PhysicsISC Class 12 · Electrostatic Potential, Potential Energy and Capacitance

Electric Potential, Potential Difference and Equipotential Surfaces

Updated 15 October 2026 · Fact-checked

The electric potential at a point is the work done per unit positive charge in bringing a test charge from infinity to that point: $V = \frac{W}{q_0}$. For a point charge, $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$. An equipotential surface has the same potential everywhere; the field is perpendicular to it, and no work is done moving a charge along it.

Field lines cross every equipotential circle at right angles, no work is done along a circle, and V falls as 1/r moving away from +q.

Illustration drawn for QuestionBuddy.

Understand Electric Potential and Equipotential Surfaces

Potential and potential difference. The electrostatic force is conservative, so the work done in moving a charge between two points does not depend on the path. The potential difference between A and B is the work done by an external agent per unit positive charge in carrying a test charge $q_{0}$ slowly from B to A:
$V_{A} - V_{B} = \frac{W _{B A}}{q _{0}} .$
Taking the potential at infinity as zero, the potential at a point is the work per unit charge in bringing $q_{0}$ from infinity to that point. Unit: volt (1 V = 1 J C⁻¹). Potential is a scalar.

Derivation for a point charge. Let $q$ be at O and P be at distance $r$ . At distance $x$ , the field is $E = \frac{q}{4 π ε _{0} x ^{2}}$ outward. To move $q_{0}$ inward by $d x$ (against the field) the external force does work $d W = - q_{0} E d x$ (dx is negative while moving in). So
$V = \frac{W}{q _{0}} = - \int_{\infty}^{r} E d x = - \frac{q}{4 π ε _{0}} \int_{\infty}^{r} \frac{d x}{x ^{2}} = \frac{q}{4 π ε _{0}} [\frac{1}{x}]_{\infty}^{r} = \frac{1}{4 π ε _{0}} \frac{q}{r} .$
V is positive near a positive charge and negative near a negative charge, and falls as $\frac{1}{r}$ (while E falls as $\frac{1}{r ^{2}}$ ). Between two points, $V_{A} - V_{B} = \frac{q}{4 π ε _{0}} (\frac{1}{r _{A}} - \frac{1}{r _{B}})$ .

System of charges. Potentials add as scalars (with sign): $V = \frac{1}{4 π ε _{0}} \sum \frac{q _{i}}{r _{i}}$ . No directions are needed, which makes potential easier than field. At the centre of a regular arrangement of equal charges the field can be zero while the potential is not.

Potential of a dipole. For a short dipole at distance r, at angle θ from the axis,
$V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}} .$
On the axis (θ = 0) $V = \frac{k p}{r ^{2}}$ ; for a dipole of length 2l exactly, $V_{a x i s} = \frac{k p}{r ^{2} - l ^{2}}$ . On the equatorial plane (θ = 90°) $V = 0$ , because every point is equidistant from $+ q$ and $- q$ . So the locus of zero potential is the equatorial plane (the perpendicular bisector), the 1-mark ISC 2024 item. The field there is not zero.

Field from potential. The field points in the direction of steepest decrease of potential:
$E = - \frac{d V}{d r} .$
In a uniform field, $E = \frac{V}{d}$ , where V is the p.d. between planes a distance d apart along the field. So the volt per metre is the same as the newton per coulomb.

Equipotential surfaces. A surface on which every point has the same potential. Properties:
1. No work is done in moving a charge along it ( $W = q Δ V = 0$ ).
2. The field is perpendicular to it at every point (otherwise a component of E along the surface would do work).
3. Two equipotential surfaces never intersect (a point cannot have two potentials).
4. Where the surfaces are drawn for equal steps of V, they are closer together where the field is stronger.
5. The surface of a conductor in electrostatics is an equipotential.

Shapes: concentric spheres for a point charge; parallel planes perpendicular to the field for a uniform field; for a dipole, closed surfaces round each charge with the equatorial plane as the zero surface.

Potential of a conducting sphere. For a charged conducting (or thin shell) sphere of radius R: outside, $V = \frac{k Q}{r}$ ; on and inside, $V = \frac{k Q}{R}$ (constant, since $E = 0$ inside).

