PhysicsISC Class 12 · Chapter 2

Electrostatic Potential, Potential Energy and Capacitance ISC Class 12: Notes and Numericals

Q: Do I need to derive the energy stored in a capacitor for ISC 2027?

No. The revised 2027 syllabus says energy stored in a capacitor is 'No derivation, formulae only'. Learn U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV and the energy density \frac{1}{2}\varepsilon_0E^2, and use them in numericals.

Q: Which derivations are required in this chapter?

Potential difference and potential due to a point charge, the potential energy of a dipole U = -\vec{p} \cdot \vec{E}, the capacitance of a parallel-plate capacitor and the equivalent capacitance in series and parallel. The partly filled capacitor formula is also listed.

Q: Why is the potential zero on the equatorial line of a dipole while the field is not?

Potential is a scalar: every equatorial point is equidistant from +q and −q, so their potentials cancel. Field is a vector: the two field vectors there add to give a field opposite to \vec{p}.

Q: Is the potential inside a charged conductor zero?

No. The field inside is zero, so the potential is constant throughout and equal to its value at the surface.

Q: How much is this chapter worth?

It shares the 14 marks of Electrostatics and Current Electricity with Chapters 1 and 3. Capacitor questions (effect of d and K, combinations, energy) appear in most papers.

Q: What is the difference between a conductor and a dielectric in a field?

A conductor has free charges that move until the field inside becomes zero. A dielectric has only bound charges; it polarises and reduces the field by a factor K but does not cancel it.

Updated 14 October 2026 · Fact-checked

Electrostatic Potential, Potential Energy and Capacitance is Chapter 2 of ISC Class 12 Physics. It covers potential and potential difference, potential of a point charge, a dipole and a system of charges, equipotential surfaces, potential energy of charges and of a dipole, capacitance of a parallel-plate capacitor, series and parallel combinations, stored energy and dielectrics.

What Electrostatic Potential, Potential Energy and Capacitance is about

Chapter 1 described charges by the force they exert; this chapter describes them by the work done in moving charges about. The potential at a point is the work done per unit positive test charge in bringing it from infinity to that point, $V = W / q_{0}$ , and the potential difference between two points is the work per unit charge in moving between them. You derive $V_{A} - V_{B} = \frac{q}{4 π ε _{0}} (\frac{1}{r _{A}} - \frac{1}{r _{B}})$ for a point charge, add potentials algebraically for a group of charges, and find the potential of a dipole on its axis, on its equatorial line and at a general point for a short dipole.

Equipotential surfaces join points at the same potential; no work is done along them and the field is perpendicular to them. Potential energy of a system is the work done in assembling it: $U = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}}$ for two charges and a sum over pairs for three. A dipole in a uniform field has $U = - p \cdot E$ , which you must derive.

The second half is capacitance. $C = Q / V$ ; you derive $C = ε_{0} A / d$ for a parallel-plate capacitor and the series and parallel formulas. The energy stored, $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ , and energy density are formulae only (no derivation) in the 2027 syllabus. A dielectric of constant K raises capacitance to KC because polarisation of the material sets up induced surface charges that weaken the field; you must handle the two cases (battery disconnected: Q constant; battery connected: V constant) and the partly filled capacitor $C^{'} = \frac{ε _{0} A}{d - t + t / K}$ .

This chapter is part of the 14 marks for Electrostatics and Current Electricity. Capacitor combinations, dielectrics and dipole potential are the usual targets.

