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CBSE · Class 12 · Physics

Electric Charges and Fields

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.
Two point charges of 2 μC and 6 μC are 15 cm apart in air. The force between them is: (k = 9 × 109 N m2/C2)
  1. 0.048 N
  2. 48 N
  3. 72 N
  4. 4.8 N
2.
Two charges attract each other with a force F. If each charge is made 3 times as large and the distance between them is made 4 times as large, the force becomes:
  1. (16/9)F
  2. (3/4)F
  3. (9/16)F
  4. (9/8)F
3.
The number of electrons removed from a body that has a charge of +6 μC is: (e = 1.6 × 10−19 C)
  1. 3.75 × 1014
  2. 9.6 × 10−25
  3. 3.75 × 1012
  4. 3.75 × 1013
4.
The electric field 20 cm from a point charge of 8 μC is: (k = 9 × 109 N m2/C2)
  1. 1.8 × 108 N/C
  2. 3.6 × 107 N/C
  3. 1.8 × 106 N/C
  4. 1.8 × 104 N/C
5.
The SI unit of electric charge is the:
  1. coulomb
  2. farad
  3. ampere
  4. volt
6.
A charge of 7 μC is at the centre of a cube. The electric flux through the whole surface of the cube is: (k = 9 × 109 N m2/C2)
  1. 2.52π × 1011 N m2/C
  2. 4.2π × 104 N m2/C
  3. 6.3π × 104 N m2/C
  4. 2.52π × 105 N m2/C
7.
A dipole made of charges ±2 μC, 9 cm apart, makes 90° with a uniform field of 9 × 104 N/C. The torque on it is:
  1. 1.62 × 10−4 N m
  2. 0.0162 N m
  3. 1.62 N m
  4. 0.0324 N m
8.
Assertion (A): The field 3 cm from a 5 μC point charge is 5 × 107 N/C. (k = 9 × 109 N m2/C2)
Reason (R): The field of a point charge is E = kq/r.
  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 −7 μC and −9 μC are at x = 0 and x = 18 cm. Find the net force on a charge of +1 μC at x = 15 cm. (k = 9 × 109 N m2/C2)
10.
Charges of 3 μC and 48 μC are 40 cm apart. Where between them is the electric field zero?
11.
A glass rod rubbed with silk loses 8 × 1013 electrons. Find the charge on the rod and its sign. (e = 1.6 × 10−19 C)
12.
A long straight wire carries a uniform charge of 9 μC/m. Find the electric field 5 cm from it. (k = 9 × 109 N m2/C2)
13.
A uniform field of 700 N/C passes through a flat surface of area 37 cm2. Find the flux if the field makes 60° with the normal to the surface.
14.
A short dipole has dipole moment 9 × 10−9 C m. Find the field 2 cm from its centre (a) on its axis, (b) on its equatorial line. (k = 9 × 109 N m2/C2)
15.
A thin spherical shell of radius 7 cm carries 4 μC. Find the electric field at 1 cm and at 15 cm from its centre. (k = 9 × 109 N m2/C2)
16.
Can a body carry a charge of 24 × 10−19 C? Explain. (e = 1.6 × 10−19 C)
Section C — Word problems4 × 4 = 16 marks
17.
Charges of 3 μC and 4 μC are placed at A and B, each 3 cm from a point C, with CA ⟂ CB. Find the magnitude of the net force on a charge of +4 μC at C. (k = 9 × 109 N m2/C2)
18.
A dipole has charges ±8 nC separated by 4 mm. Find (a) its dipole moment, (b) the field 30 cm away on its axis, (c) the greatest torque on it in a field of 9 × 104 N/C. (k = 9 × 109 N m2/C2)
19.
An oil drop of mass 1.04 × 10−13 kg stays still between two plates where the field is 500,000 N/C, directed upwards. Find the charge on the drop and how many electrons it has lost. (g = 10 m/s2; e = 1.6 × 10−19 C)
20.
A long wire carries 9 μC/m. Find the field at 5 cm and at 30 cm from it, and the force on a charge of 4 μC placed 5 cm away. (k = 9 × 109 N m2/C2)
AmbifulWorksheets
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CBSE · Class 12 · Physics

Answer key

Electric Charges and Fields · AW-12P-ECF-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 4.8 NF = kq1q2/r2 = 9 × 109 × 2 × 10−6 × 6 × 10−6 ÷ (15 × 10−2)2
2(c) (9/16)FF ∝ q1q2/r2: × 9 ÷ 16
3(d) 3.75 × 1013n = q/e = 6 × 10−6 ÷ (1.6 × 10−19)
4(c) 1.8 × 106 N/CE = kq/r2
5(a) coulomb—
6(d) 2.52π × 105 N m2/CΦ = q/ε0 = 4πkq
7(b) 0.0162 N mτ = pE sin θ; p = 2 × 10−6 × 9 × 10−2 C m
8(c) A is true, but R is false.A is true: kq/r2 = 9 × 109 × 5 × 10−6 ÷ (3 × 10−2)2 = 5 × 107 N/C. R is false.
987.2 N towards x = 18 cmfrom the first: −2.8 N; from the second: 90 N (positive means towards x = 18 cm)
108 cm from the 3 μC chargeq1/x2 = q2/(40 − x)2 ⇒ x/(40 − x) = 1/4
11+1.28 × 10−5 Cq = ne = 8 × 1013 × 1.6 × 10−19; losing electrons leaves it positive
123.24 × 106 N/CE = 2kλ/r = 2 × 9 × 109 × 9 × 10−6 ÷ (5 × 10−2)
131.295 N m2/CΦ = EA cos θ = 700 × 37 × 10−4 × cos 60°
14(a) 2.025 × 107 N/C; (b) 1.0125 × 107 N/CE(axis) = 2kp/r3; E(equator) = kp/r3
150 at 1 cm (inside); 1.6 × 106 N/C at 15 cmGauss: no charge is enclosed inside; outside, the shell acts like a point charge
16Yes: it is 15eq = ne with n a whole number
17200 N, at tan−1(4/3) to the line ACforces 120 N and 160 N at right angles; √(32 + 42) = 5
18(a) 3.2 × 10−11 C m; (b) 21.3 N/C; (c) 2.88 × 10−6 N mp = q × 2a; E = 2kp/r3; τmax = pE
19q = mg/E = 2.08 × 10−18 C (positive, since the force is upwards along the field); it has lost 13 electronsqE = mg ⇒ q = 1.04 × 10−12 ÷ 500,000
203.24 × 106 N/C; 5.4 × 105 N/C; 12.96 NE = 2kλ/r; F = qE