm. Find work.

Solution:
  1. Equipotential means ΔV = 0.
  2. W = qΔV.
  3. W = 0.
  4. Path length is irrelevant.

Answer: 0

NEET-Level Solved Examples Example 14

Potential gradient

Question: Potential decreases by 20 V over 0.5 m in a uniform field. Find field magnitude.

Solution:
  1. E = -dV/dr; magnitude = |ΔV|/Δr.
  2. E = 20/0.5.
  3. E = 40 V/m.
  4. Direction is toward decreasing potential.

Answer: 40 V/m

NEET-Level Solved Examples Example 15

Square centre

Question: Four equal charges q are at corners of square side a. Find potential at centre.

Solution:
  1. Distance from each corner to centre is a/√2.
  2. Potential from one charge is kq/(a/√2) = kq√2/a.
  3. Four charges add algebraically.
  4. V = 4√2 kq/a.

Answer: 4√2 kq/a

JEE Main-Level Solved Examples

JEE Main-Level Solved Examples Example 1

Triangle with signs

Question: Charges +q, +q and -q are at vertices of an equilateral triangle of side a. Find potential at centroid.

Solution:
  1. Distance from centroid to each vertex is a/√3.
  2. Net charge contribution algebraically is q + q - q = q.
  3. V = kq/(a/√3).
  4. V = √3 kq/a.

Answer: √3 kq/a

JEE Main-Level Solved Examples Example 2

Square alternating charges

Question: Charges +q, -q, +q, -q occupy consecutive corners of a square. Find potential at centre.

Solution:
  1. All corners are at equal distance from centre.
  2. Algebraic sum of charges is +q - q + q - q = 0.
  3. V = k(0)/r.
  4. Potential at centre is zero.

Answer: 0

JEE Main-Level Solved Examples Example 3

Potential energy pair

Question: Two charges 3 microcoulomb and 2 microcoulomb are 0.6 m apart. Find potential energy.

Solution:
  1. U = kq1q2/r.
  2. q1q2 = 6 x 10-12.
  3. U = 9 x 109 x 6 x 10-12/0.6.
  4. U = 0.09 J.

Answer: 0.09 J

JEE Main-Level Solved Examples Example 4

Field from V expression

Question: If V = 100/r - 20, find electric field magnitude at r = 5 m.

Solution:
  1. Constant term does not affect electric field.
  2. dV/dr = -100/r2.
  3. E = 100/r2.
  4. At r = 5 m, E = 4 N/C.

Answer: 4 N/C

JEE Main-Level Solved Examples Example 5

Axial exact dipole

Question: A dipole has q = 2 microcoulomb, a = 0.1 m. Find exact axial potential at r = 0.5 m.

Solution:
  1. Use V = k[2qa/(r2 - a2)].
  2. 2qa = 2 x 2 x 10-6 x 0.1 = 4 x 10-7.
  3. r2 - a2 = 0.25 - 0.01 = 0.24.
  4. V = 9 x 109 x 4 x 10-7/0.24 = 1.5 x 104 V.

Answer: 1.5 x 104 V

JEE Main-Level Solved Examples Example 6

General dipole angle

Question: For a short dipole, potential at 120 degrees is what sign relative to axial positive side?

Solution:
  1. V = kp cosθ/r2.
  2. cos 120 degrees = -1/2.
  3. Potential is negative.
  4. Its magnitude is half the positive axial value at same r.

Answer: Negative, half in magnitude

JEE Main-Level Solved Examples Example 7

Conductor inside point

Question: A conducting sphere has radius R and charge Q. Find potential at r = R/2.

Solution:
  1. Point is inside the conductor.
  2. Potential inside is constant.
  3. It equals surface value.
  4. V = kQ/R.

Answer: kQ/R

JEE Main-Level Solved Examples Example 8

Droplet inverse

Question: A large drop of potential 80 V breaks into 8 identical drops. Find potential of each small drop.

Solution:
  1. For 8 small drops combining, large potential = 82/3Vs = 4Vs.
  2. 80 = 4Vs.
  3. Vs = 20 V.
  4. This is the reverse droplet formula.

Answer: 20 V

JEE Main-Level Solved Examples Example 9

Potential zero with nonzero field

Question: Give a configuration where potential is zero at a point but electric field is nonzero.

Solution:
  1. Use an electric dipole.
  2. At equatorial point, potentials due to +q and -q cancel.
  3. Fields do not cancel because components along axis add.
  4. Thus V = 0 but E is not zero.

Answer: Equatorial point of a dipole

JEE Main-Level Solved Examples Example 10

Work by field

Question: A 1 mC positive charge moves from 100 V to 40 V. Find work done by field.

Solution:
  1. ΔV = 40 - 100 = -60 V.
  2. Wfield = -qΔV.
  3. q = 1 x 10-3 C.
  4. Wfield = 0.06 J.

Answer: 0.06 J

JEE Main-Level Solved Examples Example 11

Potential at midpoint

Question: +8 nC and -2 nC are separated by 0.6 m. Find potential at midpoint.

Solution:
  1. Midpoint distance from each charge is 0.3 m.
  2. V = k(8 nC/0.3 - 2 nC/0.3).
  3. V = k(6 x 10-9/0.3).
  4. V = 180 V.

Answer: 180 V

JEE Main-Level Solved Examples Example 12

Ring centre analogy

Question: Four charges at equal distance R from a point have charges q, 2q, -q, 3q. Find potential.

Solution:
  1. Use scalar addition.
  2. Net algebraic charge contribution = q + 2q - q + 3q = 5q.
  3. V = k(5q)/R.
  4. Directions do not matter.

Answer: 5kq/R

JEE Main-Level Solved Examples Example 13

Potential gradient sign

Question: Potential increases along +x. What is direction of electric field?