Weightage: Electrostatic Potential, Potential Energy and Capacitance is part of the unit Electrostatics; Current Electricity, which carries 14 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Potential, potential difference and the point charge: Potential at a point: $V = W / q_{0}$ , the work done per unit positive test charge in bringing it from infinity (where V = 0) to the point without acceleration. Unit volt = J C⁻¹. Moving $q_{0}$ from B to A against the field of a point charge q: $W_{B A} = \frac{q q _{0}}{4 π ε _{0}} (\frac{1}{r _{A}} - \frac{1}{r _{B}})$ , so $V_{A} - V_{B} = \frac{q}{4 π ε _{0}} (\frac{1}{r _{A}} - \frac{1}{r _{B}})$ . Letting $r_{B} \to \infty$ gives $V = \frac{1}{4 π ε _{0}} \frac{q}{r}$ : positive for q > 0, negative for q < 0. V is a scalar and falls as $1/ r$ while E falls as $1/ r^{2}$ . For a charge q at a point of potential V, $U = q V$ , and moving it from B to A changes its energy by $Δ U = q (V_{A} - V_{B})$ .; Worked example
A point charge of $+ 4 n C$ is fixed in air. Points A and B are 10 cm and 30 cm from it. Find $V_{A}$ , $V_{B}$ and the work needed to carry $+ 2 μ C$ from B to A.
Try it first, then show the solutionSolution
$V_{A} = \frac{9 \times 1 0 ^{9} \times 4 \times 1 0 ^{- 9}}{0.10} = 360 V$
$V_{B} = \frac{36}{0.30} = 120 V$
$W = q (V_{A} - V_{B}) = 2 \times 1 0^{- 6} \times 240 = 4.8 \times 1 0^{- 4} J$
Answer: $V_{A} = 360 V$ , $V_{B} = 120 V$ , work $= 4.8 \times 1 0^{- 4} J$
Potential of a system of charges and of a dipole: Potentials add algebraically (with sign, no directions): $V = \frac{1}{4 π ε _{0}} \sum \frac{q _{i}}{r _{i}}$ . For a dipole ( $\pm q$ , separation 2l, $p = 2 q l$ ): on the axis at distance r, $V = \frac{1}{4 π ε _{0}} \frac{p}{r ^{2} - l ^{2}} \approx \frac{1}{4 π ε _{0}} \frac{p}{r ^{2}}$ ; on the equatorial line every point is equidistant from $+ q$ and $- q$ , so $V = 0$ (the equatorial plane is the locus of zero potential, even though E there is not zero); at a general point for $r ≫ 2 l$ , $V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}}$ . Dipole potential falls as $1/ r^{2}$ .; Worked example
Charges of $+ 1$ , $- 2$ , $+ 3$ and $+ 4 μ C$ sit at the corners of a square of side 20 cm. Find the potential at its centre.
Try it first, then show the solutionSolution
Each corner is half a diagonal from the centre: $r = \frac{0.2 2}{2} = 0.1414 m$ .
Net charge $= 1 - 2 + 3 + 4 = 6 μ C$ (algebraic sum).
$V = \frac{9 \times 1 0 ^{9} \times 6 \times 1 0 ^{- 6}}{0.1414} = 3.82 \times 1 0^{5} V$
Answer: $3.82 \times 1 0^{5} V$
Equipotential surfaces: An equipotential surface has the same potential at every point. Properties: (1) no work is done in moving a charge along it; (2) the field is everywhere perpendicular to it (otherwise a component of E along the surface would do work); (3) two equipotentials never intersect; (4) they are closer together where the field is stronger. Examples: concentric spheres round a point charge; planes perpendicular to the field lines in a uniform field; the equatorial plane of a dipole (V = 0). The surface of a conductor in electrostatics is an equipotential, and the field inside a conductor is zero: its free charges move until they cancel any field inside, so all the charge sits on the outer surface.; Worked example
A uniform field of $500 V m^{- 1}$ points along +x. Equipotential planes are drawn every 10 V. (a) How far apart are they? (b) How much work is done in carrying $3 μ C$ 20 cm along one of them, and 20 cm along the field?
Try it first, then show the solutionSolution
In a uniform field $V = E d$ , so spacing $d = \frac{10}{500} = 0.02 m = 2 c m$ .
Along an equipotential no work is done: W = 0.
Along the field the potential falls by $500 \times 0.2 = 100 V$ ; the field does work $q E d = 3 \times 1 0^{- 6} \times 100 = 3 \times 1 0^{- 4} J$ on the charge.
Answer: (a) 2 cm (b) zero along the plane; $3 \times 1 0^{- 4} J$ done by the field along the field
Potential energy of a system of charges: The electrostatic potential energy of a system is the work done in bringing the charges from infinity to their positions. Two charges: $U_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}}$ (positive for like charges: work had to be done to push them together; negative for unlike). Three charges: add over all pairs, $U = \frac{1}{4 π ε _{0}} (\frac{q _{1} q _{2}}{r _{12}} + \frac{q _{2} q _{3}}{r _{23}} + \frac{q _{1} q _{3}}{r _{13}})$ . Insert the signs of the charges. The work needed to separate the system completely is $- U$ .; Worked example
Charges $+ 2 μ C$ , $- 4 μ C$ and $+ 3 μ C$ are at the corners of an equilateral triangle of side 10 cm. Find the potential energy of the system.
Try it first, then show the solutionSolution
$U = \frac{k}{r} (q_{1} q_{2} + q_{2} q_{3} + q_{1} q_{3})$
$= \frac{9 \times 1 0 ^{9}}{0.1} [(2) (- 4) + (- 4) (3) + (2) (3)] \times 1 0^{- 12}$
$= 9 \times 1 0^{10} \times (- 14 \times 1 0^{- 12}) = - 1.26 J$
Answer: $U = - 1.26 J$ (1.26 J of work is needed to pull the charges apart to infinity)
Potential energy of a dipole in a uniform field: The torque $pE sin θ$ tries to align the dipole. Work done in turning it slowly from $θ_{1}$ to $θ_{2}$ is $W = \int_{θ_{1}}^{θ_{2}} pE sin θ d θ = pE (cos θ_{1} - cos θ_{2})$ . Taking U = 0 at $θ = 9 0^{\circ}$ gives $U = - pE cos θ = - p \cdot E$ . Special cases: $θ = 0^{\circ}$ : $U = - pE$ (minimum, stable); $9 0^{\circ}$ : U = 0; $18 0^{\circ}$ : $U = + pE$ (maximum, unstable). Turning a dipole from stable to unstable needs $2 pE$ .; Worked example
A dipole of moment $2 \times 1 0^{- 8} C m$ is in a uniform field of $1 0^{4} N C^{- 1}$ . Find its potential energy when aligned with the field, and the work needed to turn it through $18 0^{\circ}$ from that position.
Try it first, then show the solutionSolution
$U_{0} = - pE = - 2 \times 1 0^{- 8} \times 1 0^{4} = - 2 \times 1 0^{- 4} J$
$U_{180} = + pE = + 2 \times 1 0^{- 4} J$
$W = U_{180} - U_{0} = 4 \times 1 0^{- 4} J$
Answer: $- 2 \times 1 0^{- 4} J$ ; work $4 \times 1 0^{- 4} J$
Capacitance and the parallel-plate capacitor: A capacitor stores charge and energy: two conductors with equal and opposite charges $\pm Q$ at potential difference V. Capacitance $C = Q / V$ (unit farad = C V⁻¹; practical units µF, pF). For a parallel-plate capacitor of plate area A and separation d in vacuum: the field between plates is $E = σ / ε_{0} = \frac{Q}{ε _{0} A}$ , so $V = E d = \frac{Q d}{ε _{0} A}$ and $C = \frac{ε _{0} A}{d}$ . C depends only on geometry and the medium: increase A or reduce d to increase C. Energy stored (formulae only): $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ , and energy density $u = \frac{1}{2} ε_{0} E^{2}$ .; Worked example
A parallel-plate capacitor has plates of area $0.02 m^{2}$ separated by 1 mm of air. Find its capacitance and the energy stored when it is connected to 100 V.
Try it first, then show the solutionSolution
$C = \frac{ε _{0} A}{d} = \frac{8.85 \times 1 0 ^{- 12} \times 0.02}{1 \times 1 0 ^{- 3}} = 1.77 \times 1 0^{- 10} F = 177 pF$
$U = \frac{1}{2} C V^{2} = 0.5 \times 1.77 \times 1 0^{- 10} \times 10 0^{2} = 8.85 \times 1 0^{- 7} J$
Answer: 177 pF; $8.85 \times 1 0^{- 7} J$
Capacitors in series and parallel: Series: each capacitor carries the same charge Q (charge induced from plate to plate); potential differences add, $V = V_{1} + V_{2} + V_{3} = Q (\frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}})$ , so $\frac{1}{C _{s}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$ ; $C_{s}$ is smaller than the smallest. The smallest capacitor takes the largest share of V. Parallel: each has the same V; charges add, $Q = (C_{1} + C_{2} + C_{3}) V$ , so $C_{p} = C_{1} + C_{2} + C_{3}$ . Note the opposite pattern to resistors.; Worked example
Capacitors of 2, 3 and 6 µF are connected (a) in series and (b) in parallel. Find the equivalent capacitance in each case.
Try it first, then show the solutionSolution
Series: $\frac{1}{C _{s}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = 1$ , so $C_{s} = 1 μ F$ .
Parallel: $C_{p} = 2 + 3 + 6 = 11 μ F$ .
Answer: (a) 1 µF (b) 11 µF
Dielectrics: polarisation and the change in capacitance: Conductors have free charges; a dielectric (insulator) has only bound charges. In a field its molecules are polarised (induced or aligned dipoles), producing induced surface charges of opposite sign on its faces. These set up a field opposing the applied one, so the net field falls: $E^{'} = E / K$ . Dielectric constant $K = ε_{r} = ε / ε_{0} = C^{'} / C$ . Filling the gap gives $C^{'} = \frac{K ε _{0} A}{d}$ . Battery disconnected (Q constant): $V^{'} = V / K$ , $E^{'} = E / K$ , $U^{'} = U / K$ . Battery connected (V constant): $Q^{'} = K Q$ , E unchanged, $U^{'} = K U$ . Slab of thickness t < d: $C^{'} = \frac{ε _{0} A}{d - t + t / K}$ ; a metal slab ( $K \to \infty$ ) gives $\frac{ε _{0} A}{d - t}$ .; Worked example
A 10 pF air capacitor is charged to 50 V and disconnected. A slab of K = 5 is then slid in to fill the gap. Find the new capacitance, charge, potential difference and stored energy.
Try it first, then show the solutionSolution
$C^{'} = K C = 50 pF$ ; Q stays $C V = 10 \times 1 0^{- 12} \times 50 = 500 pC$ .
$V^{'} = \frac{Q}{C ^{'}} = \frac{500}{50} = 10 V$
$U = \frac{1}{2} C V^{2} = 0.5 \times 1 0^{- 11} \times 2500 = 1.25 \times 1 0^{- 8} J$ ; $U^{'} = \frac{1}{2} C^{'} V^{'2} = 0.5 \times 5 \times 1 0^{- 11} \times 100 = 2.5 \times 1 0^{- 9} J$ (one-fifth)
Answer: 50 pF; 500 pC; 10 V; energy falls from 12.5 nJ to 2.5 nJ

