PhysicsCBSE Class 12 · Chapter 2

Electrostatic Potential and Capacitance Class 12: Notes, Derivations and Numericals

Q: Is the derivation of the energy stored in a capacitor in the 2026-27 syllabus?

No. The syllabus says 'energy stored in a capacitor (no derivation, formulae only)'. Learn U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV and use them in numericals.

Q: Which derivations should I prepare from this chapter?

Potential of a point charge, potential of a dipole, E = -dV/dr, potential energy of a dipole in a uniform field, the parallel plate capacitor (with and without a dielectric, and with a slab of thickness t), and the series and parallel formulas. In the board papers in our question bank, the slab derivation was the most frequent 5-mark item from this chapter.

Q: Can the potential be zero where the field is not zero?

Yes. On the equatorial plane of a dipole, V = 0 but E is not zero. The reverse also happens: inside a charged conducting shell, E = 0 but V equals the surface value.

Q: Why does a dielectric increase capacitance?

Polarisation charges on the dielectric's faces create a field that opposes the field of the plates. For the same charge the field and hence the potential difference fall, so C = Q/V rises (by the factor K if the dielectric fills the gap).

Q: Why is energy lost when a charged capacitor is connected to an uncharged one?

Charge flows until both are at a common potential. The total charge is conserved, but \frac{Q^2}{2C} falls because the total capacitance has increased. The difference is lost as heat in the wires and as electromagnetic radiation.

Q: What is the difference between free and bound charges?

Free charges (electrons in a conductor) can move through the material. Bound charges (in an insulator) can only shift slightly within their own atoms or molecules, which is what happens when a dielectric is polarised.

Updated 14 October 2026 · Fact-checked

Electrostatic Potential and Capacitance is Chapter 2 of CBSE Class 12 Physics. It covers electric potential of a point charge, a dipole and a system of charges, equipotential surfaces, potential energy, conductors, dielectrics, capacitors in series and parallel, the parallel plate capacitor with and without a dielectric, and the energy stored (formulae only).

What Electrostatic Potential and Capacitance is about

Chapter 1 described electrostatics with forces and fields, which are vectors. This chapter describes the same situations with potential and potential energy, which are scalars and so much easier to add. Because the electrostatic force is conservative, the work done in moving a charge depends only on where it starts and ends, so each point can be given a potential $V$ : the work done per unit positive charge in bringing it from infinity to that point.

You find the potential of a point charge, of a dipole and of a group of charges, draw equipotential surfaces and link them to the field through $E = - \frac{d V}{d r}$ . You then calculate the potential energy of a system of charges and of a dipole in an external field. The second half of the chapter is about conductors (where the field inside is zero and the whole conductor is at one potential), dielectrics (insulators that polarise and weaken the field), and capacitors: two conductors that store charge and energy.

The syllabus asks for the capacitance of a parallel plate capacitor with and without a dielectric, series and parallel combinations, and the energy stored in a capacitor as formulae only, with no derivation. In the eight CBSE board papers (2023 to 2026) in our question bank, this chapter appears in all eight, often as a 5-mark question on a capacitor with a dielectric slab or a combination, and as a 4-mark case study.

Weightage: This chapter is part of Units I and II, Electrostatics; Current Electricity, which carry 16 of the 70 theory marks together with Electric Charges and Fields and Current Electricity.Source: CBSE Physics syllabus 2026-27, Units I and II Electrostatics; Current Electricity

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6.6 marks per paper. See the questions

Key concepts

Electrostatic potential and potential difference

NCERT section 2.2

Because the electrostatic force is conservative, the work done by an external agent in moving a charge slowly (without acceleration) from R to P is stored as potential energy, and it does not depend on the path.

The potential at a point is the work done by an external force in bringing a unit positive charge, without acceleration, from infinity to that point: $V = \frac{W}{q}$ . Its SI unit is the volt: $1 V = 1 J C^{- 1}$ . Only potential differences are physically meaningful; we choose $V = 0$ at infinity for convenience.

The work done in moving a charge q from A to B is $W_{A B} = q (V_{B} - V_{A})$ . A positive charge left free moves from high to low potential; an electron moves from low to high potential.

Worked example

Point A is at a potential of 10 V and point B at 50 V. How much work must an external agent do to move a charge of $+ 2 μ C$ slowly from A to B? What if the charge were an electron? [2 marks]

Try it first, then show the solutionSolution

$W = q (V_{B} - V_{A}) = 2 \times 1 0^{- 6} \times (50 - 10)$
$W = 8 \times 1 0^{- 5} J$ (positive: work must be done against the field)
For an electron: $W = (- 1.6 \times 1 0^{- 19}) \times 40 = - 6.4 \times 1 0^{- 18} J$
Negative work means the field itself pulls the electron towards higher potential.

Answer: $8 \times 1 0^{- 5} J$ ; for an electron, $- 6.4 \times 1 0^{- 18} J$

Potential due to a point charge

NCERT section 2.3

Derivation. Bring a unit positive charge from infinity to a point P at distance r from a charge Q. At a distance r' the force on it is $\frac{Q}{4 π ε _{0} r ^{'2}}$ outward, so the external force is equal and inward. Work done in a small inward step $d r^{'}$ is $d W = - \frac{Q}{4 π ε _{0} r ^{'2}} d r^{'}$ (the step $d r^{'}$ is negative as r' decreases). Integrating from $\infty$ to r:

$V (r) = - \int_{\infty}^{r} \frac{Q}{4 π ε _{0} r ^{'2}} d r^{'} = \frac{Q}{4 π ε _{0}} [\frac{1}{r ^{'}}]_{\infty}^{r} = \frac{1}{4 π ε _{0}} \frac{Q}{r}$

This holds for Q of either sign: the potential is positive near a positive charge and negative near a negative charge. V falls as $\frac{1}{r}$ , while E falls as $\frac{1}{r ^{2}}$ . Outside a charged spherical shell, V is the same as if all the charge were at the centre; inside the shell, V stays constant at its surface value.

Worked example

A point charge of $+ 5 n C$ is in air. Find (a) the potential and (b) the field at 15 cm from it. (c) How do V and E change if the distance is doubled? [3 marks]

Try it first, then show the solutionSolution

(a) $V = \frac{k Q}{r} = \frac{9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 9}}{0.15} = \frac{45}{0.15} = 300 V$
(b) $E = \frac{k Q}{r ^{2}} = \frac{45}{0.0225} = 2000 N C^{- 1}$ , radially outward
(c) V depends on $\frac{1}{r}$ , so it halves to 150 V; E depends on $\frac{1}{r ^{2}}$ , so it becomes one quarter, $500 N C^{- 1}$

Answer: (a) 300 V (b) $2000 N C^{- 1}$ (c) V halves to 150 V; E falls to $500 N C^{- 1}$

Potential due to a dipole and a system of charges

NCERT section 2.4 and 2.5

Potentials are scalars, so for several charges you simply add: $V = \frac{1}{4 π ε _{0}} \sum_{i} \frac{q _{i}}{r _{i}}$ , keeping the sign of each charge.

Dipole. For charges $- q$ and $+ q$ at distance 2a apart, at a point P at distance r from the centre making angle θ with $p$ : $V = \frac{q}{4 π ε _{0}} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ , where $r_{1} \approx r - a cos θ$ and $r_{2} \approx r + a cos θ$ for $r ≫ a$ . Then $\frac{1}{r _{1}} - \frac{1}{r _{2}} \approx \frac{2 a c o s θ}{r ^{2}}$ , so

$V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}} = \frac{1}{4 π ε _{0}} \frac{p \cdot r ^}{r ^{2}}$

On the axis ( $θ = 0$ ): $V = \frac{k p}{r ^{2}}$ . On the equatorial plane ( $θ = 9 0^{\circ}$ ): $V = 0$ , although the field there is not zero. The dipole potential falls as $\frac{1}{r ^{2}}$ and depends on direction, unlike that of a single charge.

Worked example

Charges of $+ 6 n C$ and $- 2 n C$ are placed at $x = 0$ and $x = 12 c m$ on the x-axis. Find the points on the x-axis where the potential is zero. [3 marks]

Try it first, then show the solutionSolution

Between the charges, at distance x cm from the origin: $\frac{6}{x} = \frac{2}{12 - x} \Rightarrow 72 - 6 x = 2 x \Rightarrow x = 9 c m$
Beyond the $- 2 n C$ charge, at x > 12 cm: $\frac{6}{x} = \frac{2}{x - 12} \Rightarrow 6 x - 72 = 2 x \Rightarrow x = 18 c m$
To the left of the origin the larger positive charge is always nearer, so V cannot be zero there.

Answer: At $x = 9 c m$ and at $x = 18 c m$

Equipotential surfaces and the relation between field and potential

NCERT section 2.6

An equipotential surface has the same potential at every point. Properties:

No work is done in moving a charge along it (since $Δ V = 0$ ).
The field is always perpendicular to it, pointing towards lower potential.
Two equipotential surfaces never intersect (a point cannot have two potentials).
Where they are closer together, the field is stronger.

Shapes: concentric spheres for a point charge; parallel planes perpendicular to the field for a uniform field.

Field and potential. Moving a unit charge a small distance δl from one equipotential (V) to the next (V + δV) against the field takes work $E δ l = - δ V$ . So $E = - \frac{d V}{d l}$ : the field is the negative of the potential gradient. Its direction is the direction in which the potential decreases fastest, and its magnitude is the change in potential per unit distance along the normal. Hence $1 V m^{- 1} = 1 N C^{- 1}$ .