Solution:
  1. Ex = -dV/dx.
  2. If V increases with x, dV/dx is positive.
  3. Ex is negative.
  4. Field points along -x.

Answer: Along -x

JEE Main-Level Solved Examples Example 14

Energy assembly

Question: Three charges q, q, q are placed at vertices of equilateral triangle side a. Find total potential energy.

Solution:
  1. There are three pairs.
  2. Energy for each pair is kq2/a.
  3. Total U = 3kq2/a.
  4. Do not count pairs twice.

Answer: 3kq2/a

JEE Main-Level Solved Examples Example 15

Potential ratio

Question: For a point charge, potential at A is 30 V at distance r. Find potential at 3r.

Solution:
  1. V is inversely proportional to r.
  2. At 3r, V becomes V/3.
  3. V = 30/3.
  4. V = 10 V.

Answer: 10 V

JEE Advanced-Level Solved Examples

JEE Advanced-Level Solved Examples Example 1

Continuous idea

Question: A thin ring of radius R carries charge Q uniformly. Find potential at centre.

Solution:
  1. Each charge element is at distance R from centre.
  2. dV = kdq/R.
  3. Integrate: V = k/R ∫dq.
  4. V = kQ/R.

Answer: kQ/R

JEE Advanced-Level Solved Examples Example 2

Ring axis

Question: A uniformly charged ring of radius R and charge Q. Find potential at point on axis at distance x.

Solution:
  1. Every charge element is at same distance √(R2 + x2).
  2. dV = kdq/√(R2 + x2).
  3. Integrate dq to Q.
  4. V = kQ/√(R2 + x2).

Answer: kQ/√(R2 + x2)

JEE Advanced-Level Solved Examples Example 3

Field from ring potential

Question: Use ring axial potential V = kQ/(R2+x2)1/2 to find axial field.

Solution:
  1. Ex = -dV/dx.
  2. dV/dx = -kQx/(R2+x2)3/2.
  3. Therefore Ex = kQx/(R2+x2)3/2.
  4. Direction is along the axis for positive Q and x > 0.

Answer: kQx/(R2+x2)3/2

JEE Advanced-Level Solved Examples Example 4

Three-charge energy

Question: Charges q, -q and q are at vertices of equilateral triangle side a. Find total potential energy.

Solution:
  1. Use pair energies.
  2. Pairs: q and -q gives -kq2/a twice.
  3. Pair q and q gives +kq2/a once.
  4. Total U = -kq2/a.

Answer: -kq2/a

JEE Advanced-Level Solved Examples Example 5

Dipole from two charges

Question: Starting from exact dipole potential, justify the short dipole approximation.

Solution:
  1. Exact axial potential is kp/(r2-a2).
  2. If r >> a, then a2/r2 is very small.
  3. So r2-a2 ≈ r2.
  4. Hence V ≈ kp/r2.

Answer: Approximation valid when r >> a

JEE Advanced-Level Solved Examples Example 6

Potential and field zero

Question: Can both potential and electric field be zero at a point? Give a possible symmetric case.

Solution:
  1. Yes, possible by symmetry.
  2. For four equal charges at alternate signs at square corners, potential at centre can be zero.
  3. Depending on arrangement, vector fields can also cancel.
  4. Both conditions must be checked separately.

Answer: Yes, but both must be independently verified

JEE Advanced-Level Solved Examples Example 7

Potential along x

Question: If V(x) = Ax2 + Bx + C, find electric field.

Solution:
  1. In one dimension, E = -dV/dx.
  2. dV/dx = 2Ax + B.
  3. E = -(2Ax + B).
  4. The constant C has no effect on field.

Answer: E = -(2Ax + B)

JEE Advanced-Level Solved Examples Example 8

Equipotential proof

Question: Prove electric field is normal to equipotential surface.

Solution:
  1. On equipotential surface, dV = 0 for any tangential displacement.
  2. Work per unit charge is -dV, so tangential work is zero.
  3. If field had tangential component, it would do work.
  4. Therefore electric field must be perpendicular to the surface.

Answer: Field is normal to equipotential surface

JEE Advanced-Level Solved Examples Example 9

Conducting shell cavity

Question: Inside material of a charged conducting shell in electrostatic equilibrium, what is electric field and potential behavior?

Solution:
  1. Electric field inside conducting material is zero.
  2. Therefore potential does not vary within conductor material.
  3. The conductor is an equipotential object.
  4. Potential value depends on charge and geometry.

Answer: E = 0 and V constant

JEE Advanced-Level Solved Examples Example 10

Potential reference

Question: If every potential in a problem is increased by 50 V, what happens to electric field?

Solution:
  1. Electric field depends on gradient of potential.
  2. Adding a constant does not change gradient.
  3. Potential differences are unchanged.
  4. Electric field is unchanged.

Answer: No change

JEE Advanced-Level Solved Examples Example 11

Nonuniform field work

Question: Potential changes from 200 V to 80 V along any path. Work done by field on 3 microcoulomb positive charge?

Solution:
  1. Only endpoints matter for electrostatic force.
  2. ΔV = 80 - 200 = -120 V.
  3. Wfield = -qΔV.
  4. Wfield = -(3 x 10-6)(-120) = 3.6 x 10-4 J.

Answer: 3.6 x 10-4 J

JEE Advanced-Level Solved Examples Example 12

Dipole zero potential cone

Question: For a short dipole, at which angles is potential zero?

Solution:
  1. V = kp cosθ/r2.
  2. V = 0 when cosθ = 0.
  3. θ = 90 degrees.
  4. This is the equatorial plane.