Physics in real life: Electrostatic Potential, Potential Energy and Capacitance

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.

Energy stored in a capacitor

Suppose a hospital defibrillator charges a 100 µF capacitor before delivering a shock to restart a patient's heart.

Suppose it is charged to 2000 V. How much energy is stored, and how much charge flows if it discharges completely?

Try it first, then show the solutionSolution

$U = \frac{1}{2} C V^{2} = 0.5 \times 100 \times 1 0^{- 6} \times 200 0^{2} = 200 J$
$Q = C V = 100 \times 1 0^{- 6} \times 2000 = 0.2 C$

Answer: 200 J stored; 0.2 C flows

Memory hook: Half C V squared: double the voltage, four times the shock.

Capacitance depends on plate separation

Imagine a capacitive computer keyboard: each key has a small metal plate above a fixed plate, and pressing the key narrows the gap.

Suppose a key plate of area $50 m m^{2}$ sits 0.6 mm above the fixed plate, and pressing the key reduces the gap to 0.2 mm. Find the change in capacitance.

Try it first, then show the solutionSolution

Before: $C_{1} = \frac{8.85 \times 1 0 ^{- 12} \times 5 \times 1 0 ^{- 5}}{6 \times 1 0 ^{- 4}} = 0.74 pF$
After: $C_{2} = \frac{8.85 \times 1 0 ^{- 12} \times 5 \times 1 0 ^{- 5}}{2 \times 1 0 ^{- 4}} = 2.21 pF$
Change $= 2.21 - 0.74 = 1.47 pF$ (an increase; the circuit detects it as a key press)

Answer: Capacitance rises by about 1.47 pF (three times the original value).

Memory hook: Squeeze the gap, the C goes up.

Potential difference and energy gained by an electron

Suppose an electron gun in an old cathode-ray oscilloscope accelerates electrons from rest through a potential difference of 2000 V.

What kinetic energy and speed does each electron reach? ( $m_{e} = 9.1 \times 1 0^{- 31} k g$ )

Try it first, then show the solutionSolution

KE gained $= e V = 1.6 \times 1 0^{- 19} \times 2000 = 3.2 \times 1 0^{- 16} J$ (2000 eV)
$v = \frac{2 K E}{m} = \frac{2 \times 3.2 \times 1 0 ^{- 16}}{9.1 \times 1 0 ^{- 31}} = 2.65 \times 1 0^{7} m s^{- 1}$

Answer: $3.2 \times 1 0^{- 16} J$ ; about $2.65 \times 1 0^{7} m s^{- 1}$

Memory hook: Charge times voltage is the energy you gain.

