Electric Potential

Q: Can electric potential be negative?

Yes - the sign of the potential follows the sign of the source charge. Around a negative point charge, V is negative. In a mixed system, the net potential can be positive, negative, or zero.

Q: What is the difference between potential and potential energy?

Electric potential is a property of a point in the field (work per unit test charge). Potential energy U = qV is the energy of a specific charge q placed at that point. Potential exists whether or not a charge is there.

Q: Why is potential inside a charged conductor the same as on the surface?

Because the field is zero everywhere inside a conductor in electrostatic equilibrium, dV/dr is zero inside, so the potential is constant. Continuity across the surface then gives V_inside = V_surface.

Q: What is an equipotential surface?

A surface on which the potential V is the same at every point. Field lines are always perpendicular to equipotential surfaces. No work is done moving a charge along an equipotential.

Q: Why is the equatorial potential of a dipole zero?

On the perpendicular bisector of a dipole, both charges +q and -q are equidistant from the point. Their potentials cancel exactly. This does not mean E is zero there - the field is non-zero.

Q: When is a dipole in stable equilibrium?

When p is aligned with E (theta = 0). Here the potential energy U = -pE is a minimum. At theta = 180 degrees the equilibrium is unstable.

Q: How is potential energy related to work done in assembling a charge configuration?

The electric potential energy equals the total external work required to bring the charges from infinity to their positions one at a time. It is path-independent because electrostatic forces are conservative.

Q: What is 1 volt in fundamental units?

1 V = 1 J/C = 1 kg m^2 per (A s^3). One volt is the potential difference between two points if 1 J of work is done in moving 1 C of charge between them.

PhysicsElectrostaticsFor JEE aspirants

The electric potential at a point is the work done per unit positive test charge in bringing a charge from infinity up to that point against the electric field: $V = W / q_{0}$ . Its SI unit is the volt (V), and it is a scalar - much easier to handle than the field vector. For a point charge, $V = k q / r$ ; for a system, potentials just add algebraically. This concept underlies dipole potential, potential energy of charge systems, and the behaviour of dipoles in external fields - all core JEE and NEET topics.

Key Formulas - Quick Reference

Potential of a point charge at distance $r$ : $V = \frac{k q}{r}$ .
Superposition: $V = V_{1} + V_{2} + V_{3} + \dots$ (algebraic sum, with signs).
Relation to field: $V_{B} - V_{A} = - \int_{A}^{B} E \cdot d r$ ; equivalently $E = - d V / d r$ .
Axial potential of a short dipole (dipole moment $p$ , $r ≫ a$ ): $V = \frac{k p cos θ}{r ^{2}}$ .
Potential of a conducting sphere: $V = \frac{k q}{r}$ outside; $V = \frac{k q}{R}$ inside and on the surface.
Torque on a dipole in external field $E$ : $τ = p \times E$ , magnitude $τ = pE sin θ$ .
Potential energy of a dipole in $E$ : $U = - p \cdot E = - pE cos θ$ .
Interaction energy of two point charges: $U = \frac{k q _{1} q _{2}}{r _{12}}$ .

1. Electric Potential - Definition

The electric potential at a point in an electric field is the amount of work done in bringing a unit positive test charge from infinity to that point, against the electric field, without any acceleration.

V = \frac{W _{ext}}{q _{0}} = - \frac{W _{field}}{q _{0}}

In other words, it is the negative of the work done by the electrostatic force per unit test charge.

SI unit: volt (V) $=$ joule per coulomb (J/C). Potential is a scalar quantity.

Potential of a point charge

Starting from Coulomb's law and integrating $E \cdot d r$ from infinity to $r$ :

V (r) = \frac{k q}{r} = \frac{1}{4 π ε _{0}} \cdot \frac{q}{r}

Note that potential is positive around a positive charge and negative around a negative charge - the sign is kept, unlike in the field-magnitude formula.

Potential due to several charges (superposition)

The total potential at a point due to a system of point charges is the algebraic sum of the potentials from each charge:

V = V_{1} + V_{2} + V_{3} + \dots = k i \sum \frac{q _{i}}{r _{i}}

Because potential is a scalar, no vector resolution is needed. This makes it much easier to compute than the field.