Answer: 90 degrees, equatorial plane

JEE Advanced-Level Solved Examples Example 13

Point charge plus constant

Question: A potential is V = kQ/r + C. What is physical role of C?

Solution:
  1. C is a reference constant.
  2. Potential differences do not change if C is added everywhere.
  3. Electric field is unchanged.
  4. Only absolute numerical potential shifts.

Answer: It changes reference level only

JEE Advanced-Level Solved Examples Example 14

Droplet derivation

Question: Derive potential factor for n identical droplets merging.

Solution:
  1. Total charge Q = nq.
  2. Volume conservation gives R = n1/3r.
  3. V' = kQ/R.
  4. V' = k(nq)/(n1/3r) = n2/3V.

Answer: V' = n2/3V

JEE Advanced-Level Solved Examples Example 15

Square with centre charge

Question: Four charges q at square corners and charge -Q at centre. Find potential at a corner due to other charges.

Solution:
  1. At a corner, adjacent corner charges are distance a, opposite corner is a√2.
  2. Centre charge is distance a/√2.
  3. Potential from other corner charges is kq/a + kq/a + kq/(a√2).
  4. Centre contribution is -kQ/(a/√2) = -kQ√2/a.

Answer: k/a[2q + q/√2 - Q√2]

IB, IGCSE, ICSE and A-Level Solved Examples

IB, IGCSE, ICSE and A-Level Solved Examples Example 1

IB unit reasoning

Question: A student says volt is newton per coulomb. Correct the statement.

Solution:
  1. Newton per coulomb is electric field unit.
  2. Volt is joule per coulomb.
  3. Potential is energy per unit charge.
  4. Electric field can also be volt per metre.

Answer: 1 V = 1 J/C

IB, IGCSE, ICSE and A-Level Solved Examples Example 2

IGCSE potential difference

Question: A 2 C charge gains 10 J energy. Find potential difference.

Solution:
  1. Use ΔV = W/q.
  2. ΔV = 10/2.
  3. ΔV = 5 V.
  4. This is energy gained per coulomb.

Answer: 5 V

IB, IGCSE, ICSE and A-Level Solved Examples Example 3

ICSE equipotential

Question: Why is work done along an equipotential surface zero?

Solution:
  1. Every point has same potential.
  2. Therefore ΔV = 0.
  3. W = qΔV.
  4. So W = 0.

Answer: Because ΔV = 0

IB, IGCSE, ICSE and A-Level Solved Examples Example 4

A-Level graph

Question: For V = kQ/r, describe the shape of V against r for Q > 0.

Solution:
  1. V is positive.
  2. It decreases with r.
  3. It approaches zero asymptotically.
  4. The graph is a rectangular hyperbola.

Answer: Positive decreasing hyperbola

IB, IGCSE, ICSE and A-Level Solved Examples Example 5

IB scalar discussion

Question: Explain why potential is scalar although electric field is vector.

Solution:
  1. Potential is work per unit charge.
  2. Work is scalar.
  3. Charge is scalar.
  4. Therefore potential is scalar.

Answer: Potential is scalar

IB, IGCSE, ICSE and A-Level Solved Examples Example 6

IGCSE work

Question: A charge 0.5 C moves across 6 V. Find energy transferred.

Solution:
  1. Use W = qV.
  2. W = 0.5 x 6.
  3. W = 3 J.
  4. Energy transfer equals work done.

Answer: 3 J

IB, IGCSE, ICSE and A-Level Solved Examples Example 7

ICSE conductor

Question: What is potential inside a charged conductor?

Solution:
  1. Electric field inside is zero.
  2. No potential gradient exists.
  3. The whole conductor is equipotential.
  4. Potential is constant.

Answer: Constant

IB, IGCSE, ICSE and A-Level Solved Examples Example 8

A-Level superposition

Question: Two potentials at a point are +30 V and -45 V. Find resultant potential.

Solution:
  1. Potential is scalar.
  2. Add algebraically.
  3. V = 30 - 45.
  4. V = -15 V.

Answer: -15 V

IB, IGCSE, ICSE and A-Level Solved Examples Example 9

IB dipole equator

Question: At the equatorial point of a dipole, why is potential zero?

Solution:
  1. Distances from charges are equal.
  2. Charges have equal magnitude opposite sign.
  3. Potentials are equal and opposite.
  4. The scalar sum is zero.

Answer: Equal and opposite scalar contributions cancel

IB, IGCSE, ICSE and A-Level Solved Examples Example 10

A-Level field

Question: If potential decreases uniformly by 12 V over 0.04 m, find field strength.

Solution:
  1. Magnitude of E = ΔV/d.
  2. E = 12/0.04.
  3. E = 300 V/m.
  4. Field points toward lower potential.

Answer: 300 V/m

UP Board-Style Solved Examples

UP Board-Style Solved Examples Example 1

Definition writing

Question: Write the definition of electric potential and its formula.

Solution:
  1. Electric potential is work done per unit positive charge.
  2. The charge is brought from infinity without acceleration.
  3. Formula is V = W/q.
  4. SI unit is volt.

Answer: V = W/q

UP Board-Style Solved Examples Example 2

One volt

Question: Define one volt.

Solution:
  1. Potential is work per unit charge.
  2. If 1 J work is done to move 1 C charge, potential difference is 1 V.
  3. So 1 V = 1 J/C.
  4. This is the required definition.

Answer: 1 volt = 1 joule per coulomb

UP Board-Style Solved Examples Example 3

Point charge derivation

Question: State potential due to point charge Q at distance r.

Solution:
  1. Use standard point charge result.
  2. V = kQ/r.
  3. k = 1/(4πε0).
  4. Potential is positive for Q > 0 and negative for Q < 0.