Worked example

Two parallel equipotential surfaces at 30 V and 20 V are 5 mm apart in a uniform field. (a) Find the field strength and its direction. (b) How much work is done by the field when a $+ 3 μ C$ charge moves from the 30 V surface to the 20 V surface? [2 marks]

Try it first, then show the solutionSolution

(a) $E = \frac{∣Δ V ∣}{d} = \frac{10}{5 \times 1 0 ^{- 3}} = 2000 V m^{- 1}$ , perpendicular to the surfaces from the 30 V surface towards the 20 V surface
(b) Work by the field $= q (V_{A} - V_{B}) = 3 \times 1 0^{- 6} \times (30 - 20)$
$= 3 \times 1 0^{- 5} J$

Answer: (a) $2000 V m^{- 1}$ from the 30 V to the 20 V surface (b) $3 \times 1 0^{- 5} J$

Potential energy of a system of charges

NCERT section 2.7

The potential energy of a system of point charges is the work done in assembling them by bringing them one by one from infinity.

Bring $q_{1}$ first: no work (no field yet). Bring $q_{2}$ to distance $r_{12}$ : work $= q_{2} \times \frac{k q _{1}}{r _{12}}$ . So for two charges

$U = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}}$

For three charges, add one term for each pair: $U = k (\frac{q _{1} q _{2}}{r _{12}} + \frac{q _{1} q _{3}}{r _{13}} + \frac{q _{2} q _{3}}{r _{23}})$ . U is positive for like charges (work had to be done to push them together) and negative for unlike charges. The result does not depend on the order or the path in which the charges are assembled.

Worked example

Charges of $+ 1 μ C$ , $+ 1 μ C$ and $- 2 μ C$ are placed at the corners of an equilateral triangle of side 10 cm. Find the potential energy of the system. [2 marks]

Try it first, then show the solutionSolution

Pairs: $(+ 1) (+ 1) = + 1$ , $(+ 1) (- 2) = - 2$ , $(+ 1) (- 2) = - 2$ , all in $(μ C)^{2}$ at the same separation 0.1 m
Sum of products $= 1 - 2 - 2 = - 3 \times 1 0^{- 12} C^{2}$
$U = \frac{9 \times 1 0 ^{9} \times ( - 3 \times 1 0 ^{- 12} )}{0.1} = - 0.27 J$

Answer: $U = - 0.27 J$ (negative: the system is bound)

Potential energy in an external field, including a dipole

NCERT section 2.8

If an external field (from charges not part of our system) gives potential $V (r)$ , then:

a single charge q at $r$ has potential energy $q V (r)$ ;
two charges have $U = q_{1} V (r_{1}) + q_{2} V (r_{2}) + \frac{q _{1} q _{2}}{4 π ε _{0} r _{12}}$ .

Dipole in a uniform field. The torque $pE sin θ$ tends to align the dipole. Rotating it slowly from $θ_{0}$ to $θ_{1}$ needs work $W = \int_{θ_{0}}^{θ_{1}} pE sin θ d θ = pE (cos θ_{0} - cos θ_{1})$ . Taking $U = 0$ at $θ = 9 0^{\circ}$ :

$U (θ) = - pE cos θ = - p \cdot E$

U is minimum ( $- pE$ ) at $θ = 0^{\circ}$ : stable equilibrium. It is maximum ( $+ pE$ ) at $θ = 18 0^{\circ}$ : unstable equilibrium. The work to turn a dipole from parallel to antiparallel is $2 pE$ .

Worked example

A dipole of moment $2 \times 1 0^{- 8} C m$ is in a uniform field of $5 \times 1 0^{4} N C^{- 1}$ . Find (a) its potential energy when aligned with the field, (b) the work needed to turn it from this position to the antiparallel position. [2 marks]

Try it first, then show the solutionSolution

$pE = 2 \times 1 0^{- 8} \times 5 \times 1 0^{4} = 1 \times 1 0^{- 3} J$
(a) $U = - pE cos 0^{\circ} = - 1 \times 1 0^{- 3} J$
(b) $W = pE (cos 0^{\circ} - cos 18 0^{\circ}) = pE (1 + 1) = 2 \times 1 0^{- 3} J$

Answer: (a) $- 1 \times 1 0^{- 3} J$ (b) $2 \times 1 0^{- 3} J$

Electrostatics of conductors

NCERT section 2.9

A conductor has free electrons. In electrostatic equilibrium (charges at rest) these results hold:

1. The field inside is zero. Otherwise free electrons would move.
2. Just outside, the field is normal to the surface; a tangential part would move charges along the surface.
3. There is no net charge in the interior; any excess charge sits on the surface.
4. The potential is constant throughout the volume and equal to that on the surface.
5. **Just outside, $E = \frac{σ}{ε _{0}} \overset{n}{^}$ **, where σ is the local surface charge density.
6. Electrostatic shielding: the field inside a cavity of a conductor (with no charge in the cavity) is zero, whatever the charges and fields outside. Sensitive instruments are protected this way.

The free charges in a conductor move; the bound charges in an insulator can only shift slightly within their atoms or molecules.

Worked example

A conducting sphere of radius 10 cm carries a charge of $2 \times 1 0^{- 8} C$ . Find the electric field and the potential (a) at its centre, (b) just outside its surface, (c) at 30 cm from its centre. [3 marks]

Try it first, then show the solutionSolution

$k Q = 9 \times 1 0^{9} \times 2 \times 1 0^{- 8} = 180 V m$
(a) Inside a conductor E = 0; V equals the surface value $\frac{k Q}{R} = \frac{180}{0.1} = 1800 V$
(b) Just outside: $E = \frac{k Q}{R ^{2}} = \frac{180}{0.01} = 1.8 \times 1 0^{4} N C^{- 1}$ , $V = 1800 V$
(c) At 0.3 m: $E = \frac{180}{0.09} = 2000 N C^{- 1}$ , $V = \frac{180}{0.3} = 600 V$

Answer: (a) E = 0, V = 1800 V (b) $1.8 \times 1 0^{4} N C^{- 1}$ , 1800 V (c) $2000 N C^{- 1}$ , 600 V

Dielectrics and polarisation

NCERT section 2.10

A dielectric is an insulator. In an external field its molecules become dipoles:

In non-polar molecules (centres of positive and negative charge coincide), the field pulls the charges apart and induces a dipole moment.
In polar molecules (such as water) there are permanent dipole moments pointing randomly; the field aligns them partly.

Either way, the dielectric develops a net dipole moment per unit volume called the polarisation $P$ ; for linear isotropic dielectrics $P = χ_{e} E$ , where $χ_{e}$ is the electric susceptibility. The polarised slab behaves as if it has surface charges $\pm σ_{p}$ on its two faces; these bound charges produce a field opposing the external one. So the field inside the dielectric is reduced, by the factor K (the dielectric constant) for a slab filling the gap of a capacitor: $E = \frac{E _{0}}{K}$ .

Worked example

A uniform field of $3 \times 1 0^{4} N C^{- 1}$ exists between the plates of a charged, isolated capacitor. A slab of dielectric constant 3 is slid in to fill the gap. What is the field inside the slab, and why is it smaller? [2 marks]

Try it first, then show the solutionSolution

$E = \frac{E _{0}}{K} = \frac{3 \times 1 0 ^{4}}{3} = 1 \times 1 0^{4} N C^{- 1}$
The field polarises the slab. Bound charges appear on its faces, negative facing the positive plate and positive facing the negative plate, and their field opposes $E_{0}$ .

Answer: $1 \times 1 0^{4} N C^{- 1}$ ; the polarisation charges on the slab faces produce an opposing field.

Capacitance and the parallel plate capacitor (with and without a dielectric)

NCERT section 2.11 to 2.13

A capacitor is two conductors carrying charges $+ Q$ and $- Q$ with a potential difference V. Q is proportional to V, and the constant is the capacitance: $C = \frac{Q}{V}$ , unit farad ( $1 F = 1 C V^{- 1}$ ). C depends only on the shapes, sizes, separation and the medium, not on Q or V. The farad is a very large unit; practical values are µF, nF and pF.

Parallel plate capacitor (air). Plates of area A, separation d (small compared with the plate size), surface densities $\pm σ = \pm \frac{Q}{A}$ . Between the plates the two sheet fields add: $E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}}$ ; outside they cancel. Then $V = E d = \frac{Q d}{ε _{0} A}$ , so

$C_{0} = \frac{ε _{0} A}{d}$

Dielectric filling the gap. The field becomes $\frac{E _{0}}{K}$ , so V falls to $\frac{V _{0}}{K}$ for the same Q, and $C = \frac{K ε _{0} A}{d} = K C_{0}$ . This is the definition of the dielectric constant: $K = \frac{C}{C _{0}}$ .

Slab of thickness t < d. In the air part (thickness $d - t$ ) the field is $E_{0}$ ; in the slab it is $\frac{E _{0}}{K}$ . So $V = E_{0} (d - t) + \frac{E _{0}}{K} t$ and

$C = \frac{ε _{0} A}{d - t + \frac{t}{K}}$

For a metal slab ( $K \to \infty$ ), $C = \frac{ε _{0} A}{d - t}$ .