Topics in this chapter

Electrostatic Potential, Potential Energy and Capacitance formula sheet

Potential (definition): $V = \frac{W}{q _{0}}$; Work per unit positive charge from infinity. Unit volt = J C⁻¹.
Potential difference near a point charge: $V_{A} - V_{B} = \frac{q}{4 π ε _{0}} (\frac{1}{r _{A}} - \frac{1}{r _{B}})$; Derived by integrating $- E \cdot d r$ from B to A.
Potential of a point charge: $V = \frac{1}{4 π ε _{0}} \frac{q}{r}$; Scalar; keep the sign of q.
Potential of a system: $V = \frac{1}{4 π ε _{0}} \sum_{i} \frac{q _{i}}{r _{i}}$; Algebraic sum.
Dipole potential: $V_{a x i s} = \frac{1}{4 π ε _{0}} \frac{p}{r ^{2} - l ^{2}}; V_{e q} = 0; V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}} (r ≫ 2 l)$; Equatorial plane is the locus of zero potential.
Uniform field and potential: $E = \frac{V}{d}$; Between parallel plates; E points from high to low potential.
Potential energy of a charge: $U = q V, Δ U = q (V_{A} - V_{B})$; For an electron moved through V volts, energy = eV.
Potential energy of two or three charges: $U_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}}; U_{123} = \frac{1}{4 π ε _{0}} (\frac{q _{1} q _{2}}{r _{12}} + \frac{q _{2} q _{3}}{r _{23}} + \frac{q _{1} q _{3}}{r _{13}})$; Insert signs of the charges.
Dipole potential energy: $U = - p \cdot E = - pE cos θ$; Work to turn from $θ_{1}$ to $θ_{2}$ : $pE (cos θ_{1} - cos θ_{2})$ .
Capacitance: $C = \frac{Q}{V}, C = \frac{ε _{0} A}{d}$; Parallel-plate capacitor in vacuum (or air).
Series combination: $\frac{1}{C _{s}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$; Same Q on each; V divides in inverse ratio of C.
Parallel combination: $C_{p} = C_{1} + C_{2} + C_{3}$; Same V on each; Q divides in direct ratio of C.
Energy stored (formulae only): $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V, u = \frac{1}{2} ε_{0} E^{2}$; Derivation not required for 2027.
Dielectric constant: $K = ε_{r} = \frac{ε}{ε _{0}} = \frac{C ^{'}}{C}, C^{'} = \frac{K ε _{0} A}{d}$; Q fixed: V and E divide by K. V fixed: Q multiplies by K.
Partly filled capacitor: $C^{'} = \frac{ε _{0} A}{d - t + \frac{t}{K}}$; Slab of thickness t; metal slab: $\frac{ε _{0} A}{d - t}$ .

Electrostatic Potential, Potential Energy and Capacitance: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Equipotential surfaces with field lines

'Draw equipotential surfaces for a point charge, a dipole or a uniform field' in short answers.

Draw

Point charge: concentric dashed circles round +q with radial solid field lines crossing them at right angles; circles further apart as r increases.
Uniform field: parallel solid field lines with equally spaced dashed lines perpendicular to them.
Dipole: field lines from +q to −q; dashed equipotentials crossing them at 90°, with a straight dashed line along the perpendicular bisector (V = 0).

Label

Field lines (solid, with arrows)
Equipotentials (dashed)
V = 0 on the bisector
+q, −q

Watch out: Every crossing must be at 90°. Space the point-charge circles for equal steps of V, so they spread out with distance.

Capacitors in series and in parallel

Required with the derivation of equivalent capacitance; examiners note the circuit diagram is often missing.

Draw

Series: three capacitor symbols (two short parallel lines) in one line between A and B, with a cell across A and B.
Mark $+ Q$ and $- Q$ on the plates of each capacitor, and $V_{1}$ , $V_{2}$ , $V_{3}$ under them.
Parallel: three capacitors connected between the same two rails, each labelled $Q_{1}$ , $Q_{2}$ , $Q_{3}$ , with V across the rails.

Label

$C_{1}$ , $C_{2}$ , $C_{3}$
$\pm Q$ on plates
$V_{1}$ , $V_{2}$ , $V_{3}$ or V
Cell

Watch out: Write 'Q is the same' (series) or 'V is the same' (parallel) beside the diagram; that sentence carries a mark.

Dielectric slab between capacitor plates

Explaining polarisation and the fall in field, or deriving the partly filled capacitor.

Draw

Two horizontal plates, top +, bottom −, with field lines $E_{0}$ downwards.
A slab of thickness t between them; inside it show small dipoles aligned with the field.
Mark induced negative charge on the slab's top face and positive on its bottom face; show a weaker field $E_{0} / K$ inside the slab.

Label

$+ σ$ , $- σ$ (plates)
induced charges $- σ_{p}$ , $+ σ_{p}$
$E_{0}$ in air
$E_{0} / K$ in slab
d, t

Watch out: The induced charges are opposite to the nearby plate's charge; that is why the field inside the slab is weaker.

How the board asks Electrostatic Potential, Potential Energy and Capacitance

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
One-mark items on potential	Locus of zero potential for a dipole, shape of equipotentials, work done along an equipotential, potential inside a charged conductor.	Give the exact term: 'equatorial plane (perpendicular bisector of the dipole axis)', 'concentric spheres', 'zero', 'constant, equal to the surface value'.
Effect of changes on a capacitor (2 marks)	What happens to C, Q, V, E or U if d is increased, a dielectric is inserted, or the battery is disconnected first?	Start from $C = K ε_{0} A / d$ ; decide what stays fixed (Q if isolated, V if connected), then derive each change.
Derivations (3 marks)	Equivalent capacitance in series or parallel; $C = ε_{0} A / d$ ; potential of a point charge; $U = - p \cdot E$ ; capacitance with a slab.	Circuit or figure first, the key physical statement (Q same / V same), the algebra, and the final boxed result.
Capacitor networks	Find equivalent C, charge and voltage on each capacitor, and energy stored in a mixed network across a battery.	Reduce step by step, find total Q, then work back: series parts share Q, parallel parts share V.
Potential energy and work numericals	Energy of three charges at the corners of a triangle; work to move a charge between two points; work to rotate a dipole.	U adds over pairs with signs; $W = q Δ V$ ; dipole $W = pE (cos θ_{1} - cos θ_{2})$ .
Long answer (5 marks)	Derive C of a parallel-plate capacitor, then a slab result, then a numerical; or a network plus energy.	Mark each sub-part clearly, keep units in pF or µF consistent, and quote the energy formula without deriving it.

Tricks to remember Electrostatic Potential, Potential Energy and Capacitance

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

Capacitors are rebels
Capacitors combine the opposite way to resistors: series uses reciprocals, parallel simply adds.
Use it when: Avoiding the swapped-formula error in networks.
Q-same in series, V-same in parallel
'Series shares the Queue; Parallel shares the Voltage.'
Use it when: The key sentence in combination derivations.
Isolated keeps Q, connected keeps V
Unplugged capacitor: Q is locked, so V, E and U all divide by K. Plugged in: V is locked, so Q and U multiply by K.
Use it when: Dielectric 'what changes' questions.
Dipole powers drop by one
Point charge: V ∝ 1/r, E ∝ 1/r². Dipole: V ∝ 1/r², E ∝ 1/r³.
Use it when: Ratio and graph questions.
Minus p dot E
Energy is lowest when aligned: $U = - pE$ at 0°, zero at 90°, $+ pE$ at 180°. Turning half-way round costs 2pE.
Use it when: Dipole energy and work numericals.
Slab: d shrinks by t(1 − 1/K)
A slab acts as if the gap were reduced from d to $d - t + t / K$ .
Use it when: Partly filled capacitor numericals.

