Electrostatic Potential and Capacitance — Class 12 Physics Worksheet | AAI ATC

Before you begin

Formulas you will need

Capacitance and potential C = Q ⁄ V · C = ε₀A ⁄ d · With dielectric: C′ = kC
Series: 1⁄C = 1⁄C₁ + 1⁄C₂ · Parallel: C = C₁ + C₂
Energy: U = ½CV² = Q²⁄2C · Uniform field: E = V ⁄ d
Work: W = qΔV · ε₀ = 8.854 × 10⁻¹² F m⁻¹

How to use this worksheet. Attempt every question before checking. Tap an option to lock your answer, then open the solution. In a series combination the charge is the same on every capacitor; in parallel the voltage is the same. Those two facts settle most of Section B.

Section A · Level 1

Definitions and direct formulas

Eight questions. No calculation should take more than twenty seconds.

Q1

The SI unit of capacitance is the:

Capacitance is charge per unit potential difference, so its unit is the coulomb per volt, which is named the farad.

Q2

The SI unit of electric potential is the:

Potential is work done per unit charge, measured in joules per coulomb, which is the volt.

Q3

For a parallel plate capacitor, if the plate area A is doubled while everything else stays the same, the capacitance:

Since C = ε₀A ÷ d, capacitance is directly proportional to area. Doubling A doubles C.

Q4

Two capacitors of 3 µF and 6 µF are connected in parallel. The equivalent capacitance is:

In parallel, capacitances simply add: 3 + 6 = 9 µF. Option 1 is the series answer, which is the standard distractor.

Q5

The energy stored in a capacitor of capacitance C charged to potential V is:

The stored energy is ½CV², which can also be written as Q²÷2C or ½QV. The factor of one half comes from integrating as the capacitor charges up.

Q6

The dielectric constant of vacuum is:

The dielectric constant is a ratio, so for vacuum it is exactly 1. Option 2 is the permittivity of free space, which is a different quantity with units.

Q7

The work done in moving a charge along an equipotential surface is:

Work equals q times the potential difference, and along an equipotential surface that difference is zero. So no work is done.

Q8

The equipotential surfaces around an isolated point charge are:

Potential depends only on distance from a point charge, so all points at the same radius share the same potential. That surface is a sphere centred on the charge.

Section B · Level 2

Application and combinations

Eight questions. Series means equal charge; parallel means equal voltage.

Q9

Two capacitors of 2 µF and 3 µF are connected in series. The equivalent capacitance is:

For two capacitors in series the product over the sum gives (2 × 3) ÷ 5 = 1.2 µF. Note that the series value is always smaller than either capacitor.

Q10

Three capacitors of 6 µF each are connected in parallel. The equivalent capacitance is:

Parallel capacitances add, giving 6 + 6 + 6 = 18 µF. If they had been in series the answer would be 2 µF.

Q11

A 5 µF capacitor is charged to 100 V. The energy stored is:

Using U = ½CV², this is 0.5 × 5 × 10⁻⁶ × 100² = 0.5 × 5 × 10⁻⁶ × 10⁴ = 0.025 J. Option 1 comes from forgetting the factor of one half.

Q12

The charge stored on a 4 µF capacitor connected across 50 V is:

Q = CV = 4 µF × 50 V = 200 µC. Option 1 divides instead of multiplying, which is the commonest slip here.

Q13

Two 4 µF capacitors are joined in series, and that combination is connected in parallel with a 2 µF capacitor. The total capacitance is:

The two 4 µF in series give (4 × 4) ÷ 8 = 2 µF. Adding the parallel 2 µF gives 2 + 2 = 4 µF. Work the series part out first, then add.

Q14

The plates of a capacitor are 0.01 m apart with 100 V across them. The uniform field between the plates is:

For a uniform field, E = V ÷ d = 100 ÷ 0.01 = 10 000 V m⁻¹. Option 4 comes from mis-handling the decimal.

Q15

Moving a charge of 2 µC between two points requires 6 µJ of work. The potential difference between the points is:

From W = qΔV, the potential difference is 6 µJ ÷ 2 µC = 3 V. The micro prefixes cancel, so no conversion is needed.

Q16

If the separation between the plates of a parallel plate capacitor is doubled, the capacitance:

Capacitance is inversely proportional to plate separation, since C = ε₀A ÷ d. Doubling d halves C.

Section C · Level 3

Multi-step and conceptual

Eight questions. These need two steps or a physical insight rather than one substitution.

Q17

A parallel plate capacitor has plate area 0.2 m² and separation 0.001 m in vacuum. Its capacitance is closest to:

C = ε₀A ÷ d = 8.854 × 10⁻¹² × 0.2 ÷ 0.001 = 1.771 × 10⁻⁹ F, which is about 1.77 nF.