Worked example

A parallel plate capacitor has square plates of side 10 cm and a separation of 1 mm in air. (a) Find its capacitance. (b) Find the charge on each plate when it is connected to a 100 V supply. (c) What is the capacitance if a dielectric of K = 5 fills the gap? ( $ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}$ ) [3 marks]

Try it first, then show the solutionSolution

$A = (0.1)^{2} = 0.01 m^{2}$ , $d = 1 \times 1 0^{- 3} m$
(a) $C_{0} = \frac{ε _{0} A}{d} = \frac{8.85 \times 1 0 ^{- 12} \times 0.01}{1 0 ^{- 3}} = 8.85 \times 1 0^{- 11} F = 88.5 pF$
(b) $Q = C_{0} V = 8.85 \times 1 0^{- 11} \times 100 = 8.85 \times 1 0^{- 9} C$
(c) $C = K C_{0} = 5 \times 88.5 = 442.5 pF$

Answer: (a) 88.5 pF (b) $8.85 \times 1 0^{- 9} C$ (c) about 443 pF

Combination of capacitors

NCERT section 2.14

Series (end to end): each capacitor carries the same charge Q (charge on the inner plates is induced), and the voltages add: $V = V_{1} + V_{2} + \dots = \frac{Q}{C _{1}} + \frac{Q}{C _{2}} + \dots$ . Dividing by Q:

$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \dots$

The series combination is smaller than the smallest capacitor, and the largest share of the voltage falls on the smallest capacitor.

Parallel (across the same two points): each capacitor has the same V, and the charges add: $Q = C_{1} V + C_{2} V + \dots$ . So

$C = C_{1} + C_{2} + \dots$

Note that these rules are the reverse of those for resistors.

Worked example

A $4 μ F$ capacitor is connected in series with a parallel pair of two $2 μ F$ capacitors, and the network is joined to a 10 V battery. Find (a) the equivalent capacitance, (b) the charge drawn from the battery, (c) the voltage across and charge on each capacitor. [3 marks]

Try it first, then show the solutionSolution

Parallel pair: $2 + 2 = 4 μ F$ . In series with $4 μ F$ : $\frac{1}{C} = \frac{1}{4} + \frac{1}{4} \Rightarrow C = 2 μ F$
(b) $Q = C V = 2 μ F \times 10 V = 20 μ C$
(c) The $4 μ F$ capacitor carries $20 μ C$ , so $V = \frac{20}{4} = 5 V$
The parallel pair also has 5 V across it; each $2 μ F$ capacitor carries $2 \times 5 = 10 μ C$ (total $20 μ C$ , as it must be)

Answer: (a) $2 μ F$ (b) $20 μ C$ (c) $4 μ F$ : 5 V, $20 μ C$ ; each $2 μ F$ : 5 V, $10 μ C$

Energy stored in a capacitor (formulae only)

NCERT section 2.15

Charging a capacitor means moving charge from one plate to the other against the growing potential difference; this work is stored as electrostatic energy. The syllabus asks only for the formulae (no derivation):

$U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$

The energy can be thought of as stored in the field between the plates, with energy density (energy per unit volume) $u = \frac{1}{2} ε_{0} E^{2}$ .

Useful results: when an isolated charged capacitor (Q fixed) gets a dielectric, U falls to $\frac{U}{K}$ ; when the battery stays connected (V fixed), U rises to $K U$ . When a charged capacitor is joined to an uncharged one, charge is shared, the common potential is $\frac{C _{1} V _{1}}{C _{1} + C _{2}}$ , and some energy is always lost as heat and radiation in the connecting wires.

Worked example

A $20 μ F$ capacitor is charged to 200 V. Find (a) the charge and (b) the energy stored. (c) If the charge is doubled, by what factor does the energy change? [2 marks]

Try it first, then show the solutionSolution

(a) $Q = C V = 20 \times 1 0^{- 6} \times 200 = 4 \times 1 0^{- 3} C$
(b) $U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 20 \times 1 0^{- 6} \times (200)^{2} = 0.4 J$
(c) $U = \frac{Q ^{2}}{2 C}$ , so doubling Q multiplies U by 4

Answer: (a) $4 \times 1 0^{- 3} C$ (b) 0.4 J (c) four times

Physics in real life: Electrostatic Potential 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 camera flash unit charges a capacitor and then dumps its energy through the flash tube in a very short time.

Suppose the capacitor is $200 μ F$ and is charged to 300 V. How much energy does it store? If all of it is released in 1 ms, what is the average power?

Try it first, then show the solutionSolution

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 200 \times 1 0^{- 6} \times (300)^{2}$
$U = 100 \times 1 0^{- 6} \times 9 \times 1 0^{4} = 9 J$
Average power $= \frac{U}{t} = \frac{9}{1 \times 1 0 ^{- 3}} = 9000 W$

Answer: 9 J, released at an average power of 9000 W (9 kW)

Memory hook: A capacitor is a small energy bucket that can be emptied in a flash.

Parallel plate capacitance

Imagine a keyboard key built as a small capacitor: pressing the key pushes one plate closer to a fixed plate, and the circuit senses the change in capacitance.

Suppose each plate has area $5 \times 1 0^{- 5} m^{2}$ and the gap falls from 0.6 mm to 0.15 mm when the key is pressed (air gap). Find the capacitance before and after. ( $ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}$ )

Try it first, then show the solutionSolution

Before: $C = \frac{8.85 \times 1 0 ^{- 12} \times 5 \times 1 0 ^{- 5}}{0.6 \times 1 0 ^{- 3}} = 7.4 \times 1 0^{- 13} F \approx 0.74 pF$
After: the gap is a quarter, so C is four times: $\approx 2.95 pF$
The circuit only needs to detect that C has risen fourfold.

Answer: About 0.74 pF before and 2.95 pF after pressing.

Memory hook: Press the key, shrink the gap, grow the capacitance.

Conductors: charge sharing and sharp points

Suppose a metal sphere of radius 6 cm is joined by a long thin wire to a small metal sphere of radius 2 cm far away, and a total charge of 8 nC is shared between them.

Find the charge on each sphere, their common potential and the ratio of their surface charge densities. What does this suggest about charge on sharp, pointed parts of a conductor?

Try it first, then show the solutionSolution

Joined conductors reach the same potential: $\frac{k q _{1}}{r _{1}} = \frac{k q _{2}}{r _{2}}$ , so $q \propto r$ : $q_{1} : q_{2} = 2 : 6$
$q_{1} = 2 n C$ (small sphere), $q_{2} = 6 n C$ ; common potential $= \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 9}}{0.02} = 900 V$
$\frac{σ _{1}}{σ _{2}} = \frac{q _{1} / r _{1}^{2}}{q _{2} / r _{2}^{2}} = \frac{2/4}{6/36} = 3 = \frac{r _{2}}{r _{1}}$
The smaller sphere has the larger surface density. A sharp point is like a sphere of tiny radius, so charge density and the field there are very large.

Answer: 2 nC and 6 nC, both at 900 V; the small sphere's surface density is 3 times larger, so charge crowds at sharp points.

Memory hook: Smaller curve, crowded charge: that is why lightning conductors are pointed.

Dielectric and capacitance

Suppose you build a capacitor from two foil sheets separated by air, and then slide a thin plastic sheet of dielectric constant 2.5 into the gap while it is still connected to a 9 V battery.

If the air capacitor was 40 pF, find the charge on it before and after the plastic fills the gap, and the extra charge supplied by the battery.

Try it first, then show the solutionSolution

Before: $Q_{0} = C_{0} V = 40 \times 1 0^{- 12} \times 9 = 3.6 \times 1 0^{- 10} C$
After: $C = K C_{0} = 100 pF$ , V is still 9 V, so $Q = 9 \times 1 0^{- 10} C$
Extra charge from the battery $= 9 \times 1 0^{- 10} - 3.6 \times 1 0^{- 10} = 5.4 \times 1 0^{- 10} C$

Answer: $3.6 \times 1 0^{- 10} C$ before, $9 \times 1 0^{- 10} C$ after; the battery supplies $5.4 \times 1 0^{- 10} C$ more.

Memory hook: Battery on, voltage fixed: a dielectric lets the plates hold K times more charge.

Topics in this chapter

Electrostatic Potential and Capacitance formula sheet

Potential and work: $V = \frac{W}{q}, W_{A \to B} = q (V_{B} - V_{A})$; 1 V = 1 J C⁻¹.
Potential of a point charge: $V = \frac{1}{4 π ε _{0}} \frac{q}{r}$; Keep the sign of q.
Potential of a system of charges: $V = \frac{1}{4 π ε _{0}} \sum_{i} \frac{q _{i}}{r _{i}}$; Scalar sum, no directions.
Potential of a dipole (r ≫ a): $V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}}$; Axis: $k p / r^{2}$ ; equatorial plane: 0.
Field from potential: $E = - \frac{d V}{d r}, E = \frac{∣Δ V ∣}{d} (uniform field)$; E points towards decreasing V.
Potential energy of two charges: $U = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}}$; Three charges: sum over the three pairs.
Two charges in an external field: $U = q_{1} V (r_{1}) + q_{2} V (r_{2}) + \frac{q _{1} q _{2}}{4 π ε _{0} r _{12}}$
Dipole in a uniform field: $U = - p \cdot E = - pE cos θ, W_{θ_{0} \to θ_{1}} = pE (cos θ_{0} - cos θ_{1})$
Field just outside a conductor: $E = \frac{σ}{ε _{0}}$; Normal to the surface.
Capacitance: $C = \frac{Q}{V}$; Unit farad; 1 µF = 10⁻⁶ F, 1 pF = 10⁻¹² F.
Parallel plate capacitor: $C_{0} = \frac{ε _{0} A}{d}, C = \frac{K ε _{0} A}{d} = K C_{0}$
Partly filled with a slab of thickness t: $C = \frac{ε _{0} A}{d - t + \frac{t}{K}}$; Metal slab: $C = \frac{ε _{0} A}{d - t}$ .
Series combination: $\frac{1}{C _{s}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}} + \dots$; Same Q on each.
Parallel combination: $C_{p} = C_{1} + C_{2} + C_{3} + \dots$; Same V across each.
Energy stored (formulae only): $U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$
Energy density: $u = \frac{1}{2} ε_{0} E^{2}$
Common potential when two capacitors share charge: $V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$

Electrostatic Potential and Capacitance: diagrams to practise

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

Equipotential surfaces

"Draw equipotential surfaces for…" a point charge, a uniform field or a dipole (1-2 marks).