Electrostatic Potential, Potential Energy and Capacitance: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Equipotentials of a point charge · 1 mark · MCQBoard level

The equipotential surfaces of an isolated point charge are:
(a) planes perpendicular to the field
(b) concentric spheres centred on the charge
(c) coaxial cylinders
(d) radial lines

Try it first, then show the solutionSolution

V depends only on r, so all points at the same r have the same V: spheres.

Answer: (b)

Example 2: Work along an equipotential · 1 mark · Assertion-reasonBoard level

Assertion: No work is done in moving a charge along an equipotential surface.
Reason: The electric field is perpendicular to an equipotential surface at every point.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

$W = q Δ V = 0$ along the surface. Because E has no component along the surface, the force does no work: the reason explains the assertion.

Answer: (a)

Example 3: Points of zero potential · 3 marks · Short answerHarder board level

Charges of $+ 2 μ C$ and $- 6 μ C$ are 40 cm apart. Find the points on the line joining them where the potential is zero. [3]

Try it first, then show the solutionSolution

Zero potential needs $\frac{2}{x} = \frac{6}{d}$ , where x and d are distances from the two charges, i.e. the point is three times nearer the $+ 2 μ C$ charge.
Between the charges, x from $+ 2 μ C$ : $\frac{2}{x} = \frac{6}{0.4 - x} \Rightarrow 0.8 - 2 x = 6 x \Rightarrow x = 0.1 m$ .
Outside, beyond the $+ 2 μ C$ charge: $\frac{2}{x} = \frac{6}{0.4 + x} \Rightarrow 0.8 + 2 x = 6 x \Rightarrow x = 0.2 m$ .

Answer: 10 cm from the $+ 2 μ C$ charge (between them) and 20 cm from it on the side away from the $- 6 μ C$ charge

Example 4: Derivation: three capacitors in series · 3 marks · Short answerHarder board level

Obtain an expression for the equivalent capacitance of three capacitors $C_{1}$ , $C_{2}$ and $C_{3}$ connected in series across a source of p.d. V. Draw the circuit. [3]

Try it first, then show the solutionSolution

Circuit: $C_{1}$ , $C_{2}$ , $C_{3}$ end to end across the source. Charge Q on the first plate induces $- Q$ and $+ Q$ on the next plates, so every capacitor carries the same charge Q.
The p.d.s add: $V = V_{1} + V_{2} + V_{3} = \frac{Q}{C _{1}} + \frac{Q}{C _{2}} + \frac{Q}{C _{3}}$ .
If $C_{s}$ is the single equivalent capacitor, $V = \frac{Q}{C _{s}}$ . Dividing by Q: $\frac{1}{C _{s}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$ .

Answer: $\frac{1}{C _{s}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$

Example 5: Derivation: potential energy of a dipole · 3 marks · Short answerHarder board level

Derive an expression for the potential energy of an electric dipole of moment p placed at angle θ to a uniform field E. Hence find the work done in rotating a dipole of $2 \times 1 0^{- 8} C m$ from $0^{\circ}$ to $18 0^{\circ}$ in a field of $1 0^{4} N C^{- 1}$ . [3]

Try it first, then show the solutionSolution

Torque $τ = pE sin θ$ . Work done against it in turning through $d θ$ : $d W = pE sin θ d θ$ .
From $9 0^{\circ}$ (where U is taken as zero) to θ: $U = \int_{9 0^{\circ}}^{θ} pE sin θ d θ = - pE cos θ = - p \cdot E$ .
$W = U_{180} - U_{0} = pE - (- pE) = 2 pE = 2 \times 2 \times 1 0^{- 8} \times 1 0^{4} = 4 \times 1 0^{- 4} J$ .

Answer: $U = - pE cos θ$ ; work $= 4 \times 1 0^{- 4} J$

Example 6: Capacitor with a dielectric slab · 5 marks · Long answerHarder board level

(a) Define dielectric constant in terms of capacitance. [1]
(b) Explain, with a diagram, why a dielectric slab between the plates of a charged, isolated capacitor reduces the field between them. [2]
(c) A parallel-plate capacitor has plates of area $0.01 m^{2}$ separated by 4 mm. A slab 2 mm thick with K = 4 is placed between the plates. Calculate the capacitance with and without the slab. [2]

Try it first, then show the solutionSolution

(a) $K = C^{'} / C$ : the ratio of capacitance with the dielectric filling the space to capacitance with vacuum.
(b) The field polarises the slab's molecules; induced negative charge appears on the face near the positive plate and positive charge near the negative plate. These set up a field opposite to the applied field, so the net field inside falls to $E_{0} / K$ .
(c) Without slab: $C = \frac{8.85 \times 1 0 ^{- 12} \times 0.01}{0.004} = 2.21 \times 1 0^{- 11} F = 22.1 pF$ .
With slab: $C^{'} = \frac{ε _{0} A}{d - t + t / K} = \frac{8.85 \times 1 0 ^{- 14}}{0.004 - 0.002 + 0.0005} = \frac{8.85 \times 1 0 ^{- 14}}{0.0025} = 3.54 \times 1 0^{- 11} F = 35.4 pF$ .

Answer: (c) 22.1 pF without the slab; 35.4 pF with it

Example 7: Disconnected or connected: what changes? · 3 marks · Short answerHOTS / competency

A 10 pF air capacitor is charged to 50 V. A slab of K = 5 is inserted to fill the gap (i) after disconnecting the battery and (ii) with the battery still connected. In each case find Q, V and the stored energy. Account for the energy change in case (ii). [3]

Try it first, then show the solutionSolution

Initially $Q = 500 pC$ , $U = \frac{1}{2} \times 1 0^{- 11} \times 2500 = 12.5 n J$ . With the slab, C = 50 pF.
(i) Q fixed at 500 pC; $V = 500/50 = 10 V$ ; $U = \frac{Q ^{2}}{2 C ^{'}} = 2.5 n J$ (falls by a factor of 5).
(ii) V fixed at 50 V; $Q = 50 \times 1 0^{- 12} \times 50 = 2500 pC$ ; $U = \frac{1}{2} \times 5 \times 1 0^{- 11} \times 2500 = 62.5 n J$ . The battery supplies the extra charge $2000 pC$ at 50 V, i.e. $1 0^{- 7} J$ of work, which is more than the 50 nJ rise in stored energy; the rest goes in pulling the slab in.

