Physics · Electrostatics · chapter 2 of 14

Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance worksheet with answer key

A 20-mark worksheet on Electrostatic Potential and Capacitance for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

The equipotential surfaces of an isolated point charge are

(a) parallel planes perpendicular to a line through the charge(b) coaxial cylinders with the charge on their axis(c) planes containing the charge(d) concentric spheres centred at the charge

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Answer: (d) concentric spheres centred at the charge

V = q/(4πε₀r) depends only on r, so V is constant wherever r is constant: on spheres centred at the charge. The radial field lines cross them normally.

Question 2

Capacitors of 6 μF and 3 μF are connected in series. The capacitance of the combination is

(a) 9 μF(b) 4.5 μF(c) 18 μF(d) 2 μF

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Answer: (d) 2 μF

1/C = 1/6 + 1/3 = 1/2, so C = 2 μF. 9 μF is the parallel value; in series the combination is smaller than the smallest capacitor.

Question 3

The dimensional formula of capacitance is

(a) [M L² T⁻³ A⁻¹](b) [M⁻¹ L⁻² T² A](c) [M L² T⁻⁴ A⁻²](d) [M⁻¹ L⁻² T⁴ A²]

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Answer: (d) [M⁻¹ L⁻² T⁴ A²]

C = Q/V: charge is [A T] and potential is [M L² T⁻³ A⁻¹], so C = [A T]/[M L² T⁻³ A⁻¹] = [M⁻¹ L⁻² T⁴ A²]. [M L² T⁻³ A⁻¹] is the dimension of potential itself.

Question 4

A capacitor holds a charge of 4.0 mC when the potential difference across it is 200 V. The energy stored in it is

(a) 0.20 J(b) 0.40 J(c) 0.80 J(d) 4.0 mJ

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Answer: (b) 0.40 J

U = ½QV = ½ × 4.0 × 10⁻³ × 200 = 0.40 J. QV alone gives 0.80 J; halving twice gives 0.20 J.

Question 5

The plates of an air-filled parallel plate capacitor each have an area of 4.0 × 10⁻² m² and are 0.885 mm apart. Take ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻². Its capacitance is

(a) 400 pF(b) 40 pF(c) 4000 pF(d) 0.40 pF

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Answer: (a) 400 pF

C = ε₀A/d = 8.85 × 10⁻¹² × 4.0 × 10⁻² / (0.885 × 10⁻³) = 4.0 × 10⁻¹⁰ F = 400 pF.

Question 6 · 2 marks

Distinguish between polar and non-polar molecules, giving one example of each, and state how each kind develops a dipole moment in an external electric field.

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In a non-polar molecule (O₂, H₂) the centres of positive and negative charge coincide, so it has no permanent dipole moment; a field displaces the charges in opposite directions and induces a dipole moment. In a polar molecule (H₂O, HCl) the centres are separated even without a field, giving a permanent dipole moment; a field tends to align these randomly oriented dipoles, against thermal agitation.

definitions with one example of each (1)
induced dipole for non-polar; alignment of permanent dipoles for polar (1)

Question 7 · 2 marks

Capacitors of capacitance C and 2C are connected in series across a battery of emf V. Find the charge on each capacitor and the potential difference across each.

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Series: 1/C_s = 1/C + 1/2C = 3/2C, so C_s = 2C/3, and each carries the same charge Q = 2CV/3. Then V₁ = Q/C = 2V/3 across C and V₂ = Q/2C = V/3 across 2C; they add to V. The smaller capacitor takes the larger share of the voltage.

equivalent 2C/3 and the same charge 2CV/3 on each (1)
2V/3 across C and V/3 across 2C (1)

Question 8 · 2 marks

An electron and a proton, each starting from rest, are accelerated through a potential difference of 2.0 kV. (a) Find the kinetic energy each gains, in eV and in joule (e = 1.6 × 10⁻¹⁹ C). (b) Taking the proton's mass as 1836 times the electron's, find the ratio of the electron's speed to the proton's.

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(a) Each has charge of magnitude e and falls through 2.0 kV, so each gains eΔV = 2.0 keV = 2.0 × 10³ × 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹⁶ J. (b) Equal kinetic energies: ½m_e v_e² = ½m_p v_p², so v_e/v_p = √(m_p/m_e) = √1836 ≈ 43.

2.0 keV = 3.2 × 10⁻¹⁶ J for each (1)
v_e/v_p = √1836 ≈ 43 (1)

Question 9 · 3 marks

A 10 μF parallel plate capacitor is charged to 100 V and kept connected to the 100 V battery. A slab of dielectric constant 3 is slid in to fill the gap completely. Find (a) the extra charge that flows from the battery, (b) the increase in the energy stored in the capacitor, and (c) the work done by the battery in sending this charge round, and compare it with (b).

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(a) Q₀ = CV = 10 × 10⁻⁶ × 100 = 1.0 × 10⁻³ C. With the battery connected V is fixed and C becomes 30 μF, so Q = 3.0 × 10⁻³ C: an extra 2.0 × 10⁻³ C flows. (b) U₀ = ½ × 10 × 10⁻⁶ × 100² = 0.050 J; U = ½ × 30 × 10⁻⁶ × 100² = 0.15 J; increase 0.10 J. (c) The battery moves 2.0 × 10⁻³ C through a potential difference of 100 V: W = qV = 0.20 J, twice the increase in stored energy; the other 0.10 J does not stay in the capacitor's field.

extra charge 2.0 × 10⁻³ C (1)
stored energy rises from 0.050 J to 0.15 J, by 0.10 J (1)
battery work 0.20 J, twice the increase (1)

Question 10 · 3 marks

An air-filled parallel plate capacitor has plates of area 2.0 × 10⁻² m², 2.0 mm apart. Air breaks down in a field of about 3 × 10⁶ V m⁻¹. Take ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻². Find (a) its capacitance, (b) the largest potential difference it can safely take, and (c) the largest charge it can then hold.

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(a) C = ε₀A/d = 8.85 × 10⁻¹² × 2.0 × 10⁻²/2.0 × 10⁻³ = 8.85 × 10⁻¹¹ F ≈ 89 pF. (b) V_max = E_max d = 3 × 10⁶ × 2.0 × 10⁻³ = 6.0 × 10³ V. (c) Q_max = CV_max = 8.85 × 10⁻¹¹ × 6.0 × 10³ ≈ 5.3 × 10⁻⁷ C. Beyond this the air ionises and the charge leaks away.

C ≈ 8.9 × 10⁻¹¹ F (1)
V_max = 6.0 × 10³ V (1)
Q_max ≈ 5.3 × 10⁻⁷ C (1)

Question 11 · 3 marks

The gap d between the plates of a parallel plate capacitor (air capacitance C₀) is filled with two dielectric layers, each of thickness d/2 and parallel to the plates, of dielectric constants 2 and 4. Show that the capacitance becomes (8/3)C₀.

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Each layer acts as a capacitor of plate area A and separation d/2, and the two are in series (the same charge sits on each). C₁ = 2ε₀A/(d/2) = 4C₀ and C₂ = 4ε₀A/(d/2) = 8C₀. Then 1/C = 1/4C₀ + 1/8C₀ = 3/8C₀, so C = (8/3)C₀.

layers treated as two capacitors in series (1)
C₁ = 4C₀ and C₂ = 8C₀ (1)
1/C = 3/8C₀, so C = 8C₀/3 (1)

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