Draw

Point charge: concentric circles round the charge, closer together near it; add a few radial field lines crossing them at right angles.
Uniform field: equally spaced parallel field lines with arrows; draw equipotentials as equally spaced straight lines perpendicular to them.
Dipole: curved equipotentials closed round each charge, and the straight perpendicular bisector line as the V = 0 surface.

Label

Equipotential surfaces (dashed)
Field lines (solid, with arrows)
Higher and lower V
V = 0 plane (dipole)

Watch out: Field lines must cross equipotentials at 90°. Use dashed lines for equipotentials so the examiner can tell them apart.

Parallel plate capacitor with a dielectric slab

Slab derivation (3-5 marks) and polarisation questions.

Draw

Draw two long parallel plates, the top marked + (charge +Q) and the bottom − (charge −Q), separation d.
Draw a slab of thickness t between them, leaving air gaps.
Show the induced charges on the slab: − on the face nearer the + plate and + on the face nearer the − plate.
Draw field arrows: longer arrows ( $E_{0}$ ) in the air gaps, shorter arrows ( $E_{0} / K$ ) in the slab.

Label

+Q and −Q plates
Area A
d
t
K
$E_{0}$ in air
$E_{0} / K$ in slab
Induced charges ±σ_p

Watch out: The figure earns a mark; label t and d clearly so the examiner can follow $V = E_{0} (d - t) + E_{0} t / K$ .

How the board asks Electrostatic Potential and Capacitance

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

Question type	How it is asked	Method
Potential and work numericals	Potential at a point due to several charges (square, triangle, hexagon), points of zero potential, work to bring a charge from infinity or between two points.	Add potentials as scalars with signs; work = q × (final − initial potential). For zero potential on a line, check both the region between and outside the charges.
Field-potential relation	V given as a function of x; or a V-x graph, and the field or force on a charge is asked; equipotential spacing.	Differentiate: $E = - d V / d x$ . On a graph, E is minus the slope; a flat part means zero field.
Potential energy of a system and of a dipole	Work to assemble three or four charges, two charges in an external field, work to rotate a dipole, stable and unstable equilibrium.	Write one $k q_{i} q_{j} / r_{ij}$ term per pair; for a dipole use $U = - pE cos θ$ and $W = pE (cos θ_{0} - cos θ_{1})$ .
Conductors	MCQ and reasoning items: field and potential inside a conductor, charge on two spheres joined by a wire, why charge collects at sharp points, shielding.	Use the six properties; connected spheres share a common potential, so $q \propto r$ and $σ \propto 1/ r$ .
Capacitor derivations	5-mark: derive $C = ε_{0} A / d$ , then with a dielectric or metal slab of thickness t; derive series and parallel formulas.	State the field between the plates, find V by adding $E \times$ thickness in each region, then C = Q/V.
Battery connected or disconnected	A dielectric is inserted or removed with the battery on or off; say what happens to C, Q, V, E and U.	Battery disconnected: Q fixed, so V, E and U divide by K. Battery connected: V fixed, so Q and U multiply by K, E unchanged.
Networks and charge sharing	Equivalent capacitance of a network, charge and voltage on each capacitor, energy lost when two capacitors are joined.	Reduce step by step, find total Q, then work back out. For sharing, find the common potential, then compare energies before and after.

Tricks to remember Electrostatic Potential and Capacitance

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

V goes 1/r, E goes 1/r²
Point charge: potential has one r downstairs, field has two. Dipole adds one more to each: V as $1/ r^{2}$ , E as $1/ r^{3}$ .
Use it when: Graph and ratio MCQs.
Capacitors are opposite to resistors
Capacitors in parallel ADD like resistors in series; capacitors in series combine by reciprocals like resistors in parallel.
Use it when: Avoids swapping the formulas in networks.
On: V fixed. Off: Q fixed.
Battery ON keeps V fixed, so Q and U go up K times. Battery OFF keeps Q fixed, so V, E and U go down K times.
Use it when: Every 'dielectric inserted' question.
d minus t plus t over K
For a slab, only the slab's thickness is 'shrunk' by K: effective gap $= (d - t) + t / K$ . A metal slab shrinks its thickness to zero.
Use it when: Slab capacitance derivations and numericals.
Downhill is the field
Field lines run downhill on the potential map, at right angles to the contour lines (equipotentials). Steeper slope, closer contours, stronger field.
Use it when: Equipotential and V-x graph questions.
Half of QV
Energy is $\frac{1}{2} Q V$ , not QV: the battery supplies QV, and half of it is lost as heat while charging.
Use it when: Energy formulas and battery-energy questions.

Electrostatic Potential 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: Energy in a capacitor · 1 mark · MCQBoard level

A $4 μ F$ capacitor is charged to 50 V. The energy stored is:
(a) $1 \times 1 0^{- 4} J$
(b) $5 \times 1 0^{- 3} J$
(c) $1 \times 1 0^{- 2} J$
(d) $2 \times 1 0^{- 4} J$

Try it first, then show the solutionSolution

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 4 \times 1 0^{- 6} \times 2500 = 5 \times 1 0^{- 3} J$

Answer: (b) $5 \times 1 0^{- 3} J$

Example 2: Equipotentials and the field · 1 mark · Assertion-reasonBoard level

Assertion (A): No work is done in moving a charge from one point to another on an equipotential surface.
Reason (R): The electric field is perpendicular to an equipotential surface at every point.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

A is true because $Δ V = 0$ . R is true. R explains A: the force qE is perpendicular to any displacement along the surface, so it does no work.

Answer: (a)

Example 3: Work to complete a square · 2 marks · Very short answerHarder board level

Charges of $+ 1 μ C$ each sit at three corners of a square of side 10 cm. How much work is needed to bring a fourth $+ 1 μ C$ charge from infinity to the empty corner? ( $2 = 1.414$ ) [2 marks]

Try it first, then show the solutionSolution

Potential at the empty corner: two charges at 0.1 m and one at $0.1 2 = 0.1414 m$
$V = 9 \times 1 0^{9} \times 1 0^{- 6} (\frac{1}{0.1} + \frac{1}{0.1} + \frac{1}{0.1414}) = 9000 \times 27.07 \approx 2.44 \times 1 0^{5} V$
$W = q V = 1 0^{- 6} \times 2.44 \times 1 0^{5} \approx 0.24 J$

Answer: About 0.24 J

Example 4: Field from a potential function · 2 marks · Very short answerHarder board level

The potential in a region is $V = 6 x^{2} - 4 x$ volts, with x in metres. Find the electric field at x = 1 m, and the force on a charge of $- 2 μ C$ there. [2 marks]

Try it first, then show the solutionSolution

$E_{x} = - \frac{d V}{d x} = - (12 x - 4)$ ; at x = 1 m, $E_{x} = - 8 V m^{- 1}$
The field is $8 V m^{- 1}$ along the negative x-direction
$F = q E_{x} = (- 2 \times 1 0^{- 6}) (- 8) = + 1.6 \times 1 0^{- 5} N$ , along $+ x$

Answer: $8 V m^{- 1}$ along $- x$ ; force $1.6 \times 1 0^{- 5} N$ along $+ x$

Example 5: Derivation: parallel plate capacitor with a dielectric slab · 5 marks · Long answerHarder board level

(a) Derive the capacitance of a parallel plate capacitor (plate area A, separation d) in air. (b) A slab of thickness t and dielectric constant K is introduced. Derive the new capacitance. (c) For $A = 0.02 m^{2}$ , d = 6 mm, t = 4 mm and K = 4, find the capacitance with and without the slab. [5 marks]

Try it first, then show the solutionSolution

(a) Plates carry $\pm Q$ , so $σ = Q / A$ . Between them the fields of the two sheets add: $E_{0} = \frac{σ}{ε _{0}}$ ; outside they cancel.
$V = E_{0} d = \frac{Q d}{ε _{0} A}$ , so $C_{0} = \frac{Q}{V} = \frac{ε _{0} A}{d}$ .
(b) The slab is polarised, so inside it $E = \frac{E _{0}}{K}$ ; the air gaps (total $d - t$ ) still have $E_{0}$ .
$V = E_{0} (d - t) + \frac{E _{0} t}{K} = \frac{Q}{ε _{0} A} (d - t + \frac{t}{K})$ , so $C = \frac{ε _{0} A}{d - t + t / K}$ .
(c) $C_{0} = \frac{8.85 \times 1 0 ^{- 12} \times 0.02}{0.006} = 2.95 \times 1 0^{- 11} F = 29.5 pF$
With slab: $d - t + \frac{t}{K} = 2 + 1 = 3 mm$ , so $C = \frac{1.77 \times 1 0 ^{- 13}}{0.003} = 5.9 \times 1 0^{- 11} F = 59 pF$ (twice $C_{0}$ ).