Relation between $V$ and $E$

The potential difference between two points $A$ and $B$ :

V_{B} - V_{A} = - \int_{A}^{B} E \cdot d r

Equivalently, the field is the negative gradient of the potential. In one dimension:

E = - \frac{d V}{d r}

The field points from high potential to low potential. Along a field line, $V$ decreases.

2. Potential due to an Electric Dipole

An electric dipole consists of two equal and opposite charges $+ q$ and $- q$ separated by a small distance $2 a$ . The dipole moment is a vector:

p = q \cdot 2 a, ∣ p ∣ = q \cdot 2 a

directed from the negative charge to the positive charge.

Figure 1: Dipole geometry - moment

p

points from

- q

+ q

2.1 General point (distance $r$ from centre, angle $θ$ from axis)

For $r ≫ a$ (short-dipole approximation):

V = \frac{k p cos θ}{r ^{2}} = \frac{p \cdot r ^}{4 π ε _{0} r ^{2}}

2.2 Special cases

Axial point (

θ = 0

, on the +q side):

V_{axial} = \frac{k p}{r ^{2}}

.
Axial point (

θ = 180°

, on the -q side):

V = - \frac{k p}{r ^{2}}

.
Equatorial (perpendicular bisector, $θ = 90°$ ):

V = 0

- the two charges contribute equal and opposite potentials.

Compare with the point-charge potential $V \sim 1/ r$ : a dipole potential falls faster, as $1/ r^{2}$ , because the two opposite charges partially cancel at large distances.

3. Potential due to Standard Distributions

3.1 Uniformly charged ring (radius $R$ , total charge $Q$ )

On the axis, at distance $x$ from the centre:

V = \frac{k Q}{R ^{2} + x ^{2}}

At the centre ( $x = 0$ ): $V = k Q / R$ . Far away ( $x ≫ R$ ): $V \approx k Q / x$ (like a point charge).

3.2 Uniformly charged disc (radius $R$ , surface density $σ$ )

(i) On the axis, at distance $x$ from the centre:

V = \frac{σ}{2 ε _{0}} [R^{2} + x^{2} - x]

(ii) At the centre ( $x = 0$ ):

V_{centre} = \frac{σ R}{2 ε _{0}}

(iii) At the edge of the disc (on the disc itself, at distance $R$ from the centre):

V_{edge} = \frac{σ R}{π ε _{0}}

3.3 Charged conducting sphere (radius $R$ , charge $q$ )

All the charge on a conductor resides on its outer surface, and the interior is field-free.

Outside (

r > R

V = \frac{k q}{r}

- behaves like a point charge.
On the surface (

r = R

V = \frac{k q}{R}

.
Inside (

r < R

V = \frac{k q}{R}

- constant, equal to the surface value (since

E = 0

inside).

Figure 2: Potential is constant inside a charged conducting sphere, then falls as

1/ r

outside.

3.4 Non-conducting (dielectric) sphere with uniform volume charge density (radius $R$ , total charge $q$ )

Outside (

r > R

V = \frac{k q}{r}

.
On the surface (

r = R

V = \frac{k q}{R}

.
Inside (

r < R

V = \frac{k q ( 3 R ^{2} - r ^{2} )}{2 R ^{3}}

At the centre ( $r = 0$ ): $V_{centre} = \frac{3 k q}{2 R} = 1.5 \times V_{surface}$ . The potential is highest at the centre.

3.5 Uniformly charged cone (charge $Q$ on curved surface, slant length $L$ )

Potential at the apex:

V_{apex} = \frac{k Q}{L}

4. Dipole in an External Electric Field

Place a dipole in a uniform external field $E$ , with the dipole moment $p$ making an angle $θ$ with $E$ .

Figure 3: A dipole in a uniform field experiences a torque but zero net force.

4.1 Net force

In a uniform field, the forces on $+ q$ ( $+ q E$ ) and on $- q$ ( $- q E$ ) are equal and opposite. The net force is zero - the dipole does not translate.

4.2 Torque

Although the net force is zero, the two forces form a couple. The torque magnitude:

τ = (q E) \cdot (2 a sin θ) = pE sin θ

In vector form:

τ = p \times E

The torque tries to align $p$ with $E$ .