Answer: (i) 500 pC, 10 V, 2.5 nJ (ii) 2500 pC, 50 V, 62.5 nJ

Example 8: Work to assemble three equal charges · 2 marks · Short answerHOTS / competency

How much work is needed to bring three charges of $+ 1 μ C$ each from infinity to the corners of an equilateral triangle of side 10 cm? [2]

Try it first, then show the solutionSolution

Three pairs, each separated by 0.1 m: $U = 3 \times \frac{k q ^{2}}{r}$ .
$U = 3 \times \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 12}}{0.1} = 0.27 J$

Answer: 0.27 J

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

A capacitive fuel gauge

Suppose a fuel tank has two vertical metal plates, 0.50 m tall and 0.10 m wide, held 5 mm apart. Fuel (dielectric constant 2) fills the gap up to the fuel level and air fills the rest. The gauge reads the capacitance. Take $ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}$ .

(i) · 1 mark
Find the capacitance when the tank is empty.
Answer
Answer: 88.5 pF
$C_{0} = \frac{ε _{0} A}{d} = \frac{8.85 \times 1 0 ^{- 12} \times 0.05}{0.005} = 8.85 \times 1 0^{- 11} F$
(ii) · 1 mark
Find the capacitance when it is full.
Answer
Answer: 177 pF
$C = K C_{0} = 2 \times 88.5 = 177 pF$
(iii) · 1 mark
Explain why the partly filled gauge behaves like two capacitors in parallel.
Answer
Answer: The fuel part and the air part each span the full gap between the same two plates.
Both parts have the same p.d. across them (same two plates), and their areas add side by side, so $C = C_{ai r} + C_{f u e l}$ .
(iv) · 2 marks
The gauge reads 115.05 pF. Find the height of the fuel.
Answer
Answer: 0.15 m
With fuel height h: $C = \frac{ε _{0} w}{d} [(0.5 - h) + 2 h] = 177 \times 1 0^{- 12} \times (0.5 + h)$ (since $\frac{ε _{0} w}{d} = 1.77 \times 1 0^{- 10} F m^{- 1}$ ). $115.05 = 177 (0.5 + h) \Rightarrow 0.5 + h = 0.65 \Rightarrow h = 0.15 m$ .

Sharing charge between two capacitors

Suppose a 4 µF capacitor is charged to 100 V, removed from the supply and then connected across an uncharged 6 µF capacitor (positive plate to one plate, negative to the other).

(i) · 1 mark
What is the total charge before connection?
Answer
Answer: 400 µC
$Q = C V = 4 \times 1 0^{- 6} \times 100 = 4 \times 1 0^{- 4} C$
(ii) · 1 mark
Find the common potential after connection.
Answer
Answer: 40 V
Charge is conserved and the capacitors are now in parallel: $V = \frac{400 μ C}{( 4 + 6 ) μ F} = 40 V$ .
(iii) · 2 marks
Find the energy before and after connection.
Answer
Answer: 0.02 J before; 0.008 J after
Before: $\frac{1}{2} \times 4 \times 1 0^{- 6} \times 10 0^{2} = 0.02 J$ . After: $\frac{1}{2} \times 10 \times 1 0^{- 6} \times 4 0^{2} = 0.008 J$ .
(iv) · 1 mark
Where has the missing energy gone?
Answer
Answer: 0.012 J is lost as heat in the connecting wires (and a little as electromagnetic radiation in the spark).
Charge is conserved but energy in the capacitors is not: current flows through the wires' resistance while the charges redistribute.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

How does the capacitance of a parallel-plate capacitor change if (a) the distance between the plates is doubled, and (b) a dielectric slab of constant K is introduced to fill the gap? (2 marks)

Step	Marks
(a) Uses $C = ε_{0} A / d$ : C is halved	1
(b) $C^{'} = K ε_{0} A / d$ : C becomes K times	1

Derive an expression for the equivalent capacitance of capacitors $C_{1}$ , $C_{2}$ , $C_{3}$ in parallel. (3 marks)

Step	Marks
Circuit diagram with the three capacitors across the same two points and the source	1
Same V across each; $Q_{1} = C_{1} V$ , $Q_{2} = C_{2} V$ , $Q_{3} = C_{3} V$ and $Q = Q_{1} + Q_{2} + Q_{3}$	1
$C_{p} V = (C_{1} + C_{2} + C_{3}) V \Rightarrow C_{p} = C_{1} + C_{2} + C_{3}$	1

Capacitors of 2 µF and 3 µF are in parallel, and this pair is in series with a 5 µF capacitor across a 100 V supply. Find the equivalent capacitance, the charge on each capacitor and the total energy stored. (5 marks)

Step	Marks
Parallel pair: 2 + 3 = 5 µF	1
With 5 µF in series: $\frac{1}{C} = \frac{1}{5} + \frac{1}{5} \Rightarrow C = 2.5 μ F$	1
Total charge $Q = 2.5 \times 100 = 250 μ C$ , which is the charge on the 5 µF capacitor	1
p.d. across the pair $= 250/5 = 50 V$ : 2 µF holds 100 µC, 3 µF holds 150 µC	1
$U = \frac{1}{2} C V^{2} = 0.5 \times 2.5 \times 1 0^{- 6} \times 10 0^{2} = 0.0125 J$	1

Where students lose marks in Electrostatic Potential, Potential Energy and Capacitance