Answer: $C_{0} = \frac{ε _{0} A}{d}$ ; $C = \frac{ε _{0} A}{d - t + t / K}$ ; 29.5 pF without and 59 pF with the slab.

Example 6: Network of capacitors · 3 marks · Short answerHarder board level

A $2 μ F$ and a $3 μ F$ capacitor are connected in series, and this pair is connected in parallel with a $1.8 μ F$ capacitor. The combination is connected to a 12 V battery. Find (a) the equivalent capacitance, (b) the charge on each capacitor, (c) the total energy stored. [3 marks]

Try it first, then show the solutionSolution

(a) Series pair: $\frac{2 \times 3}{2 + 3} = 1.2 μ F$ ; with $1.8 μ F$ in parallel: $C = 3 μ F$
(b) Each branch has 12 V. The $1.8 μ F$ carries $1.8 \times 12 = 21.6 μ C$ . The series pair carries $1.2 \times 12 = 14.4 μ C$ on each of the 2 µF and 3 µF capacitors.
Check: total $= 21.6 + 14.4 = 36 μ C = 3 μ F \times 12 V$
(c) $U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 3 \times 1 0^{- 6} \times 144 = 2.16 \times 1 0^{- 4} J$

Answer: (a) $3 μ F$ (b) $1.8 μ F$ : $21.6 μ C$ ; 2 µF and 3 µF: $14.4 μ C$ each (c) $2.16 \times 1 0^{- 4} J$

Example 7: Dielectric with the battery on and off · 3 marks · Short answerHOTS / competency

A 10 pF air capacitor is charged by a 100 V battery. A dielectric of K = 5 is then made to fill the gap. Find the new charge, potential difference and energy (a) if the battery is first disconnected, (b) if the battery stays connected. Explain the change in energy in (a). [3 marks]

Try it first, then show the solutionSolution

Initially $Q_{0} = 1 0^{- 9} C$ and $U_{0} = \frac{1}{2} \times 1 0^{- 11} \times 1 0^{4} = 5 \times 1 0^{- 8} J$ ; with the dielectric, C = 50 pF
(a) Q fixed at $1 0^{- 9} C$ ; $V = \frac{Q}{C} = 20 V$ ; $U = \frac{Q ^{2}}{2 C} = \frac{1 0 ^{- 18}}{1 0 ^{- 10}} = 1 \times 1 0^{- 8} J$
The energy falls by $4 \times 1 0^{- 8} J$ : the field does work pulling the slab into the gap.
(b) V fixed at 100 V; $Q = 50 \times 1 0^{- 12} \times 100 = 5 \times 1 0^{- 9} C$ ; $U = \frac{1}{2} \times 5 \times 1 0^{- 11} \times 1 0^{4} = 2.5 \times 1 0^{- 7} J$

Answer: (a) $1 0^{- 9} C$ , 20 V, $1 \times 1 0^{- 8} J$ (energy falls as the slab is pulled in) (b) $5 \times 1 0^{- 9} C$ , 100 V, $2.5 \times 1 0^{- 7} J$

Example 8: Two spheres joined by a wire · 3 marks · Short answerHOTS / competency

Two conducting spheres of radii 2 cm and 6 cm, far apart, are joined by a thin wire and given a total charge of 8 nC. (a) Find the charge on each. (b) Find the ratio of the electric fields just outside their surfaces. (c) Use this to explain why charge leaks away from sharp points. [3 marks]

Try it first, then show the solutionSolution

(a) Same potential: $\frac{q _{1}}{r _{1}} = \frac{q _{2}}{r _{2}}$ , so $q_{1} : q_{2} = 2 : 6$ , giving 2 nC and 6 nC
(b) $E = \frac{k q}{r ^{2}}$ at each surface: $\frac{E _{1}}{E _{2}} = \frac{q _{1}}{q _{2}} \cdot \frac{r _{2}^{2}}{r _{1}^{2}} = \frac{1}{3} \times 9 = 3$ , so $E_{1} : E_{2} = 3 : 1$ (equivalently $\frac{r _{2}}{r _{1}}$ )
(c) The field at the surface varies as $\frac{1}{r}$ . At a sharp point r is tiny, so the field is very large and can ionise the surrounding air, letting charge leak away.

Answer: (a) 2 nC on the small and 6 nC on the large sphere (b) 3 : 1 (c) the field ∝ 1/r is huge at a point, ionising the air.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Sliding a slab into a charged capacitor

Suppose a student charges a 50 pF air-filled parallel plate capacitor using a 12 V battery, then disconnects the battery. She then slides a slab of dielectric constant 4 into the gap so that it fills it completely.

(i) · 1 mark
What is the charge on the capacitor before the slab is inserted?
Answer
Answer: $6 \times 1 0^{- 10} C$
$Q = C V = 50 \times 1 0^{- 12} \times 12 = 6 \times 1 0^{- 10} C$
(ii) · 1 mark
What is the potential difference after the slab is inserted?
Answer
Answer: 3 V
Q is fixed, C becomes 200 pF: $V = \frac{6 \times 1 0 ^{- 10}}{2 \times 1 0 ^{- 10}} = 3 V$
(iii) · 1 mark
By how much does the stored energy change?
Answer
Answer: It falls by $2.7 \times 1 0^{- 9} J$ .
$U_{i} = \frac{1}{2} \times 50 \times 1 0^{- 12} \times 144 = 3.6 \times 1 0^{- 9} J$ ; $U_{f} = \frac{U _{i}}{4} = 0.9 \times 1 0^{- 9} J$ ; change $= - 2.7 \times 1 0^{- 9} J$
(iv) · 1 mark
Where does this energy go?
Answer
Answer: The fringe field pulls the slab into the gap and does work on it.
The slab is attracted into the capacitor; if it is not held back, the work appears as its kinetic energy, finally as heat.

Equipotential planes in a uniform field

Suppose a region has a uniform electric field. Parallel equipotential planes are drawn every 10 V, and neighbouring planes are 2 cm apart. Plane P is at 40 V and plane S, three planes further along the field, is at 10 V.

(i) · 1 mark
What is the magnitude of the field?
Answer
Answer: $500 V m^{- 1}$
$E = \frac{10 V}{0.02 m} = 500 V m^{- 1}$
(ii) · 1 mark
In which direction does the field point?
Answer
Answer: Perpendicular to the planes, from P (40 V) towards S (10 V).
The field points in the direction of decreasing potential.
(iii) · 1 mark
How much work does an external agent do in moving $+ 2 μ C$ slowly from P to S?
Answer
Answer: $- 6 \times 1 0^{- 5} J$
$W = q (V_{S} - V_{P}) = 2 \times 1 0^{- 6} \times (10 - 40) = - 6 \times 1 0^{- 5} J$ (the field does $+ 6 \times 1 0^{- 5} J$ )
(iv) · 1 mark
How much work is done in moving the same charge 5 cm along plane P?
Answer
Answer: Zero
All points of P are at 40 V, so $Δ V = 0$ .

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.

A uniform field is along the x-axis. The potential at x = 0 is 50 V and at x = 0.25 m it is 30 V. Find the magnitude and direction of the field. [2 marks] (2 marks)

Step	Marks
$E = - \frac{Δ V}{Δ x} = - \frac{30 - 50}{0.25}$	1
$E = 80 V m^{- 1}$ along $+ x$ (towards decreasing potential)	1

Derive an expression for the potential at a point due to an electric dipole, at distance r from its centre and at angle θ with the dipole axis ( $r ≫ a$ ). [3 marks] (3 marks)

Step	Marks
Figure and $V = \frac{q}{4 π ε _{0}} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ with $r_{1}, r_{2}$ the distances from $+ q$ and $- q$	1
For $r ≫ a$ : $r_{1} \approx r - a cos θ$ , $r_{2} \approx r + a cos θ$ , so $\frac{1}{r _{1}} - \frac{1}{r _{2}} \approx \frac{2 a c o s θ}{r ^{2}}$	1
$V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}}$ , with $p = 2 q a$ ; axis: $\frac{k p}{r ^{2}}$ , equatorial: 0	1

(a) Derive the capacitance of a parallel plate capacitor when a dielectric slab of thickness t (t < d) and dielectric constant K is placed between its plates. (b) Hence write the result for a metal slab. (c) A capacitor with plate area $0.02 m^{2}$ and separation 6 mm has a slab of K = 4 and thickness 4 mm inserted. Find its capacitance. [5 marks] (5 marks)