4.3 Potential energy

Work done by the field when $p$ rotates from angle $θ_{1}$ to $θ_{2}$ : integrating $τ d θ$ we get

U (θ) = - pE cos θ = - p \cdot E

Choosing the reference $U = 0$ at $θ = 90°$ (perpendicular position).

Stable equilibrium (

θ = 0°

p

parallel to

E

U_{m i n} = - pE

.
Perpendicular (

θ = 90°

U = 0

.
Unstable equilibrium (

θ = 180°

p

anti-parallel to

E

U_{m a x} = + pE

Work done by an external agent to rotate the dipole from $θ_{1}$ to $θ_{2}$ :

W_{ext} = U (θ_{2}) - U (θ_{1}) = pE (cos θ_{1} - cos θ_{2})

5. Electric Potential Energy of a System of Charges

The electric potential energy of a system of point charges is the work done in assembling the system by bringing each charge from infinity, one at a time, to its final position (against the field of the ones already in place).

5.1 Two-particle system

For point charges $q_{1}$ and $q_{2}$ separated by distance $r_{12}$ :

U_{12} = \frac{k q _{1} q _{2}}{r _{12}}

Figure 4: Interaction energy of a two-charge system.

5.2 Three-particle system

Sum over all distinct pairs $(i, j)$ with $i < j$ :

U = k (\frac{q _{1} q _{2}}{r _{12}} + \frac{q _{2} q _{3}}{r _{23}} + \frac{q _{1} q _{3}}{r _{13}})

5.3 General $n$ -particle system

U = k i < j \sum \frac{q _{i} q _{j}}{r _{ij}} = \frac{1}{2} i \neq = j \sum \frac{k q _{i} q _{j}}{r _{ij}}

The factor of $1/2$ in the unrestricted double sum avoids double-counting pairs.

5.4 Potential energy in an external field

For a single test charge $q_{0}$ placed at a point where the potential is $V_{P}$ :

U_{P} = q_{0} V_{P}

Equivalently, the potential can be defined as $V_{P} = U_{P} / q_{0}$ - the potential energy per unit test charge.

Solved Example 1

Two conducting spheres of radii

R_{1}

and

R_{2}

carry charges

Q_{1}

and

Q_{2}

respectively. They are far apart (so each is unaffected by the other's field). Find the potentials on the two surfaces. If they are then connected by a thin conducting wire, what is the final common potential, and what is the ratio of the surface charge densities?

Solution:

Before connecting, the potential on the surface of sphere 1 is

V_{1} = \frac{k Q _{1}}{R _{1}} \dots (A)

The potential on the surface of sphere 2 is

V_{2} = \frac{k Q _{2}}{R _{2}} \dots (B)

When connected by a wire, charge flows from higher to lower potential until both surfaces are at the same potential. Let the final charges be $Q_{1}^{'}$ and $Q_{2}^{'}$ . Charge conservation:

Q_{1}^{'} + Q_{2}^{'} = Q_{1} + Q_{2}

Equal potentials:

\frac{k Q _{1}^{'}}{R _{1}} = \frac{k Q _{2}^{'}}{R _{2}} ⟹ \frac{Q _{1}^{'}}{Q _{2}^{'}} = \frac{R _{1}}{R _{2}}

Solving:

Q_{1}^{'} = \frac{R _{1} ( Q _{1} + Q _{2} )}{R _{1} + R _{2}}, Q_{2}^{'} = \frac{R _{2} ( Q _{1} + Q _{2} )}{R _{1} + R _{2}}

The common final potential is

V_{common} = \frac{k ( Q _{1} + Q _{2} )}{R _{1} + R _{2}}

Surface charge densities: $σ = Q^{'} / (4 π R^{2})$ . So

\frac{σ _{1}}{σ _{2}} = \frac{Q _{1}^{'} / R _{1}^{2}}{Q _{2}^{'} / R _{2}^{2}} = \frac{R _{1} R _{2}^{2}}{R _{2} R _{1}^{2}} \cdot \frac{R _{2}}{R _{1}} \cdot \frac{1}{1} = \frac{R _{2}}{R _{1}}

The smaller sphere has the higher surface charge density - the classic reason why lightning discharges from sharp points.