Answer: V = (1/4πε0)Q/r

UP Board-Style Solved Examples Example 4

Potential energy

Question: Write formula for potential energy of two point charges.

Solution:
  1. Potential due to q1 at q2 is kq1/r.
  2. Energy is q2V.
  3. Therefore U = kq1q2/r.
  4. Sign depends on product of charges.

Answer: U = kq1q2/r

UP Board-Style Solved Examples Example 5

Equipotential surface

Question: State two properties of equipotential surfaces.

Solution:
  1. Potential is same at all points.
  2. Work done along it is zero.
  3. Electric field is perpendicular to it.
  4. These are standard properties.

Answer: Same potential; zero work; perpendicular field

UP Board-Style Solved Examples Example 6

Dipole moment

Question: Define electric dipole moment.

Solution:
  1. Dipole moment equals charge times separation.
  2. For charges +q and -q separated by 2a, p = q(2a).
  3. Direction is from negative to positive charge.
  4. SI unit is C m.

Answer: p = q(2a), from -q to +q

UP Board-Style Solved Examples Example 7

Axial dipole

Question: Write short dipole potential on axial line.

Solution:
  1. For short dipole, r >> a.
  2. Axial potential is V = kp/r2.
  3. This is for the side of positive charge.
  4. Opposite side has opposite sign.

Answer: V = kp/r2

UP Board-Style Solved Examples Example 8

Equatorial dipole

Question: What is potential on equatorial line of a dipole?

Solution:
  1. Distances from both charges are equal.
  2. Potentials are equal and opposite.
  3. Sum is zero.
  4. So equatorial potential is zero.

Answer: 0

UP Board-Style Solved Examples Example 9

Conducting sphere

Question: Write potential inside a charged conducting sphere.

Solution:
  1. Inside conductor, E = 0.
  2. Potential is constant.
  3. It equals surface value.
  4. V = kQ/R.

Answer: V = kQ/R

UP Board-Style Solved Examples Example 10

Droplets

Question: If n identical charged drops combine, write new potential.

Solution:
  1. R = n1/3r.
  2. Q = nq.
  3. V' = kQ/R.
  4. Therefore V' = n2/3V.

Answer: V' = n2/3V

Section 16

Exam Question Bank with Clickable Answers

This question bank contains exam-style questions. No fake year labels are used. Each answer is clickable and includes the correct answer with explanation.

NEET Questions

NEET 1. Definition MCQ

Electric potential at a point is best described as:

  • A. Force on a unit charge
  • B. Work done per unit positive charge
  • C. Work done per unit mass
  • D. Charge per unit work
Show Answer

Correct Answer: B. Work done per unit positive charge

Solution: Electric potential is defined as work done per unit positive test charge in bringing it from infinity to the point without acceleration.

NEET 2. Point charge sign

A negative point charge produces which type of potential at a nearby point?

  • A. Always positive
  • B. Always zero
  • C. Negative
  • D. Infinite only
Show Answer

Correct Answer: C. Negative

Solution: V = kQ/r, and r is positive. If Q is negative, V is negative.

NEET 3. Dipole equator

Potential at the equatorial point of an electric dipole is:

  • A. kp/r2
  • B. 2kp/r2
  • C. Zero
  • D. Infinite
Show Answer

Correct Answer: C. Zero

Solution: The point is equally distant from both charges, so equal and opposite scalar potentials cancel.

NEET 4. Droplet potential

If 64 identical droplets merge, final potential becomes:

  • A. 4V
  • B. 8V
  • C. 16V
  • D. 64V
Show Answer

Correct Answer: C. 16V

Solution: V' = n2/3V = 642/3V = 16V.

NEET 5. Work on equipotential

A charge moves along an equipotential surface. Work done is:

  • A. qV
  • B. q/V
  • C. zero
  • D. depends on path length
Show Answer

Correct Answer: C. zero

Solution: On an equipotential surface, ΔV = 0, so W = qΔV = 0.

NEET 6. Potential gradient

Electric field points in the direction of:

  • A. increasing potential
  • B. decreasing potential
  • C. constant potential
  • D. increasing charge
Show Answer

Correct Answer: B. decreasing potential

Solution: E = -∇V, so field is opposite to the direction of increasing potential.

NEET 7. Conducting sphere

Potential inside a charged conducting sphere is:

  • A. zero everywhere
  • B. proportional to r
  • C. constant
  • D. inversely proportional to r
Show Answer

Correct Answer: C. constant

Solution: Electric field inside conductor is zero, so potential cannot vary with position.

JEE Main Questions

JEE Main 1. Square centre

Four equal charges q are placed at the corners of a square side a. Potential at centre is:

  • A. 4kq/a
  • B. 4√2 kq/a
  • C. kq/a
  • D. zero
Show Answer

Correct Answer: B. 4√2 kq/a

Solution: Distance of each charge from centre is a/√2, so total potential is 4kq/(a/√2).

JEE Main 2. Opposite square

Charges +q, -q, +q, -q are placed alternately at square corners. Potential at centre is:

  • A. zero
  • B. 4kq/a
  • C. 2kq/a
  • D. depends on field direction
Show Answer

Correct Answer: A. zero

Solution: All distances are equal and algebraic sum of charges is zero.

JEE Main 3. Potential energy

Potential energy of charges q and -2q separated by r is:

  • A. kq2/r
  • B. -2kq2/r
  • C. 2kq2/r
  • D. zero
Show Answer

Correct Answer: B. -2kq2/r

Solution: Use U = kq1q2/r = k(q)(-2q)/r.

JEE Main 4. Field from V

If V = A/r, the electric field magnitude is:

  • A. A/r
  • B. A/r2
  • C. -A/r
  • D. Ar
Show Answer

Correct Answer: B. A/r2

Solution: E = -dV/dr = -(-A/r2) = A/r2.