Step	Marks
Figure; field in air $E_{0} = \frac{σ}{ε _{0}} = \frac{Q}{A ε _{0}}$ ; field in slab $\frac{E _{0}}{K}$	1
$V = E_{0} (d - t) + \frac{E _{0}}{K} t = \frac{Q}{A ε _{0}} (d - t + \frac{t}{K})$	1
$C = \frac{Q}{V} = \frac{ε _{0} A}{d - t + \frac{t}{K}}$	1
(b) Metal slab: $K \to \infty$ , so $C = \frac{ε _{0} A}{d - t}$	1
(c) $C = \frac{8.85 \times 1 0 ^{- 12} \times 0.02}{0.002 + 0.001} = 5.9 \times 1 0^{- 11} F = 59 pF$	1

Where students lose marks in Electrostatic Potential and Capacitance

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

Mistake: Adding potentials as vectors or dropping the sign of a negative charge
Habit from field questions.
Wrong working
At the centre of a square with +q, +q, −q, −q: $V = 4 \frac{k q}{r}$
Right working
$V = \frac{k}{r} (q + q - q - q) = 0$
Do this: Potential is a scalar: add with signs, no components.
Usually costs: Full marks of a 1-2 mark question.
Mistake: Saying the field is zero where the potential is zero
Confusing V and E.
Do this: E depends on how V changes, not on its value. On the equatorial plane of a dipole V = 0 but E ≠ 0; inside a charged shell E = 0 but V ≠ 0.
Usually costs: 1 mark in assertion-reason items.
Mistake: Using $C = K ε_{0} A / d$ when the slab does not fill the gap
Memorising only the full-gap result.
Wrong working
Slab of thickness d/2: $C = K ε_{0} A / d$
Right working
$C = \frac{ε _{0} A}{d - \frac{d}{2} + \frac{d}{2 K}} = \frac{2 K ε _{0} A}{d ( K + 1 )}$
Do this: Check whether t = d. If not, use $d - t + t / K$ .
Usually costs: 1-2 marks.
Mistake: Mixing up what stays constant when a dielectric is inserted
Not reading whether the battery is connected.
Do this: Write the constant quantity first ('battery disconnected, so Q is constant'), then derive the rest.
Usually costs: Usually 1 mark per wrong quantity in a 3-mark question.
Mistake: Wrong sign for work in moving a charge
Forgetting which work is asked: by the field or by an external agent.
Wrong working
Work by the field from A to B $= q (V_{B} - V_{A})$
Right working
External agent: $q (V_{B} - V_{A})$ . Field: $q (V_{A} - V_{B})$ .
Do this: Name the agent before writing the formula.
Usually costs: ½-1 mark.
Mistake: Writing a derivation for the energy stored in a capacitor
Older books include it.
Do this: The 2026-27 syllabus asks for formulae only; spend the time on the parallel plate and slab derivations instead.
Usually costs: Time lost, not marks, but it costs you elsewhere.

NCERT exercise guide: Electrostatic Potential and Capacitance

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 2.1 to 2.3 (potential of charges and equipotentials)

What it tests: Points of zero potential for two unlike charges, potential at the centre of a symmetric arrangement, and the equipotential surface of a pair of equal and opposite charges.

How to approach it: Write V as a scalar sum and set it to zero; test both 'between' and 'outside'. For symmetric shapes, every charge is at the same distance from the centre, so multiply. For ±q, the perpendicular bisector plane is the V = 0 equipotential, and the field on it is parallel to the line joining the charges.

Same kind as Exercise 2.2

A square of side 20 cm has a charge of $+ 2 n C$ at each corner. Find the potential at the centre. ( $2 \approx 1.414$ )

Show the solutionSolution

Distance from each corner to the centre $= \frac{0.2 2}{2} = 0.1414 m$
$V = 4 \times \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 9}}{0.1414} = 4 \times 127.3$
$V \approx 509 V$

Answer: About 509 V

Exercise 2.4 (charged spherical conductor)

What it tests: Field inside, just outside and away from a charged conducting sphere.

How to approach it: Inside: E = 0. Outside: treat the charge as concentrated at the centre, $E = k Q / r^{2}$ .

Same kind as Exercise 2.4

A conducting sphere of radius 5 cm carries $5 \times 1 0^{- 9} C$ . Find the field (a) at 2 cm from the centre, (b) just outside, (c) at 15 cm from the centre.

Show the solutionSolution

(a) Inside a conductor: E = 0
(b) $E = \frac{9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 9}}{( 0.05 ) ^{2}} = \frac{45}{0.0025} = 1.8 \times 1 0^{4} N C^{- 1}$
(c) $E = \frac{45}{( 0.15 ) ^{2}} = 2000 N C^{- 1}$

Answer: (a) 0 (b) $1.8 \times 1 0^{4} N C^{- 1}$ (c) $2000 N C^{- 1}$

Exercises 2.5 to 2.8 (capacitance and combinations)

What it tests: Effect of changing d and K together, capacitors in series and parallel with the voltage or charge on each, and the capacitance and charge of a given parallel plate capacitor.

How to approach it: Use $C \propto \frac{K A}{d}$ for scaling. Series: same Q, voltages split inversely to C. Parallel: same V, charges split in proportion to C.

Same kind as Exercises 2.5 and 2.6

(a) An air capacitor of 10 pF has its plate separation tripled and the gap filled with a dielectric of K = 6. Find the new capacitance. (b) Three 6 pF capacitors in series are connected to 90 V. Find the voltage across each.

Show the solutionSolution

(a) $C^{'} = C \frac{K}{3} = 10 \times \frac{6}{3} = 20 pF$
(b) Series: $C_{s} = \frac{6}{3} = 2 pF$ ; $Q = 2 \times 1 0^{- 12} \times 90 = 1.8 \times 1 0^{- 10} C$
Each: $V = \frac{Q}{C} = \frac{1.8 \times 1 0 ^{- 10}}{6 \times 1 0 ^{- 12}} = 30 V$

Answer: (a) 20 pF (b) 30 V each

Exercises 2.9 to 2.11 (dielectric insertion and energy)

What it tests: What happens when a dielectric is inserted with the supply connected or disconnected, energy stored, and energy lost when a charged capacitor shares charge.

How to approach it: Decide what is fixed (V or Q). For sharing, find the common potential and compare $\frac{1}{2} C V^{2}$ before and after.

Same kind as Exercise 2.11

A $3 μ F$ capacitor charged to 120 V is disconnected and joined to an uncharged $6 μ F$ capacitor. Find the common potential and the energy lost.

Show the solutionSolution

Charge $Q = 3 \times 1 0^{- 6} \times 120 = 360 μ C$ , conserved
Common potential $V = \frac{360 μ C}{9 μ F} = 40 V$
$U_{i} = \frac{1}{2} \times 3 \times 1 0^{- 6} \times 12 0^{2} = 0.0216 J$ ; $U_{f} = \frac{1}{2} \times 9 \times 1 0^{- 6} \times 4 0^{2} = 0.0072 J$
Energy lost $= 0.0216 - 0.0072 = 0.0144 J$

Answer: 40 V; $1.44 \times 1 0^{- 2} J$ lost

Electrostatic Potential and Capacitance: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6.6 marks per paper; 17 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Electrostatic Potential and Capacitance

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 55/1/1, Q1 · 1 mark · MCQ

In a region, the electric potential varies as $V = 10 - 50 x$ , where V is in volts and $x$ in meters. The electric field in the region is :

(A)10 N/C along $+ x$
(B)10 N/C along $- x$
(C)50 N/C along $+ x$
(D)50 N/C along $- x$

Try it first, then show our solutionSolution

$E = - \frac{d V}{d x} = - \frac{d}{d x} (10 - 50 x) = + 50$ V/m.
1 mark
The positive sign means $E$ is $50$ N/C along $+ x$ .

Answer: 50 N/C along +x, option (C)

Where the marks are

Uses $E = - d V / d x$ to get 50 N/C along +x (1 mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/1/1, Q23 · 3 marks · Short answer

A parallel plate capacitor of capacitance C has a dielectric slab between its plates. It is charged to a potential difference V by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected ?
(a) Capacitance of the capacitor,
(b) Energy stored in the capacitor, and
(c) The potential difference between the plates of the capacitor.
Justify your answer in each case.

Try it first, then show our solutionSolution

After the battery is disconnected the charge $Q = C V$ on the plates stays constant, because it has no path to flow away.
(a) Capacitance: with the slab $C = \frac{K ε _{0} A}{d}$ . Without it, $C^{'} = \frac{ε _{0} A}{d} = \frac{C}{K}$ . The capacitance decreases by a factor $K$ .
1 mark
(b) Energy: $U = \frac{Q ^{2}}{2 C}$ with $Q$ constant, so $U^{'} = \frac{Q ^{2}}{2 ( C / K )} = K U$ . The energy increases $K$ times. The extra energy comes from the work done in pulling the slab out against the attractive force.
1 mark
(c) Potential difference: $V^{'} = \frac{Q}{C ^{'}} = \frac{K Q}{C} = K V$ . The p.d. increases $K$ times.
1 mark

Answer: (a) Decreases to C/K; (b) increases to K times; (c) increases to KV

Where the marks are

Capacitance decreases to C/K, with reason (1 mark)
Energy increases K times, since U = Q²/2C at constant Q (1 mark)
Potential difference increases to KV, since V = Q/C (1 mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/5/1, Q29 · 4 marks · Case-basedCompetency-based

The electric potential (V) and electric field (E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the force per unit charge at a given point in space, whereas electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in space. Electric field and electric potential are related by the equations $E_{r} = \frac{- d V}{d r}$ and $E = E_{r} \overset{r}{^}$ , i.e., electric field is the negative gradient of the electric potential. This means that electric field points in the direction of decreasing potential and its magnitude is the rate of change of potential with distance. The electric field is the force that drives a unit charge to move from higher potential region to lower potential region and electric potential difference between the two points determines the work done in moving a unit charge from one point to the other point.