Solved Example 2

Determine the interaction energy of the point charges of the following set-up: four point charges

+ q

+ q

+ q

, and

+ q

are placed at the four corners of a square of side

a

Solution:

There are $(2 4) = 6$ pairs. Of these:

4 pairs are along the sides of the square (distance $a$ ).
2 pairs are along the diagonals (distance $a 2$ ).

Total potential energy:

U = 4 \cdot \frac{k q ^{2}}{a} + 2 \cdot \frac{k q ^{2}}{a 2} = \frac{k q ^{2}}{a} (4 + 2)

U = \frac{k q ^{2}}{a} (4 + 2) \approx \frac{5.414 k q ^{2}}{a}

The energy is positive because all four charges are like (repulsive) - external work must be done to assemble the configuration.

Common Mistakes to Avoid

Watch out

Confusing $V$ and $E$ : potential is a scalar and adds algebraically (with sign of the charge); field is a vector and requires component-wise addition. Never add field magnitudes and never resolve potentials into components.
Sign of $q$ in potential formula: for $V = k q / r$ , use the algebraic sign of $q$ . A negative charge gives a negative potential.
Dipole formula only for $r ≫ a$ : the "short-dipole" formulas $V = k p cos θ / r^{2}$ etc. hold only when the field/observation point is much farther than the dipole separation. For close distances, use the two-charge formulas directly.
Inside vs outside a conductor: potential inside a conductor equals the surface value (constant), not zero. Only the field is zero inside.
Dipole PE reference: the formula $U = - pE cos θ$ takes $θ = 90°$ as $U = 0$ . Do not accidentally take $θ = 0$ as zero energy - that gives the wrong sign convention.
Double-counting pairs in $n$ -charge energy: for the interaction energy of $n$ point charges, sum over distinct pairs only. Do not include $q_{1} q_{2}$ and $q_{2} q_{1}$ as two separate terms.
Work-energy sign: work done by the field equals $- Δ U$ . Work done against the field (by an external agent) equals $+ Δ U$ . Missing a sign here reverses your answer.

Frequently Asked Questions

Q1. Can electric potential be negative?

Yes - the sign of the potential simply follows the sign of the source charge. Around a negative point charge, $V = k q / r < 0$ . In a system of positive and negative charges, the net potential at a point can be positive, negative, or zero depending on the algebraic sum.

Q2. What is the difference between potential and potential energy?

Electric potential $V$ is a property of a point in the field (work per unit test charge). Potential energy $U = q V$ is the energy of a specific charge $q$ placed at that point. Potential exists whether or not a charge is there; potential energy needs an actual charge.

Q3. Why is potential inside a charged conductor the same as on the surface?

Because the field is zero everywhere inside a conductor in electrostatic equilibrium, $d V / d r = 0$ inside - so the potential is constant. Continuity across the surface then gives $V_{inside} = V_{surface}$ .

Q4. What is an equipotential surface?

A surface on which the potential $V$ is the same at every point. Field lines are always perpendicular to equipotential surfaces. No work is done in moving a charge along an equipotential. For a point charge, equipotentials are concentric spheres.

Q5. Why is the equatorial potential of a dipole zero?

On the perpendicular bisector of a dipole, both charges $+ q$ and $- q$ are equidistant from the observation point. Their potentials $+ k q / r$ and $- k q / r$ cancel exactly, giving $V = 0$ . This does not mean $E = 0$ there - the field is non-zero and points opposite to $p$ .

Q6. When is a dipole in stable equilibrium?

When $p$ is aligned with $E$ ( $θ = 0$ ). Here the potential energy $U = - pE$ is a minimum, and any small displacement produces a restoring torque. At $θ = 180°$ the equilibrium is unstable ( $U$ is at a maximum).

Q7. How is potential energy related to the work done in assembling a charge configuration?

The electric potential energy of a system equals the total external work required to bring the charges from infinity to their positions, one at a time, against the electrostatic forces of the previously placed charges. It is path-independent because electrostatic forces are conservative.

Q8. What is 1 volt in fundamental units?

$1 V = 1 J/C = 1 kg\cdotm^{2} / (A\cdots^{3})$ . One volt is the potential difference between two points if 1 joule of work is done in moving 1 coulomb of charge between them.

Previous year questions on Electric Potential

40 questions from past papers, each with a step-by-step solution.

Show all 40 questions

Ready to master Electrostatics?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test