Electric Potential and Equipotential Surfaces: formulas and key facts

Definition: $V_{A} - V_{B} = \frac{W _{B A}}{q _{0}}$; Work per unit positive charge; 1 V = 1 J C⁻¹.
Point charge: $V = \frac{1}{4 π ε _{0}} \frac{q}{r}$; Sign of q kept.
System of charges: $V = \frac{1}{4 π ε _{0}} \sum_{i} \frac{q _{i}}{r _{i}}$; Scalar addition.
Short dipole: $V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}}$; Zero on the equatorial plane.
Field and potential: $E = - \frac{d V}{d r}; E = \frac{V}{d} (uniform)$; E points towards lower V.
Work between points: $W = q (V_{A} - V_{B})$; Zero along an equipotential.
Conducting sphere: $V = \frac{k Q}{R} (r \leq R), \frac{k Q}{r} (r > R)$; Constant inside.

How to solve Electric Potential and Equipotential Surfaces questions

Potential problems are mostly scalar sums and work calculations.

1Write each charge with its sign and its distance from the point (in metres).
2Add $\frac{k q _{i}}{r _{i}}$ algebraically; no vector diagram is needed.
3For work, find the p.d. between the end points and use $W = q (V_{f ina l} - V_{ini t ia l})$ for an external agent.
4For zero-potential points on a line, solve $\frac{q _{1}}{x} + \frac{q _{2}}{r _{2} ( x )} = 0$ separately between the charges and outside.
5For field from potential, differentiate: $E = - \frac{d V}{d x}$ , or use $E = \frac{Δ V}{d}$ in a uniform field.
6Check signs: potential falls along the direction of the field.

Electric Potential and Equipotential Surfaces in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Equipotentials and safety

Suppose a bird perches on a single bare power line at 11 kV, with its feet 5 cm apart along the wire, and suppose the potential falls by only 0.02 V per metre along the wire because of the current.

What p.d. acts across the bird? Why is it safe?

Try it first, then show the solutionSolution

p.d. between its feet: $0.02 \times 0.05 = 0.001$ V.
Both feet are almost at the same potential (11 kV), so practically no current passes through the bird.

Answer: About 1 mV: the bird sits on what is nearly an equipotential, so it is safe. Touching the line and an earthed pole at the same time would not be.

Memory hook: Current needs a potential difference, not a high potential.

Energy from potential difference

Imagine a thundercloud whose base is at a potential difference of $1 \times 1 0^{8}$ V from the ground, and a lightning flash that carries 10 C to the ground.

How much energy does the flash release?

Try it first, then show the solutionSolution

$W = q V = 10 \times 1 0^{8} = 1 0^{9}$ J.
That is enough to run a 1 kW heater for $1 0^{6}$ s, about 11.6 days.

Answer: $1 \times 1 0^{9}$ J.

Memory hook: Energy = charge × p.d.

Tricks to remember Electric Potential and Equipotential Surfaces

Short memory aids. Say each one aloud once, then use it on a question the same day.

Potential is a scalar
"Add potentials, add fields as arrows."
Use it when: Systems of charges.
One power less
Point charge: E ∝ $\frac{1}{r ^{2}}$ , V ∝ $\frac{1}{r}$ . Dipole: E ∝ $\frac{1}{r ^{3}}$ , V ∝ $\frac{1}{r ^{2}}$ .
Use it when: Comparing E and V.
Downhill field
Field lines run from high to low potential, at right angles to equipotentials.
Use it when: Direction of E; sketches.
Equator is zero
Dipole: V = 0 on the equatorial plane, but E is not.
Use it when: Locus questions.

Electric Potential and Equipotential Surfaces: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Equipotentials of a uniform field · 1 mark · MCQBoard level

The equipotential surfaces of a uniform electric field are:
(a) concentric spheres
(b) planes parallel to the field
(c) planes perpendicular to the field
(d) coaxial cylinders

Try it first, then show the solutionSolution

Equipotentials are always perpendicular to the field lines.

Answer: (c).

Example 2: Potential and work · 2 marks · Short answerBoard level

A +6 nC point charge is fixed in air. Find the potential at points A and B, 20 cm and 60 cm from it, and the work needed to carry a +3 µC charge from B to A.

Try it first, then show the solutionSolution

$V_{A} = \frac{9 \times 1 0 ^{9} \times 6 \times 1 0 ^{- 9}}{0.2} = 270$ V; $V_{B} = 90$ V. (1)
$W = q (V_{A} - V_{B}) = 3 \times 1 0^{- 6} \times 180 = 5.4 \times 1 0^{- 4}$ J. (1)

Answer: 270 V, 90 V; $5.4 \times 1 0^{- 4}$ J.