A pair of square conducting plates having sides of length 0·05 m are arranged parallel to each other in x-y plane. They are 0·01 m apart along z-axis and are connected to a 200 V power supply as shown in the figure. An electron enters with a speed of $3 \times 1 0^{7} ms^{- 1}$ horizontally and symmetrically in the space between the two plates. Neglect the effect of gravity on the electron.

(i)
The electric field $E$ in the region between the plates is :
1. (A) $(2 \times 1 0^{2} \frac{V}{m}) \hat{k}$
2. (B) $- (2 \times 1 0^{2} \frac{V}{m}) \hat{k}$
3. (C) $(2 \times 1 0^{4} \frac{V}{m}) \hat{k}$
4. (D) $- (2 \times 1 0^{4} \frac{V}{m}) \hat{k}$
1 mark
(ii)
In the region between the plates, the electron moves with an acceleration $a$ given by :
1. (A) $- (3 \cdot 5 \times 1 0^{15} ms^{- 2}) \hat{k}$
2. (B) $(3 \cdot 5 \times 1 0^{15} ms^{- 2}) \hat{k}$
3. (C) $(3 \cdot 5 \times 1 0^{13} ms^{- 2}) \hat{i}$
4. (D) $- (3 \cdot 5 \times 1 0^{13} ms^{- 2}) \hat{i}$
1 mark
(iii)(a)
Time interval during which an electron moves through the region between the plates is :
1. (A) $9 \cdot 0 \times 1 0^{- 9}$ s
2. (B) $1 \cdot 67 \times 1 0^{- 8}$ s
3. (C) $1 \cdot 67 \times 1 0^{- 9}$ s
4. (D) $2 \cdot 17 \times 1 0^{- 9}$ s
1 mark
OR
(iii)(b)
The vertical displacement of the electron which travels through the region between the plates is :
1. (A)10 mm
2. (B)4·9 mm
3. (C)5·9 mm
4. (D)3·0 mm
1 mark
(iv)
Which one of the following is the path traced by the electron in between the two plates ?
1. (A)a
2. (B)b
3. (C)c
4. (D)d
1 mark

Figure in the paper: Two parallel horizontal plates of length 0·05 m, separated by 0·01 m, connected to a 200 V battery (+ terminal to the upper plate, − to the lower). An electron enters midway between the plates moving horizontally with speed 3×10^7 m/s. Axes shown: x to the right (i), y up (j), z out of page/toward viewer (k). For (iv): plates B (upper) and A (lower), 0·05 m long, with electron entering along line e; possible paths after plates: a (straight, undeflected), b (curving upward steeply), c (curving downward), d (upward at a smaller angle, straight line). See it in the official paper.

Try it first, then show our solutionSolution

(i) $E = \frac{V}{d} = \frac{200}{0.01} = 2 \times 1 0^{4}$ V/m, directed from the positive (upper) plate to the negative (lower) plate, i.e. along $- \hat{k}$ . $E = - (2 \times 1 0^{4} V/m) \hat{k}$ — option (D).
1 mark
(ii) Force on electron is opposite to $E$ , i.e. along $+ \hat{k}$ . $a = \frac{e E}{m} = \frac{1.6 \times 1 0 ^{- 19} \times 2 \times 1 0 ^{4}}{9.1 \times 1 0 ^{- 31}} \approx 3.5 \times 1 0^{15}$ m s⁻². $a = (3.5 \times 1 0^{15} m s^{- 2}) \hat{k}$ — option (B).
1 mark
(iii)(a) Horizontal velocity is unchanged: $t = \frac{l}{v} = \frac{0.05}{3 \times 1 0 ^{7}} = 1.67 \times 1 0^{- 9}$ s — option (C).
1 mark
(iii)(b) [Alternative to (iii)(a), 1 mark] $y = \frac{1}{2} a t^{2} = \frac{1}{2} \times 3.5 \times 1 0^{15} \times (1.67 \times 1 0^{- 9})^{2} \approx 4.9 \times 1 0^{- 3}$ m $= 4.9$ mm — option (B). (Just less than 5 mm, so the electron just leaves the plates.)
(iv) Inside the plates the electron has uniform velocity horizontally and uniform acceleration towards the positive (upper) plate, so it follows a parabola curving upward — path b, option (B).
1 mark

Answer: (i) D; (ii) B; (iii)(a) C; (iii)(b) B; (iv) B

Where the marks are

(i) E = 2 × 10⁴ V/m along −k̂, option D (1 mark)
(ii) a = 3.5 × 10¹⁵ m/s² along +k̂, option B (1 mark)
(iii)(a) t = 1.67 × 10⁻⁹ s, option C; or (iii)(b) y ≈ 4.9 mm, option B (1 mark)
(iv) Parabolic path bending towards positive plate, path b, option B (1 mark)

See it in the full 2026 board paper, set 55/5/1

Every Electrostatic Potential and Capacitance question, by year

For 2027

Sample paper 2026-27: Q19 (2 marks, solved), Q31(A) (5 marks, solved), Q31(B) (5 marks, solved)

2026

Set 55/1/1: Q1 (1 mark, solved), Q2 (1 mark, solved), Q23 (3 marks, solved), Q24 (3 marks, solved)

Set 55/5/1: Q2 (1 mark, solved), Q22 (3 marks, solved), Q29 (4 marks, solved)

Sample paper 2025-26: Q31(I) (5 marks, solved), Q31(II) (5 marks, solved)

2025

Set 55/1/1: Q29 (4 marks, solved), Q31(a) (5 marks, solved)

Set 55/7/1: Q31(a) (5 marks, solved)

Sample paper 2024-25: Q1 (1 mark, solved), Q2 (1 mark, solved), Q23 (3 marks, solved)

2024

Set 55/1/1: Q2 (1 mark, solved), Q31(b) (5 marks, solved)

Set 55/5/1: Q29 (4 marks, solved), Q31(a) (5 marks, solved)

Sample paper 2023-24: Q1 (1 mark, solved), Q15 (1 mark, solved), Q23 (3 marks, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

2023

Set 55/1/1: Q26 (3 marks, solved)

Set 55/5/1: Q12 (1 mark, solved), Q35 (4 marks, solved)