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

Mistake: Calling the zero-potential locus 'an equipotential surface' or 'broadside position'
CISCE examiners' comments, 2024: asked for the locus of zero potential of a dipole, candidates could not distinguish a point from a locus, wrote 'equipotential surfaces', or used 'broadside position', which was not accepted.
Wrong working
The broadside position.
Right working
The equatorial plane, i.e. the perpendicular bisector of the dipole axis.
Do this: A locus is a set of points: name the plane or line.
Usually costs: 1 mark.
Mistake: Inverting the final step of the series formula
CISCE examiners' comments, 2024: in the series derivation, candidates drew no circuit, did not state that the charge stays the same, wrote $C = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$ instead of $\frac{1}{C}$ , or swapped series and parallel.
Wrong working
$C = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$
Right working
$\frac{1}{C _{s}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$ , after stating Q is the same and V adds.
Do this: Diagram, then 'Q same, V adds', then the formula with $1/ C_{s}$ on the left.
Usually costs: 1-2 marks in a 3-mark derivation.
Mistake: Saying only that capacitance 'changes'
CISCE examiners' comments, 2025: on increasing plate separation or inserting a dielectric, candidates said C 'changes' without the direction or factor, used the parallel-combination formula, or could not link C to K.
Wrong working
The capacitance will change.
Right working
$C = \frac{ε _{0} A}{d}$ : increasing d decreases C (in inverse proportion). With a dielectric, $C^{'} = K C$ : it increases K times.
Do this: Write the formula, then the conclusion with 'increases/decreases' and the factor.
Usually costs: 1 mark per part.
Mistake: Adding potentials as vectors
A frequent slip in system-of-charges numericals: candidates resolve potentials into components or ignore the sign of a negative charge.
Wrong working
At the centre of a square with +q, −q, +q, −q: $V = 4 \frac{k q}{r}$ .
Right working
V is a scalar: $V = \frac{k}{r} (q - q + q - q) = 0$ (E, a vector, must be found separately by adding components).
Do this: Add potentials with signs only; save vector addition for E and F.
Usually costs: 1-2 marks.
Mistake: Assuming the charge changes when a dielectric is inserted into an isolated capacitor
Candidates treat every case as if the battery were connected.
Wrong working
Q becomes KQ, so the energy rises.
Right working
Isolated: Q is constant, V and U fall by K. Only if the battery stays connected does Q rise to KQ.
Do this: Underline 'disconnected' or 'connected' in the question before you start.
Usually costs: 1-2 marks.

Practice questions on Electrostatic Potential, Potential Energy and Capacitance

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 SI unit of potential gradient is:
(a) $V m$
(b) $V m^{- 1}$
(c) $J C^{- 1}$
(d) $C V^{- 1}$
Hint
Potential gradient is change of potential per unit distance.
Answer
(b) $V m^{- 1}$ (the same as N C⁻¹)
Q2 · 1 mark · MCQBoard level
When capacitors are connected in series, each capacitor has the same:
(a) potential difference
(b) charge
(c) energy
(d) capacitance
Hint
Think of induced charges from plate to plate.
Answer
(b) charge
Q3 · 1 mark · MCQBoard level
A hollow metal sphere of radius R carries charge Q. The potential at its centre is:
(a) zero
(b) $\frac{k Q}{R}$
(c) $\frac{k Q}{2 R}$
(d) infinite
Hint
E inside is zero, so V does not change from the surface inwards.
Answer
(b) $\frac{k Q}{R}$
Q4 · 1 mark · Very short answerBoard level
Name the locus of points of zero potential due to an electric dipole.
Hint
Every point on it is equidistant from +q and −q.
Answer
The equatorial plane (perpendicular bisector plane of the dipole axis).
Q5 · 2 marks · Short answerBoard level
A parallel-plate air capacitor has capacitance C. What is its capacitance if (a) the plate separation is halved, and (b) the space is then filled with mica of K = 6?
Hint
$C = K ε_{0} A / d$ .
Answer
(a) 2C (b) $6 \times 2 C = 12 C$
Q6 · 2 marks · Short answerHarder board level
Capacitors of 2 µF, 3 µF and 6 µF are available. Find the largest and smallest capacitance obtainable using all three.
Hint
Largest: all parallel. Smallest: all series.
Answer
Largest: 2 + 3 + 6 = 11 µF. Smallest: $\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ , so 1 µF.
Q7 · 2 marks · Short answerHarder board level
Charges of $+ 1$ , $- 2$ , $+ 3$ and $+ 4 μ C$ are placed at the corners of a square of side 20 cm. Find the potential at the centre. Is the field at the centre necessarily zero?
Hint
Potentials add as scalars.
Answer
$V = \frac{9 \times 1 0 ^{9} \times 6 \times 1 0 ^{- 6}}{0.1414} = 3.82 \times 1 0^{5} V$ . No: the field is a vector sum and the charges are unequal, so E at the centre is not zero.
Q8 · 2 marks · Short answerHOTS / competency
Capacitors of 3 µF and 6 µF are joined in series across 90 V. Find the charge on each, the p.d. across each, and show which one stores more energy.
Hint
In series Q is the same; $U = Q^{2} /2 C$ .
Answer
$C_{s} = 2 μ F$ , $Q = 180 μ C$ on each. $V_{3} = 60 V$ , $V_{6} = 30 V$ . $U_{3} = \frac{1}{2} \times 3 \times 1 0^{- 6} \times 6 0^{2} = 5.4 m J$ , $U_{6} = 2.7 m J$ : the smaller capacitor stores more, because with Q fixed $U = Q^{2} /2 C$ .
Q9 · 3 marks · Short answerHarder board level
A short dipole has moment $2 \times 1 0^{- 9} C m$ . Find the potential at 10 cm from its centre (a) on the axis, (b) on the equatorial line, (c) on a line at $6 0^{\circ}$ to the axis.
Hint
$V = \frac{k p c o s θ}{r ^{2}}$ .
Answer
(a) $\frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 9}}{0.01} = 1800 V$ (b) 0 (c) $1800 \times cos 6 0^{\circ} = 900 V$
Q10 · 3 marks · Short answerHarder board level
Find the electrostatic potential energy of charges $+ 2 μ C$ , $- 4 μ C$ and $+ 3 μ C$ placed at the corners of an equilateral triangle of side 10 cm. What does the sign of your answer tell you?
Hint
Sum over the three pairs.
Answer
$U = \frac{9 \times 1 0 ^{9}}{0.1} (- 8 - 12 + 6) \times 1 0^{- 12} = - 1.26 J$ . Negative: the system is bound; 1.26 J must be supplied to separate the charges to infinity.
Q11 · 3 marks · Short answerHarder board level
A 177 pF parallel-plate capacitor (plates $0.02 m^{2}$ , gap 1 mm) is connected to 100 V. Find (a) the field between the plates, (b) the energy stored, (c) the energy density, and check (c) with $u = \frac{1}{2} ε_{0} E^{2}$ .
Hint
Volume of the field region is A × d.
Answer
(a) $E = V / d = 1 0^{5} V m^{- 1}$ (b) $U = \frac{1}{2} \times 1.77 \times 1 0^{- 10} \times 1 0^{4} = 8.85 \times 1 0^{- 7} J$ (c) $u = \frac{8.85 \times 1 0 ^{- 7}}{0.02 \times 1 0 ^{- 3}} = 0.0443 J m^{- 3}$ ; $\frac{1}{2} \times 8.85 \times 1 0^{- 12} \times 1 0^{10} = 0.0443 J m^{- 3}$ , which agrees.
Q12 · 5 marks · Long answerHarder board level
(a) Derive the expression for the capacitance of a parallel-plate capacitor with vacuum between the plates. [2]
(b) Write the formula for the energy stored in a capacitor in three forms. [1]
(c) Two capacitors of 2 µF and 3 µF in parallel are joined in series with a 5 µF capacitor across 100 V. Find the charge on the 3 µF capacitor. [2]
Hint
For (a), use $E = σ / ε_{0}$ between the plates.
Answer
(a) $E = \frac{Q}{ε _{0} A}$ , $V = E d = \frac{Q d}{ε _{0} A}$ , $C = Q / V = \frac{ε _{0} A}{d}$ . (b) $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ . (c) Equivalent 2.5 µF, Q = 250 µC, p.d. across the pair 50 V, so the 3 µF capacitor holds $3 \times 50 = 150 μ C$ .
Q13 · 2 marks · Short answerHarder board level
An electron starts from rest and is accelerated through 2000 V. Find its kinetic energy in joules and electron-volts, and its final speed. ( $m_{e} = 9.1 \times 1 0^{- 31} k g$ )
Hint
Energy gained = eV.
Answer
KE $= 2000 e V = 3.2 \times 1 0^{- 16} J$ ; $v = \frac{2 \times 3.2 \times 1 0 ^{- 16}}{9.1 \times 1 0 ^{- 31}} = 2.65 \times 1 0^{7} m s^{- 1}$