Example 3: Derivation: potential of a point charge · 3 marks · Long answerHarder board level

Derive an expression for the electric potential at a distance r from a point charge q. Sketch V and E against r.

Try it first, then show the solutionSolution

Potential = work done per unit positive charge in bringing a test charge from infinity to P; at distance x, $E = \frac{q}{4 π ε _{0} x ^{2}}$ . (1)
$V = - \int_{\infty}^{r} E d x = - \frac{q}{4 π ε _{0}} \int_{\infty}^{r} x^{- 2} d x = \frac{q}{4 π ε _{0} r}$ . (1)
Graphs: V falls as $\frac{1}{r}$ , E as $\frac{1}{r ^{2}}$ (E falls faster); both positive for q > 0. (1)

Answer: $V = \frac{1}{4 π ε _{0}} \frac{q}{r}$ .

Example 4: Potential and field at a centre · 2 marks · Short answerHarder board level

Six charges of +2 nC each are placed at the corners of a regular hexagon of side 10 cm. Find the potential and the field at its centre.

Try it first, then show the solutionSolution

Each corner is 10 cm from the centre (side = radius for a regular hexagon): $V = 6 \times \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 9}}{0.1} = 1080$ V. (1)
Fields from opposite corners are equal and opposite, so E = 0. (1)

Answer: V = 1080 V; E = 0.

Example 5: Field from potential · 3 marks · Short answerHarder board level

In a region, the potential varies as $V = 4 x^{2} - 2 x$ volt (x in metres). (a) Find the field at x = 1 m. (b) Two parallel plates 4 cm apart have a p.d. of 200 V. Find the field and the spacing of equipotential surfaces drawn every 50 V.

Try it first, then show the solutionSolution

(a) $E = - \frac{d V}{d x} = - (8 x - 2) = - 6$ V m⁻¹ at x = 1 m: magnitude 6 V m⁻¹, along −x. (1)
(b) $E = \frac{200}{0.04} = 5000$ V m⁻¹. (1)
Spacing $= \frac{50}{5000} = 0.01$ m = 1 cm, planes parallel to the plates. (1)

Answer: (a) 6 V m⁻¹ along −x; (b) 5000 V m⁻¹, 1 cm apart.

Example 6: A charged conducting sphere · 4 marks · Case-basedHOTS / competency

Suppose a hollow conducting sphere of radius 10 cm is charged to a potential of 9000 V.
(i) Find its charge. (1 mark)
(ii) What is the potential at its centre? (1 mark)
(iii) Find the potential 30 cm from the centre. (1 mark)
(iv) Find the field just outside the surface and inside it. (1 mark)

Try it first, then show the solutionSolution

(i) $Q = \frac{V R}{k} = \frac{9000 \times 0.1}{9 \times 1 0 ^{9}} = 1 \times 1 0^{- 7}$ C.
(ii) 9000 V: inside a conductor E = 0, so V is the same as on the surface.
(iii) $V = \frac{k Q}{r} = \frac{900}{0.3} = 3000$ V.
(iv) Outside: $E = \frac{k Q}{R ^{2}} = \frac{V}{R} = 9 \times 1 0^{4}$ V m⁻¹; inside: 0.

Answer: (i) 0.1 µC; (ii) 9000 V; (iii) 3000 V; (iv) $9 \times 1 0^{4}$ V m⁻¹ outside, zero inside.

Where students lose marks in Electric Potential and Equipotential Surfaces

The slips examiners see most often, and what to write instead.

Mistake: Point instead of locus
In ISC 2024 candidates could not distinguish a point from a locus, and some answered "equipotential surfaces".
Do this: Name the locus: the perpendicular bisector (equatorial plane) of the dipole.
Usually costs: 1 mark.
Mistake: Using "broadside position"
Also noted in 2024: this magnetism term was not accepted.
Do this: For an electric dipole say "equatorial line or plane".
Usually costs: 1 mark.
Mistake: Adding potentials as vectors
Students try to resolve potentials into components.
Wrong working
Resolving $\frac{k q}{r}$ along x and y.
Right working
Add $\frac{k q _{i}}{r _{i}}$ algebraically with signs.
Do this: Remember potential is a scalar.
Usually costs: 1 to 2 marks.
Mistake: Sign of work
Confusing work done by the field with work done by an external agent.
Do this: External agent: $W = q (V_{f ina l} - V_{ini t ia l})$ ; field: the negative of this.
Usually costs: 1 mark.