Practice questions on Electrostatic Potential 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
One farad is equal to:
(a) 1 C V
(b) 1 C V⁻¹
(c) 1 V C⁻¹
(d) 1 J C⁻¹
Hint
C = Q/V.
Answer
(b) 1 C V⁻¹
Q2 · 1 mark · MCQBoard level
Two 6 pF capacitors are joined in parallel, and this pair is joined in series with a third 6 pF capacitor. The equivalent capacitance is:
(a) 18 pF
(b) 4 pF
(c) 9 pF
(d) 2 pF
Hint
Combine the parallel pair first, then use reciprocals for the series step.
Answer
(b) 4 pF: the parallel pair is 12 pF, and $\frac{12 \times 6}{12 + 6} = 4 pF$ .
Q3 · 1 mark · MCQBoard level
A charge Q is fixed at the centre of a circle of radius r. The work done in carrying a charge q once round the circle is:
(a) $\frac{k Qq}{r}$
(b) $\frac{k Qq}{r} \times 2 π r$
(c) zero
(d) $\frac{k Qq}{r ^{2}}$
Hint
The circle is an equipotential.
Answer
(c) zero, since every point of the circle is at the same potential.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored decreases.
Reason (R): The charge on the plates stays the same while the capacitance increases.
(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
Use $U = Q^{2} /2 C$ .
Answer
(a). With Q fixed, $U = Q^{2} /2 C$ falls as C rises.
Q5 · 2 marks · Very short answerHarder board level
Four charges $+ q$ , $+ q$ , $- q$ , $- q$ sit at the corners of a square, with the two positive charges at adjacent corners. Is the potential at the centre zero? Is the field at the centre zero? Explain.
Hint
V is a scalar; E is a vector.
Answer
V = 0, because all four charges are equidistant and their charges add to zero. E is not zero: the fields of the opposite pairs add up and point from the side with the positive charges towards the side with the negative charges.
Q6 · 2 marks · Very short answerHarder board level
Charges of $+ 2 μ C$ and $- 3 μ C$ are 30 cm apart. How much work must be done to bring a charge of $+ 1 μ C$ from infinity to the midpoint of the line joining them? Interpret the sign.
Hint
Find V at the midpoint first.
Answer
$V = \frac{9 \times 1 0 ^{9} ( 2 - 3 ) \times 1 0 ^{- 6}}{0.15} = - 6 \times 1 0^{4} V$ . $W = q V = 1 0^{- 6} \times (- 6 \times 1 0^{4}) = - 0.06 J$ . Negative: the field pulls the charge in, so the agent must hold it back.
Q7 · 3 marks · Short answerHarder board level
A $4 μ F$ capacitor charged to 150 V is disconnected from the supply and joined to an uncharged $2 μ F$ capacitor. (a) Find the common potential and the final charge on each. (b) Find the energy lost. (c) Show that the loss equals $\frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} V_{1}^{2}$ .
Hint
Total charge is conserved; both end at the same potential.
Answer
(a) $Q = 4 \times 150 = 600 μ C$ is conserved; $V = \frac{600}{6} = 100 V$ ; charges $4 \times 100 = 400 μ C$ and $2 \times 100 = 200 μ C$ . (b) Before: $\frac{1}{2} \times 4 \times 1 0^{- 6} \times 15 0^{2} = 0.045 J$ ; after: $\frac{1}{2} \times 6 \times 1 0^{- 6} \times 10 0^{2} = 0.030 J$ ; lost 0.015 J. (c) $\frac{1}{2} \times \frac{4 \times 2}{6} \times 1 0^{- 6} \times 15 0^{2} = 0.015 J$ , the same.
Q8 · 2 marks · Very short answerHarder board level
A parallel plate capacitor has plate area A and separation d. A metal slab of thickness d/2 is inserted parallel to the plates. Show that the capacitance doubles. What would it be with a dielectric slab of the same thickness and K = 3?
Hint
Use $d - t + t / K$ .
Answer
Metal: $C = \frac{ε _{0} A}{d - d /2} = \frac{2 ε _{0} A}{d} = 2 C_{0}$ . Dielectric: $C = \frac{ε _{0} A}{d /2 + d /6} = \frac{ε _{0} A}{2 d /3} = 1.5 C_{0}$ .
Q9 · 5 marks · Long answerHarder board level
(a) Derive the expression for the equivalent capacitance of three capacitors in series. (b) Three capacitors of 2, 3 and 6 µF are connected in series to a 12 V battery. Find the charge on each and the voltage across each. (c) If they are instead connected in parallel to the same battery, find the total energy stored.
Hint
Series: same Q; parallel: same V.
Answer
(a) Same Q on each; $V = V_{1} + V_{2} + V_{3} = Q (\frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}})$ , so $\frac{1}{C} = \sum \frac{1}{C _{i}}$ . (b) $\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ , C = 1 µF; Q = 12 µC on each; voltages 6 V, 4 V and 2 V (sum 12 V). (c) Parallel C = 11 µF; $U = \frac{1}{2} \times 11 \times 1 0^{- 6} \times 144 = 7.92 \times 1 0^{- 4} J$ .
Q10 · 2 marks · Very short answerHarder board level
A dipole of moment $4 \times 1 0^{- 9} C m$ is aligned with a uniform field of $2 \times 1 0^{4} N C^{- 1}$ . How much work is needed to rotate it through (a) $9 0^{\circ}$ , (b) $18 0^{\circ}$ ? In which position is it in unstable equilibrium?
Hint
$W = pE (cos θ_{0} - cos θ_{1})$ .
Answer
$pE = 8 \times 1 0^{- 5} J$ . (a) $W = pE (1 - 0) = 8 \times 1 0^{- 5} J$ . (b) $W = pE (1 + 1) = 1.6 \times 1 0^{- 4} J$ . Unstable equilibrium at θ = 180° (antiparallel).
Q11 · 3 marks · Short answerHOTS / competency
A capacitor is half filled with a dielectric of constant K, but this time the slab fills the full gap d and covers only half the plate area (side by side, not stacked). Find the capacitance in terms of $C_{0}$ . Compare it with a slab of thickness d/2 covering the whole area, for K = 3.
Hint
Side by side behaves like two capacitors in parallel.
Answer
Two halves in parallel: $C = \frac{ε _{0} ( A /2 )}{d} + \frac{K ε _{0} ( A /2 )}{d} = \frac{C _{0} ( 1 + K )}{2}$ ; for K = 3, $C = 2 C_{0}$ . Stacked slab of thickness d/2 (series-like): $C = \frac{ε _{0} A}{d /2 + d /6} = 1.5 C_{0}$ . Side by side gives more.
Q12 · 2 marks · Very short answerHOTS / competency
The V-x graph for a region is a straight line rising from 0 V at x = 0 to 20 V at x = 4 m, then flat at 20 V up to x = 6 m. Find the electric field in each part and say where a free electron would accelerate.
Hint
E is minus the slope.
Answer
0 to 4 m: slope $= 5 V m^{- 1}$ , so $E = 5 V m^{- 1}$ along $- x$ . 4 to 6 m: slope 0, so E = 0. An electron feels a force opposite to E, along $+ x$ , only in the first part, so it accelerates there and moves at constant velocity in the flat part.
Q13 · 2 marks · Very short answerHarder board level
Two point charges $+ 4 μ C$ and $- 4 μ C$ are 20 cm apart in air. Find the potential energy of the system. How much work is needed to separate them to infinity?
Hint
$U = k q_{1} q_{2} / r$ .
Answer
$U = \frac{9 \times 1 0 ^{9} \times ( 4 \times 1 0 ^{- 6} ) ( - 4 \times 1 0 ^{- 6} )}{0.2} = - 0.72 J$ . Work needed to separate them $= 0 - (- 0.72) = 0.72 J$ .

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 electrostatic potential at a point.Show answer
Work done by an external force in bringing a unit positive charge, without acceleration, from infinity to that point.
2.Write the potential of a dipole at distance r, angle θ, and its value on the equatorial plane.Show answer
$V = \frac{p c o s θ}{4 π ε _{0} r ^{2}}$ ; zero on the equatorial plane.
3.Why is no work done in moving a charge on an equipotential surface?Show answer
The potential difference between any two points on it is zero, so $W = q Δ V = 0$ ; equivalently the field is perpendicular to the motion.
4.Write the relation between E and V.Show answer
$E = - \frac{d V}{d r}$ ; the field points in the direction of steepest decrease of potential.
5.What is the potential energy of a dipole at angle θ to a uniform field? When is it minimum?Show answer
$U = - pE cos θ$ ; minimum ( $- pE$ ) at θ = 0, stable equilibrium.
6.Write five properties of a conductor in electrostatic equilibrium.Show answer
E = 0 inside; E normal to the surface just outside; no charge inside, excess on the surface; V constant throughout; E just outside = σ/ε₀ (and a cavity is shielded).
7.Capacitance of a parallel plate capacitor with a slab of thickness t and dielectric constant K?Show answer
$C = \frac{ε _{0} A}{d - t + t / K}$ .
8.Three 6 pF capacitors: series and parallel values?Show answer
Series 2 pF; parallel 18 pF.
9.Write three forms of the energy stored in a capacitor.Show answer
$\frac{Q ^{2}}{2 C}$ , $\frac{1}{2} C V^{2}$ , $\frac{1}{2} Q V$ .

Best order to study it

Start with potential and potential difference as work per unit charge, then derive the potential of a point charge.
Learn the dipole potential and how to add potentials of several charges, including points of zero potential.
Study equipotential surfaces and the relation $E = - d V / d r$ , with V-x graph questions.
Do potential energy of a system of charges, then in an external field, ending with the dipole ( $U = - pE cos θ$ ).
Learn the properties of conductors and shielding, then dielectrics and polarisation.
Finish with capacitors: definition, parallel plate with and without dielectric or slab, series and parallel, and energy formulas with battery on/off cases.

Quick revision checklist

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

Can you derive $V = \frac{q}{4 π ε _{0} r}$ from the definition of potential?
Can you derive the dipole potential and state its value on the axis and the equatorial plane?
Can you find points of zero potential for two unlike charges, both between and outside them?
Can you draw equipotentials for a point charge, a uniform field and a dipole?
Can you get E from a V(x) formula or a V-x graph?
Can you write the PE of three charges and of a dipole in a uniform field?
Can you list the properties of a conductor in electrostatic equilibrium?
Can you explain polarisation in polar and non-polar dielectrics?
Can you derive C for a parallel plate capacitor with no slab, a dielectric slab and a metal slab?
Can you say how Q, V, E, C and U change when a dielectric is inserted with the battery on and off?

Electrostatic Potential and Capacitance: frequently asked questions

Is the derivation of the energy stored in a capacitor in the 2026-27 syllabus?

No. The syllabus says 'energy stored in a capacitor (no derivation, formulae only)'. Learn U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV and use them in numericals.

Which derivations should I prepare from this chapter?

Potential of a point charge, potential of a dipole, E = -dV/dr, potential energy of a dipole in a uniform field, the parallel plate capacitor (with and without a dielectric, and with a slab of thickness t), and the series and parallel formulas. In the board papers in our question bank, the slab derivation was the most frequent 5-mark item from this chapter.

Can the potential be zero where the field is not zero?

Yes. On the equatorial plane of a dipole, V = 0 but E is not zero. The reverse also happens: inside a charged conducting shell, E = 0 but V equals the surface value.

Why does a dielectric increase capacitance?

Polarisation charges on the dielectric's faces create a field that opposes the field of the plates. For the same charge the field and hence the potential difference fall, so C = Q/V rises (by the factor K if the dielectric fills the gap).

Why is energy lost when a charged capacitor is connected to an uncharged one?

Charge flows until both are at a common potential. The total charge is conserved, but \frac{Q^2}{2C} falls because the total capacitance has increased. The difference is lost as heat in the wires and as electromagnetic radiation.

What is the difference between free and bound charges?

Free charges (electrons in a conductor) can move through the material. Bound charges (in an insulator) can only shift slightly within their own atoms or molecules, which is what happens when a dielectric is polarised.

Based on: CBSE Physics (042) syllabus 2026-27; NCERT Physics Part I, Textbook for Class XII, Chapter 2 (Reprint 2026-27)