JEE Main 5. Short dipole

For a short dipole, axial potential varies with distance as:

  • A. 1/r
  • B. 1/r2
  • C. r
  • D. r2
Show Answer

Correct Answer: B. 1/r2

Solution: Short dipole axial potential is V = kp/r2.

JEE Main 6. Angle dependence

Short dipole potential at angle 90 degrees is:

  • A. maximum
  • B. minimum positive
  • C. zero
  • D. kp/r
Show Answer

Correct Answer: C. zero

Solution: V = kp cosθ/r2, and cos 90 degrees is zero.

JEE Main 7. Reference level

Adding a constant to potential everywhere changes:

  • A. electric field
  • B. force
  • C. potential differences
  • D. absolute potential values only
Show Answer

Correct Answer: D. absolute potential values only

Solution: Field and work depend on potential difference or gradient, not absolute reference.

JEE Advanced Questions

JEE Advanced 1. Ring potential

Potential at centre of a uniformly charged ring is:

  • A. kQ/R
  • B. kQ/R2
  • C. zero
  • D. kQR
Show Answer

Correct Answer: A. kQ/R

Solution: Every charge element is at the same distance R, so integrate kdq/R.

JEE Advanced 2. Ring axis

Potential on axis of charged ring at distance x is:

  • A. kQ/x
  • B. kQ/√(R2+x2)
  • C. kQ/(R+x)
  • D. zero
Show Answer

Correct Answer: B. kQ/√(R2+x2)

Solution: All ring elements are at distance √(R2 + x2) from the axial point.

JEE Advanced 3. Zero V and E

If V = 0 at a point, electric field at that point:

  • A. must be zero
  • B. must be maximum
  • C. may or may not be zero
  • D. is infinite
Show Answer

Correct Answer: C. may or may not be zero

Solution: Potential is scalar; electric field depends on gradient. Zero potential alone is insufficient.

JEE Advanced 4. Equipotential proof

Electric field at an equipotential surface is:

  • A. tangential
  • B. perpendicular
  • C. zero always
  • D. parallel only
Show Answer

Correct Answer: B. perpendicular

Solution: Any tangential field would do work along the surface, contradicting ΔV = 0.

JEE Advanced 5. Potential energy assembly

Total potential energy of three equal charges q at equilateral triangle side a is:

  • A. kq2/a
  • B. 2kq2/a
  • C. 3kq2/a
  • D. zero
Show Answer

Correct Answer: C. 3kq2/a

Solution: There are three pairs, each with energy kq2/a.

JEE Advanced 6. Gradient

If V = Ax2, Ex equals:

  • A. Ax
  • B. -2Ax
  • C. 2Ax
  • D. -A/x
Show Answer

Correct Answer: B. -2Ax

Solution: Ex = -dV/dx = -2Ax.

JEE Advanced 7. Dipole sign

For a short dipole, potential is negative when:

  • A. cosθ > 0
  • B. cosθ = 0
  • C. cosθ < 0
  • D. r is large
Show Answer

Correct Answer: C. cosθ < 0

Solution: V = kp cosθ/r2; sign follows cosθ.

CBSE Questions

CBSE 1. Volt definition

One volt is equal to:

  • A. 1 N/C
  • B. 1 J/C
  • C. 1 C/J
  • D. 1 J m
Show Answer

Correct Answer: B. 1 J/C

Solution: V = W/q, so SI unit is joule per coulomb.

CBSE 2. Potential difference

If 20 J work is done to move 4 C charge, potential difference is:

  • A. 5 V
  • B. 80 V
  • C. 16 V
  • D. 24 V
Show Answer

Correct Answer: A. 5 V

Solution: ΔV = W/q = 20/4 = 5 V.

CBSE 3. Scalar

Electric potential is a:

  • A. vector
  • B. scalar
  • C. tensor
  • D. field line
Show Answer

Correct Answer: B. scalar

Solution: It is work per unit charge, and both work and charge are scalar.

CBSE 4. Point charge

Potential at distance r from point charge Q is:

  • A. kQr
  • B. kQ/r
  • C. kQ/r2
  • D. kr/Q
Show Answer

Correct Answer: B. kQ/r

Solution: This is the standard point charge potential with zero at infinity.

CBSE 5. Conductor

Electric field inside conductor in electrostatic equilibrium is:

  • A. zero
  • B. infinite
  • C. kQ/R
  • D. V/R always
Show Answer

Correct Answer: A. zero

Solution: Charges rearrange until internal field becomes zero.

CBSE 6. Potential energy

Potential energy of charge q at potential V is:

  • A. V/q
  • B. q/V
  • C. qV
  • D. q+V
Show Answer

Correct Answer: C. qV

Solution: Use U = qV.

CBSE 7. Dipole moment

Direction of electric dipole moment is:

  • A. positive to negative
  • B. negative to positive
  • C. perpendicular to axis
  • D. random
Show Answer

Correct Answer: B. negative to positive

Solution: By convention, dipole moment points from -q to +q.

IB Physics Questions

IB Physics 1. Concept

Which quantity is energy transferred per unit charge?

  • A. Electric field
  • B. Electric potential difference
  • C. Electric force
  • D. Electric flux
Show Answer

Correct Answer: B. Electric potential difference

Solution: Potential difference is work or energy transferred per unit charge.

IB Physics 2. Graph

For an isolated positive point charge, V-r graph is:

  • A. linear increasing
  • B. linear decreasing
  • C. positive inverse curve
  • D. zero line
Show Answer

Correct Answer: C. positive inverse curve

Solution: V = kQ/r, so it decreases as an inverse curve.