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 test charge in bringing it from infinity to that point without acceleration: $V = W / q_{0}$ .
2.Potential due to a point charge?Show answer
$V = \frac{1}{4 π ε _{0}} \frac{q}{r}$ , positive for positive q.
3.Give two properties of equipotential surfaces.Show answer
No work is done in moving a charge along one; the field is perpendicular to it at every point (also: they never intersect).
4.Potential energy of a dipole in a uniform field?Show answer
$U = - p \cdot E = - pE cos θ$ .
5.Capacitance of a parallel-plate capacitor?Show answer
$C = ε_{0} A / d$ in vacuum; $K ε_{0} A / d$ with a dielectric filling the gap.
6.Energy stored in a capacitor?Show answer
$\frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ (formulae only).
7.What is polarisation of a dielectric?Show answer
Alignment or induction of molecular dipoles by an applied field, giving induced surface charges that reduce the field inside.
8.Capacitance with a slab of thickness t and constant K?Show answer
$C^{'} = \frac{ε _{0} A}{d - t + t / K}$ .

Best order to study it

Define potential and derive the potential difference near a point charge; practise $W = q Δ V$ .
Learn potential of a system and of a dipole (axis, equator, general point).
Study equipotential surfaces and conductors (E = 0 inside, charge on the surface).
Learn potential energy of two and three charges, then derive $U = - p \cdot E$ .
Derive $C = ε_{0} A / d$ and the series and parallel formulas; memorise the energy formulas.
Finish with dielectrics: polarisation, the connected and disconnected cases, and the partly filled capacitor.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you derive $V_{A} - V_{B}$ for a point charge?
Can you write the dipole potential on the axis, on the equator and at angle θ?
Can you state three properties of equipotential surfaces and sketch them for a point charge and a dipole?
Can you find U for three charges with signs?
Can you derive $U = - pE cos θ$ and find the work to turn a dipole?
Can you derive C for a parallel-plate capacitor and the series and parallel formulas with a diagram?
Can you say what happens to Q, V, E and U when a dielectric is inserted, battery connected or disconnected?
Can you use $C^{'} = \frac{ε _{0} A}{d - t + t / K}$ ?

Electrostatic Potential, Potential Energy and Capacitance: frequently asked questions

Do I need to derive the energy stored in a capacitor for ISC 2027?

No. The revised 2027 syllabus says energy stored in a capacitor is 'No derivation, formulae only'. Learn U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV and the energy density \frac{1}{2}\varepsilon_0E^2, and use them in numericals.

Which derivations are required in this chapter?

Potential difference and potential due to a point charge, the potential energy of a dipole U = -\vec{p} \cdot \vec{E}, the capacitance of a parallel-plate capacitor and the equivalent capacitance in series and parallel. The partly filled capacitor formula is also listed.

Why is the potential zero on the equatorial line of a dipole while the field is not?

Potential is a scalar: every equatorial point is equidistant from +q and −q, so their potentials cancel. Field is a vector: the two field vectors there add to give a field opposite to \vec{p}.

Is the potential inside a charged conductor zero?

No. The field inside is zero, so the potential is constant throughout and equal to its value at the surface.

How much is this chapter worth?

It shares the 14 marks of Electrostatics and Current Electricity with Chapters 1 and 3. Capacitor questions (effect of d and K, combinations, energy) appear in most papers.

What is the difference between a conductor and a dielectric in a field?

A conductor has free charges that move until the field inside becomes zero. A dielectric has only bound charges; it polarises and reduces the field by a factor K but does not cancel it.

Based on: CISCE ISC 2027 Physics syllabus (revised, uploaded February 2026); CISCE ISC 2027 Physics specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Physics, 2025; CISCE Analysis of Pupil Performance, ISC Physics, 2024