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CBSE · Class 12 · Physics

Electrostatic Potential and Capacitance

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The potential 15 cm from a point charge of −2 μC is: (k = 9 × 109 N m2/C2)
  1. 1.2 × 105 V
  2. −8 × 105 V
  3. −1,200 V
  4. −1.2 × 105 V
2.
The work needed to carry a charge of 9 μC from a point at 470 V to a point at 500 V is:
  1. 2.7 × 10−4 J
  2. −2.7 × 10−4 J
  3. 8.73 × 10−3 J
  4. 0.27 J
3.
The equivalent capacitance of 1 μF, 10 μF joined in series is:
  1. 10 μF
  2. 11 μF
  3. 0.91 μF
  4. 1.82 μF
4.
A 1 μF capacitor is charged to 270 V. The energy stored in it is:
  1. 0.03645 J
  2. 0.0729 J
  3. 1.35 × 10−4 J
  4. 36.45 J
5.
The SI unit of electric potential is the:
  1. volt
  2. farad
  3. newton
  4. coulomb
6.
A 12 μF air capacitor is filled with a dielectric of constant 4. Its capacitance becomes:
  1. 192 μF
  2. 3 μF
  3. 48 μF
  4. 16 μF
7.
The electrostatic potential energy of charges −5 μC and 5 μC, 3 cm apart, is: (k = 9 × 109 N m2/C2)
  1. −7.5 J
  2. −75 J
  3. 7.5 J
  4. −2.5 J
8.
Assertion (A): 7 μF and 2 μF capacitors in series have an equivalent capacitance of 9 μF.
Reason (R): For capacitors in series, 1/C = 1/C1 + 1/C2.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Charges of −2 μC and −8 μC are at x = 0 and x = 24 cm. Find the potential at x = 16 cm. (k = 9 × 109 N m2/C2)
10.
9 μF and 5 μF capacitors in parallel are joined in series with a 7 μF capacitor across 190 V. Find the equivalent capacitance and the charge on the 7 μF capacitor.
11.
A 16 μF capacitor is charged to 170 V. A dielectric of constant 6 is then slid in after the battery is removed. Find the energy stored before and after.
12.
How much work is done in moving a charge of 3 μC from 3 cm to 15 cm from a fixed charge of 4 μC? (k = 9 × 109 N m2/C2)
13.
A parallel-plate capacitor has plates of area 17 × 10−2 m2, 2 mm apart, in air. Find its capacitance. (ε0 = 8.85 × 10−12 C2/N m2)
14.
A 12 μF capacitor charged to 140 V is joined to an uncharged 10 μF capacitor. Find the common potential and the energy lost.
15.
Charges of +8 μC and −3 μC are 44 cm apart. Where on the line between them is the potential zero?
16.
Two parallel plates 4 mm apart are kept at a potential difference of 150 V. Find the field between them and the force on a charge of 2 μC there.
Section C — Word problems4 × 4 = 16 marks
17.
11 μF and 12 μF capacitors in series are joined in parallel with 8 μF; this group is in series with 11 μF, across 160 V. Find the equivalent capacitance, the total charge and the energy stored.
18.
A parallel-plate capacitor (plate area 18 × 10−2 m2, gap 3 mm, air) is charged to 180 V and the battery is removed. A dielectric of constant 5 then fills the gap. Find the capacitance, charge and potential difference after this. (ε0 = 8.85 × 10−12 C2/N m2)
19.
Charges of 5 μC, 9 μC and −7 μC sit at the corners of an equilateral triangle of side 9 cm. Find the potential energy of the system. (k = 9 × 109 N m2/C2)
20.
A 5 μF capacitor at 20 V and a 1 μF capacitor at 50 V are joined positive to positive. Find the common potential, the charge on each, and the energy lost.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Physics

Answer key

Electrostatic Potential and Capacitance · AW-12P-EPC-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) −1.2 × 105 VV = kq/r
2(a) 2.7 × 10−4 JW = q(V_B − V_A) = 9 × 10−6 × (30)
3(c) 0.91 μF1/C = Σ 1/Ci
4(a) 0.03645 JU = ½CV2
5(a) volt—
6(c) 48 μFC = KC0
7(a) −7.5 JU = kq1q2/r = 9 × 109 × −5 × 10−6 × 5 × 10−6 ÷ (3 × 10−2)
8(d) A is false, but R is true.A is false: 1/C = 1/7 + 1/2, so C = 1.56 μF. R is true.
9−1.0125 × 106 VV = k(q1/r1 + q2/r2), with r1 = 16 cm and r2 = 8 cm
104.67 μF; 8.87 × 10−4 Cparallel pair 14 μF; series with 7 μF; Q = CV
11before 0.2312 J; after 0.0385 JQ stays 2720 μC, C becomes 96 μF, so U = Q2/2C falls 6 times
12−2.88 JW = q(V_B − V_A); V = kQ/r
137.5225 × 10−10 FC = ε0A/d = 8.85 × 10−12 × 17 × 10−2 ÷ (2 × 10−3)
1476.36 V; 0.0535 JV = C1V1/(C1 + C2); loss = ½C1V12 − ½(C1 + C2)V2
1532 cm from the +8 μC charge8/x = 3/(44 − x)
163.75 × 104 V/m; 0.075 NE = V/d; F = qE
176.11 μF; 9.77 × 10−4 C; 0.0782 Jgroup = 5.74 + 8 = 13.74 μF
18C = 2.655 × 10−9 F; Q = 9.558 × 10−8 C (unchanged); V = 36 VC0 = 5.31 × 10−10 F; Q = C0V; C = KC0; V = Q/C
19−5.3 JU = k(q1q2 + q2q3 + q3q1)/a; q1q2 + q2q3 + q3q1 = −53
2025 V; 125 μC and 25 μC; 3.75 × 10−4 JV = (C1V1 + C2V2)/(C1 + C2)