Exam tips

In the derivation, state the definition of potential before integrating.
Keep signs of charges inside the potential formula; drop them only for field magnitudes.
Draw equipotentials as dashed lines crossing field lines at right angles.
Quote units: volt for potential, V m⁻¹ (= N C⁻¹) for field.

Practice questions on Electric Potential and Equipotential Surfaces

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The work done in moving a 5 µC charge between two points on a 100 V equipotential surface is:
(a) 500 µJ
(b) 5 µJ
(c) zero
(d) 20 J
Hint
ΔV = 0.
Answer
(c).
Q2 · 1 mark · Very short answerBoard level
Charges of +2 μC and −2 μC are 6 cm apart. Name the locus of points where the potential is zero, and find the potential at a point 4 cm from each charge. (A variation on ISC 2024.)
Hint
Equidistant from both charges.
Answer
The equatorial plane (perpendicular bisector) of the pair; the point 4 cm from each charge lies on it, so its potential is zero.
Q3 · 3 marks · Short answerHarder board level
Charges of +3 nC and −1 nC are at x = 0 and x = 40 cm. Find the points on the x-axis where the potential is zero.
Hint
One point between the charges, one beyond −1 nC.
Answer
Between: $\frac{3}{x} = \frac{1}{0.4 - x}$ , x = 30 cm. Beyond: $\frac{3}{x} = \frac{1}{x - 0.4}$ , x = 60 cm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The electric field is always perpendicular to an equipotential surface.
Reason (R): No work is done in moving a charge along an equipotential surface, so E has no component along it.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
W = qE·d.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
Equipotential surfaces of 30 V and 20 V in a uniform field are 5 mm apart. Find the field and its direction.
Hint
$E = \frac{Δ V}{d}$ .
Answer
2000 V m⁻¹, perpendicular to the surfaces, from the 30 V surface towards the 20 V surface.
Q6 · 2 marks · Short answerHOTS / competency
A uniform field $E = 200 \hat{i}$ V m⁻¹ exists in a region. Find $V_{B} - V_{A}$ for A(1 m, 2 m) and B(4 m, 6 m). Which point is at the higher potential?
Hint
Only the x-displacement matters.
Answer
$V_{B} - V_{A} = - 200 \times 3 = - 600$ V; A is at the higher potential.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a short dipole of moment $4 \times 1 0^{- 10}$ C m points along +x.
(i) Find the potential 20 cm from its centre on the axis (on the +q side). (1 mark)
(ii) Find the potential 20 cm away on the equatorial line. (1 mark)
(iii) Find the potential 20 cm away along a line at 60° to the axis. (1 mark)
(iv) Find the work done in carrying a +2 nC charge from the equatorial point to the axial point. (1 mark)
Hint
$V = \frac{k p c o s θ}{r ^{2}}$ .
Answer
(i) 90 V. (ii) 0. (iii) 45 V. (iv) $W = q Δ V = 2 \times 1 0^{- 9} \times 90 = 1.8 \times 1 0^{- 7}$ J.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define electric potential at a point.Show answer
The work done per unit positive charge in bringing a test charge from infinity to that point.
2.Potential 30 cm from a 3 nC charge?Show answer
90 V.
3.Relation between E and V?Show answer
$E = - \frac{d V}{d r}$ .
4.Why can two equipotential surfaces not intersect?Show answer
The point of intersection would have two different potentials.
5.Potential inside a charged conducting sphere?Show answer
Constant, equal to $\frac{k Q}{R}$ (its surface value).

Electric Potential and Equipotential Surfaces: frequently asked questions

Can the potential be zero where the field is not?

Yes. On the equatorial plane of a dipole, V = 0 but E is not zero. Potential depends on a sum of terms that can cancel; the field depends on how the potential changes.

Can the field be zero where the potential is not?

Yes. At the centre of a ring or hexagon of equal charges, E = 0 but V is the sum of all the positive terms. Inside a charged conducting sphere, E = 0 while V equals its surface value.

Why is potential defined using a test charge?

A small test charge does not disturb the charges producing the field, so the work per unit charge measures a property of the field alone.

Is the dipole potential in the ISC syllabus?

Yes. The 2027 syllabus asks for the potential of a dipole on its axial line, its equatorial line, and at any point for a short dipole.

How was this examined recently?

In 2024 a 1-mark item asked for the locus of zero potential of a dipole.