Q18

Two 10 µF capacitors are connected in series across 20 V. The charge on each capacitor is:

The series equivalent is 5 µF, so the total charge is 5 × 20 = 100 µC. In a series combination that same charge sits on every capacitor, so each carries 100 µC.

Q19

Capacitors of 3 µF and 6 µF are connected in series across 18 V. The total energy stored is:

The series equivalent is (3 × 6) ÷ 9 = 2 µF. Then U = ½CV² = 0.5 × 2 × 10⁻⁶ × 324 = 3.24 × 10⁻⁴ J.

Q20

Capacitors of 2 µF and 3 µF are connected in parallel across 10 V. The total charge drawn is:

In parallel the capacitances add to 5 µF, and both see the full 10 V. So Q = 5 × 10 = 50 µC. Equivalently 20 µC plus 30 µC.

Q21

A capacitor is charged and then disconnected from the battery. A dielectric of constant k is now inserted. The energy stored:

Once disconnected the charge Q is fixed. Capacitance rises to kC, and since U = Q² ÷ 2C, the energy falls to one k-th of its former value. The dielectric is pulled in, and that work comes out of the stored energy.

Q22

A capacitor stays connected to a battery while a dielectric of constant k is inserted. The charge on the plates:

With the battery still attached the voltage V is fixed. Capacitance becomes kC, so from Q = CV the charge rises k times. Contrast this with the previous question, where the battery was removed and the charge was fixed instead.

Q23

Three capacitors of 2 µF, 3 µF and 6 µF are connected in series. The equivalent capacitance is:

The reciprocals add: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1. So 1/C = 1 and C = 1 µF. These three values are chosen to give a clean result, which is a common exam trick.

Q24

The potential at a distance r from an isolated point charge q in vacuum is proportional to:

Potential goes as one over r, while the electric field goes as one over r squared. Confusing the two is the most frequent conceptual error in this chapter.

Section D · Previous year questions

Straight from the AAI ATC papers

Six questions, every one exactly as it appeared. Five come from the two 2025 shifts, which makes capacitance the most heavily tested physics chapter in that year's domain section.

Q25PYQ 2025 · Shift 2

Two capacitors of capacitance 3 μF and 6 μF are connected in series combination. The equivalent capacitance of the system is equal to:

Product over sum gives (3 × 6) ÷ 9 = 2 µF. Option 1 is the parallel answer, which sits there for anyone who reads the word "series" too quickly.

Q26PYQ 2025 · Shift 2

Two capacitors, 6 µF and 12 µF, are connected in series across a potential difference of 30 V. What is the voltage across the 12 µF capacitor?

The series equivalent is (6 × 12) ÷ 18 = 4 µF, so the total charge is 4 × 30 = 120 µC. That same charge sits on the 12 µF capacitor, giving V = 120 ÷ 12 = 10 V. The larger capacitor always takes the smaller share of the voltage.

Q27PYQ 2025 · Shift 2

The capacitance of a parallel plate capacitor with air between the plates is C. The area of the plates is A and the distance between them is d. If the distance is reduced to d/4, the capacitance becomes C′. What is the ratio C : C′, keeping every other parameter the same?

Since C is inversely proportional to d, quartering the separation makes the new capacitance four times the old one. So C′ = 4C and the ratio C : C′ is 1 : 4. Option 1 is the same numbers written the wrong way round, which is exactly what the question is testing.

Q28PYQ 2025 · Shift 1

A parallel plate capacitor with air between the plates has capacitance C. When a dielectric of dielectric constant k is introduced between the plates, its capacitance becomes C′. Which relation between C and C′ is correct?

Inserting a dielectric multiplies the capacitance by the dielectric constant, so C′ = kC. The two options that divide by C are dimensionally impossible, which is a quick way to eliminate them.

Q29PYQ 2025 · Shift 1

A parallel plate capacitor has plate area 0.25 m² and plate separation 0.002 m, in vacuum, with 15 V across the plates. Taking ε₀ = 8.854 × 10⁻¹² F m⁻¹, if the distance between the plates is reduced to half, calculate the capacitance and the energy stored at the same voltage.

Halving the separation gives d = 0.001 m, so C = 8.854 × 10⁻¹² × 0.25 ÷ 0.001 = 2.21 × 10⁻⁹ F, which is 2.2 nF. The energy is ½CV² = 0.5 × 2.21 × 10⁻⁹ × 225 = 2.49 × 10⁻⁷ J. Both parts must match, and only one option gets the order of magnitude right on each.

Q30PYQ 2021 · Shift 2

When a charge of 3 µC is carried from point A to point B, the work done by the electric field is 75 µJ. Determine the potential difference.

From W = qΔV, the potential difference is 75 µJ ÷ 3 µC = 25 V. Option 1 multiplies instead of dividing, giving 225, which is why the units are worth writing out.

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