IB Physics 3. Field relation

The SI unit V/m is equivalent to:

  • A. J/C
  • B. N/C
  • C. C/J
  • D. C/m
Show Answer

Correct Answer: B. N/C

Solution: Electric field may be measured in V/m or N/C.

IB Physics 4. Work by field

A positive charge moves naturally in the direction of:

  • A. increasing potential
  • B. decreasing potential
  • C. constant charge
  • D. increasing mass
Show Answer

Correct Answer: B. decreasing potential

Solution: Positive charges move in the direction of electric field, which points toward decreasing potential.

IB Physics 5. Potential reference

Why can potential at infinity be chosen as zero?

  • A. Because field is always zero everywhere
  • B. Because only potential differences matter
  • C. Because charge disappears
  • D. Because potential is vector
Show Answer

Correct Answer: B. Because only potential differences matter

Solution: The zero of potential is a reference choice; physical results depend on differences.

IB Physics 6. Dipole

At the equatorial point of dipole, V = 0 because:

  • A. charges are absent
  • B. fields cancel completely
  • C. scalar potentials cancel
  • D. distance is infinite
Show Answer

Correct Answer: C. scalar potentials cancel

Solution: Equal distances and opposite charges make the potential contributions equal and opposite.

IB Physics 7. Energy

If potential at a point is -10 V, potential energy of +2 C charge is:

  • A. -20 J
  • B. 20 J
  • C. -5 J
  • D. 5 J
Show Answer

Correct Answer: A. -20 J

Solution: U = qV = 2 x (-10) = -20 J.

IGCSE Questions

IGCSE 1. Charge energy

A charge of 3 C moves through 4 V. Energy transferred is:

  • A. 7 J
  • B. 12 J
  • C. 1.33 J
  • D. 0.75 J
Show Answer

Correct Answer: B. 12 J

Solution: W = qV = 3 x 4 = 12 J.

IGCSE 2. Unit

Potential difference is measured in:

  • A. volt
  • B. ampere
  • C. ohm
  • D. tesla
Show Answer

Correct Answer: A. volt

Solution: The SI unit of potential difference is volt.

IGCSE 3. Definition

Potential difference is energy transferred per:

  • A. second
  • B. kilogram
  • C. coulomb
  • D. metre
Show Answer

Correct Answer: C. coulomb

Solution: It is work or energy per unit charge.

IGCSE 4. Simple work

If 30 J are transferred by 6 C of charge, p.d. is:

  • A. 5 V
  • B. 36 V
  • C. 180 V
  • D. 24 V
Show Answer

Correct Answer: A. 5 V

Solution: V = W/q = 30/6 = 5 V.

IGCSE 5. Equipotential

Two points at same potential have potential difference:

  • A. zero
  • B. one volt
  • C. infinite
  • D. negative only
Show Answer

Correct Answer: A. zero

Solution: Potential difference is the difference between potentials of two points.

IGCSE 6. Point charge trend

As distance from a positive charge increases, potential:

  • A. increases linearly
  • B. decreases toward zero
  • C. stays constant
  • D. becomes infinite
Show Answer

Correct Answer: B. decreases toward zero

Solution: For a point charge, V = kQ/r.

IGCSE 7. Scalar addition

Potentials +4 V and +7 V combine to give:

  • A. 3 V
  • B. 11 V
  • C. 28 V
  • D. 0
Show Answer

Correct Answer: B. 11 V

Solution: Potential is scalar, so add algebraically.

ICSE Questions

ICSE 1. Formula

Electric potential due to charge Q at r is:

  • A. Q/kr
  • B. kQ/r
  • C. kr/Q
  • D. kQr2
Show Answer

Correct Answer: B. kQ/r

Solution: The point-charge potential is kQ/r.

ICSE 2. Work

A 5 C charge moved through 2 V requires work:

  • A. 2.5 J
  • B. 7 J
  • C. 10 J
  • D. 0.4 J
Show Answer

Correct Answer: C. 10 J

Solution: W = qΔV = 5 x 2 = 10 J.

ICSE 3. Conductor

A conductor in electrostatic equilibrium is:

  • A. equipotential
  • B. never charged
  • C. always at zero potential
  • D. nonconducting inside
Show Answer

Correct Answer: A. equipotential

Solution: Since E = 0 inside, potential is constant throughout the conductor.

ICSE 4. Potential energy sign

Two like charges have potential energy:

  • A. negative
  • B. positive
  • C. zero always
  • D. imaginary
Show Answer

Correct Answer: B. positive

Solution: Product q1q2 is positive for like charges.

ICSE 5. Dipole moment unit

Unit of dipole moment is:

  • A. C/m
  • B. C m
  • C. N/C
  • D. J/C
Show Answer

Correct Answer: B. C m

Solution: p = charge x distance, so unit is coulomb metre.

ICSE 6. Potential difference

If work by external agent per unit charge is 8 J/C, potential difference is:

  • A. 8 V
  • B. 1/8 V
  • C. 8 C
  • D. 8 N
Show Answer

Correct Answer: A. 8 V

Solution: 1 J/C equals 1 volt.

ICSE 7. Equatorial

Equatorial potential of a dipole is:

  • A. zero
  • B. kp/r2
  • C. 2kp/r2
  • D. kq/r
Show Answer

Correct Answer: A. zero

Solution: The two scalar potentials cancel exactly.

UP Board Questions

UP Board 1. Definition

Electric potential equals:

  • A. Wq
  • B. W/q
  • C. q/W
  • D. F/q
Show Answer

Correct Answer: B. W/q

Solution: It is work done per unit positive test charge.

UP Board 2. Unit

SI unit of electric potential is:

  • A. volt
  • B. coulomb
  • C. joule
  • D. newton
Show Answer

Correct Answer: A. volt

Solution: The SI unit is volt.

UP Board 3. Potential energy

Formula U = qV represents:

  • A. electric field
  • B. potential energy
  • C. capacitance
  • D. current
Show Answer

Correct Answer: B. potential energy

Solution: Potential energy of charge q at potential V is qV.

UP Board 4. Dipole axial

Short dipole axial potential is:

  • A. kp/r
  • B. kp/r2
  • C. zero
  • D. kr/p
Show Answer

Correct Answer: B. kp/r2

Solution: For short dipole on axial line, V = kp/r2.

UP Board 5. Droplet formula

If n droplets combine, new potential is:

  • A. nV
  • B. n1/3V
  • C. n2/3V
  • D. V/n
Show Answer

Correct Answer: C. n2/3V

Solution: Use V' = n2/3V.

UP Board 6. Field relation

Relation between E and V in one dimension is:

  • A. E = dV/dr
  • B. E = -dV/dr
  • C. E = Vr
  • D. E = V/r2
Show Answer

Correct Answer: B. E = -dV/dr

Solution: Field points in direction of decreasing potential.

UP Board 7. Equipotential work

Work done on equipotential surface is:

  • A. maximum
  • B. minimum nonzero
  • C. zero
  • D. infinite
Show Answer

Correct Answer: C. zero

Solution: ΔV = 0 on equipotential surface, so W = qΔV = 0.

Section 17

CBSE Case Studies

Each case study includes a passage and five clickable answer questions in CBSE-style format.

Case Study 1: Potential Due to a Point Charge

A positive point charge Q is fixed in space. Points A, B and C lie on the same radial line at distances r, 2r and 3r respectively from the charge. Potential at infinity is taken as zero.

Case Study 1. Question 1

Potential at A is proportional to:

  • A. r
  • B. 1/r
  • C. r2
  • D. zero
Show Answer

Correct Answer: B. 1/r

Solution: For a point charge, V = kQ/r.

Case Study 2. Question 2

Potential at B compared with A is:

  • A. same
  • B. half
  • C. double
  • D. four times
Show Answer

Correct Answer: B. half

Solution: At 2r, potential becomes kQ/(2r), which is half.

Case Study 3. Question 3

Potential at C compared with A is:

  • A. one-third
  • B. three times
  • C. nine times
  • D. zero
Show Answer

Correct Answer: A. one-third

Solution: At 3r, potential is one-third of potential at r.

Case Study 4. Question 4

If Q were negative, the potential at A would be:

  • A. positive
  • B. negative
  • C. zero
  • D. undefined
Show Answer

Correct Answer: B. negative

Solution: Sign of point charge potential follows sign of Q.

Case Study 5. Question 5

Electric field direction for positive Q is:

  • A. radially outward
  • B. radially inward
  • C. tangential
  • D. random
Show Answer

Correct Answer: A. radially outward

Solution: A positive source charge repels a positive test charge.

Case Study 2: Dipole Potential

An electric dipole consists of charges +q and -q separated by 2a. A point P lies far away at distance r from the centre, making an angle theta with the dipole axis.

Case Study 1. Question 1

Short dipole potential at P is:

  • A. kp/r
  • B. kp cosθ/r2
  • C. kq/r2
  • D. zero always
Show Answer

Correct Answer: B. kp cosθ/r2

Solution: This is the general short dipole formula.

Case Study 2. Question 2

At theta = 0 degrees, potential is:

  • A. zero
  • B. kp/r2
  • C. -kp/r2
  • D. kq/r
Show Answer

Correct Answer: B. kp/r2

Solution: cos 0 degrees = 1.

Case Study 3. Question 3

At theta = 90 degrees, potential is:

  • A. zero
  • B. maximum
  • C. negative maximum
  • D. infinite
Show Answer

Correct Answer: A. zero

Solution: cos 90 degrees = 0.

Case Study 4. Question 4

Dipole moment direction is:

  • A. +q to -q
  • B. -q to +q
  • C. perpendicular to axis
  • D. toward centre
Show Answer

Correct Answer: B. -q to +q

Solution: This is the convention for electric dipole moment.

Case Study 5. Question 5

If r is doubled at same theta, potential becomes:

  • A. half
  • B. one-fourth
  • C. double
  • D. four times
Show Answer

Correct Answer: B. one-fourth

Solution: Short dipole potential varies as 1/r2.

Case Study 3: Equipotential Surfaces

A charge is moved slowly on a surface where electric potential is constant. The electric field near the surface is not necessarily zero.

Case Study 1. Question 1

Potential difference between two points on the surface is:

  • A. zero
  • B. positive
  • C. negative
  • D. infinite
Show Answer

Correct Answer: A. zero

Solution: All points on an equipotential surface have the same potential.

Case Study 2. Question 2

Work done in moving charge q on the surface is:

  • A. qV
  • B. q/V
  • C. zero
  • D. kq/r
Show Answer

Correct Answer: C. zero

Solution: W = qΔV = 0.

Case Study 3. Question 3

Electric field is directed:

  • A. tangent to surface
  • B. perpendicular to surface
  • C. randomly
  • D. only upward
Show Answer

Correct Answer: B. perpendicular to surface

Solution: Tangential electric field would do work along the surface.

Case Study 4. Question 4

Equipotential surfaces for a point charge are:

  • A. planes
  • B. concentric spheres
  • C. cylinders only
  • D. straight lines only
Show Answer

Correct Answer: B. concentric spheres

Solution: Distance from the point charge is constant on a sphere.

Case Study 5. Question 5

If V is constant, potential gradient along the surface is:

  • A. zero
  • B. infinite
  • C. negative
  • D. positive
Show Answer

Correct Answer: A. zero

Solution: No change in potential along the surface means zero tangential gradient.

Case Study 4: Conducting Sphere

A conducting sphere of radius R carries charge Q and is in electrostatic equilibrium. The potential at infinity is zero.

Case Study 1. Question 1

Potential on surface is:

  • A. kQ/R
  • B. kQ/R2
  • C. zero
  • D. kQR
Show Answer

Correct Answer: A. kQ/R

Solution: A charged conducting sphere behaves like a point charge for outside points and surface value is kQ/R.

Case Study 2. Question 2

Potential at centre is:

  • A. zero
  • B. kQ/R
  • C. kQ/(R/2)
  • D. kQ/R2
Show Answer

Correct Answer: B. kQ/R

Solution: Inside a conductor, potential is constant and equal to surface potential.

Case Study 3. Question 3

Electric field inside is:

  • A. zero
  • B. kQ/R2
  • C. infinite
  • D. kQ/R
Show Answer

Correct Answer: A. zero

Solution: Electrostatic field inside conductor is zero.

Case Study 4. Question 4

Outside at distance 2R, potential is:

  • A. kQ/R
  • B. kQ/(2R)
  • C. 2kQ/R
  • D. zero
Show Answer

Correct Answer: B. kQ/(2R)

Solution: Outside formula is V = kQ/r.

Case Study 5. Question 5

The conductor is:

  • A. not equipotential
  • B. equipotential
  • C. potential-free
  • D. always neutral
Show Answer

Correct Answer: B. equipotential

Solution: No internal electric field means no potential gradient inside the conductor.

Case Study 5: Charged Droplets

A large charged drop is formed by combining n identical small droplets. Each small droplet has radius r, charge q and potential V.

Case Study 1. Question 1

Final radius is:

  • A. nr
  • B. n1/3r
  • C. n2/3r
  • D. r/n
Show Answer

Correct Answer: B. n1/3r

Solution: Volume conservation gives R = n1/3r.

Case Study 2. Question 2

Final charge is:

  • A. q/n
  • B. nq
  • C. n1/3q
  • D. zero
Show Answer

Correct Answer: B. nq

Solution: Charges add directly.

Case Study 3. Question 3

Final potential is:

  • A. nV
  • B. n1/3V
  • C. n2/3V
  • D. V/n
Show Answer

Correct Answer: C. n2/3V

Solution: V' = k(nq)/(n1/3r) = n2/3V.

Case Study 4. Question 4

For n = 8, final potential is:

  • A. 2V
  • B. 4V
  • C. 8V
  • D. 16V
Show Answer

Correct Answer: B. 4V

Solution: 82/3 = 4.

Case Study 5. Question 5

For n = 27, final radius is:

  • A. 3r
  • B. 9r
  • C. 27r
  • D. r/3
Show Answer

Correct Answer: A. 3r

Solution: 271/3 = 3.

Section 18

Common Student Mistakes

Confusing potential with field

Potential is energy per unit charge; electric field is force per unit charge. Their units, nature and formulas differ.

Treating potential as vector

Do not resolve potential into components. Add signed scalar values directly.

Forgetting sign of charge

A negative charge creates negative potential when zero is taken at infinity.

Wrong dipole distance

For axial exact formula, distances are r - a and r + a from the two charges, not both r.

Zero potential means zero field

The equatorial point of a dipole has zero potential but nonzero electric field.

Axial and equatorial confusion

Axial short dipole potential is kp/r2; equatorial potential is zero.

Droplet formula error

Use V' = n2/3V, not nV, because radius also changes.

Conductor inside error

Inside a conductor, potential is constant. Do not use V = kQ/r for r < R.

Work sign confusion

External work for slow motion is qΔV; work by field is -qΔV.

Taking infinity casually

For isolated charges zero at infinity is standard, but potential reference can be shifted by a constant.

Section 19

Exam Strategy

CBSE

Write definitions precisely, show derivation steps for point charge and dipole, and state units. Practice conductor and equipotential explanations in words.

NEET

Memorize formulas with sign and distance dependence. Focus on quick MCQs: point charge, dipole equator, droplet factor, conducting sphere and work relation.

JEE Main

Practice mixed numerical problems using superposition, potential energy, potential gradient and conductor graphs. Be careful with algebraic signs.

JEE Advanced

Strengthen derivations, approximations, continuous charge distributions, field from potential, and multi-charge potential energy.

IB Physics

Emphasize energy transfer per unit charge, graphical interpretation, equipotential reasoning and clear explanations in complete sentences.

IGCSE

Focus on voltage as energy per coulomb, simple work calculations, units, and interpretation of potential difference.

ICSE

Prepare definitions, properties, point charge formula, conductor potential, and short-answer reasoning about equipotential surfaces.

A-Level

Practice potential graphs, field as negative gradient, ring potential, and potential energy sign conventions.

UP Board

Write formula statements cleanly, memorize derivations in sequence, and practice standard numerical examples with units.

Section 20

Final Quick Revision

Definition

Electric potential is work done per unit positive test charge brought from infinity without acceleration.

Unit

SI unit is volt. 1 V = 1 J C-1.

Point charge

V = kQ/r. Sign depends on Q.

Dipole axial

Exact: kp/(r2-a2), short dipole: kp/r2.

Dipole equatorial

V = 0 because equal and opposite scalar potentials cancel.

General dipole

V = kp cosθ/r2 for a short dipole.

Work relation

External work W = qΔV; field work = -qΔV.

Potential energy

U = qV and U = kq1q2/r.

Conducting sphere

Inside: constant kQ/R. Outside: kQ/r.

Droplet formula

R = n1/3r and V' = n2/3V.

Exam trap

Zero potential does not necessarily mean zero electric field.

Superposition

Potential adds algebraically; electric field adds vectorially.

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