Electrostatics and Current Electricity — Practice

Question 1

The force between two point charges of 1 μC each, separated by 1 m in vacuum, is (k = 9 × 10⁹):
(a) 9 × 10⁻³ N (b) 9 × 10⁹ N (c) 9 × 10⁻⁶ N (d) 9 N

Accepted Answer

(a) 9 × 10⁻³ N.

F = kq₁q₂/r² = 9 × 10⁹ × (10⁻⁶)² / 1² = 9 × 10⁹ × 10⁻¹² = 9 × 10⁻³ N.

Question 2

If the distance between two charges is doubled, the electrostatic force between them becomes:
(a) one-quarter (b) one-half (c) double (d) four times

Accepted Answer

(a) one-quarter.

F ∝ 1/r². Doubling r multiplies the force by 1/2² = 1/4.

Question 3

The electric field at 1 m from a point charge of 1 μC is (k = 9 × 10⁹):
(a) 9000 N/C (b) 9 × 10⁻³ N/C (c) 9 × 10⁹ N/C (d) 900 N/C

Accepted Answer

(a) 9000 N/C.

E = kQ/r² = 9 × 10⁹ × 10⁻⁶ / 1² = 9 × 10³ = 9000 N/C, directed away from a positive charge.

Question 4

Two capacitors of 2 μF each are connected in parallel. The equivalent capacitance is:
(a) 4 μF (b) 1 μF (c) 2 μF (d) 0.5 μF

Accepted Answer

(a) 4 μF.

In parallel, capacitances add: C = 2 + 2 = 4 μF.

Question 5

Two capacitors of 2 μF each are connected in series. The equivalent capacitance is:
(a) 1 μF (b) 4 μF (c) 2 μF (d) 0.5 μF

Accepted Answer

(a) 1 μF.

In series, 1/C = 1/2 + 1/2 = 1, so C = 1 μF — capacitors combine oppositely to resistors.

Question 6

The energy stored in a 2 μF capacitor charged to 10 V is:
(a) 1 × 10⁻⁴ J (b) 2 × 10⁻⁴ J (c) 1 × 10⁻² J (d) 20 J

Accepted Answer

(a) 1 × 10⁻⁴ J.

E = ½CV² = ½ × 2 × 10⁻⁶ × 10² = ½ × 2 × 10⁻⁶ × 100 = 1 × 10⁻⁴ J.

Question 7

A 12 V battery drives current through a 4 Ω resistor. The current is:
(a) 3 A (b) 48 A (c) 0.33 A (d) 8 A

Accepted Answer

(a) 3 A.

I = V/R = 12/4 = 3 A.

Question 8

If the length of a wire is doubled (with the cross-sectional area unchanged), its resistance:
(a) doubles (b) halves (c) is unchanged (d) quadruples

Accepted Answer

(a) doubles.

R = ρL/A ∝ L at fixed area, so doubling the length doubles the resistance.

Question 9

Two 6 Ω resistors are connected in parallel. The equivalent resistance is:
(a) 3 Ω (b) 12 Ω (c) 6 Ω (d) 1.5 Ω

Accepted Answer

(a) 3 Ω.

For two equal resistors in parallel, R = R/2 = 6/2 = 3 Ω (or 1/R = 1/6 + 1/6 = 1/3).

Question 10

A 100 W bulb is rated for 200 V. Its resistance (at operating conditions) is:
(a) 400 Ω (b) 200 Ω (c) 2 Ω (d) 100 Ω

Accepted Answer

(a) 400 Ω.

R = V²/P = 200²/100 = 40 000/100 = 400 Ω. It draws I = P/V = 100/200 = 0.5 A.

Question 11

The power dissipated in a 5 Ω resistor carrying 2 A is:
(a) 20 W (b) 10 W (c) 2.5 W (d) 40 W

Accepted Answer

(a) 20 W.

P = I²R = 2² × 5 = 4 × 5 = 20 W.

Question 12

A cell of EMF 6 V and internal resistance 1 Ω delivers a current of 2 A. Its terminal voltage is:
(a) 4 V (b) 6 V (c) 8 V (d) 2 V

Accepted Answer

(a) 4 V.

V = EMF − Ir = 6 − (2 × 1) = 4 V. The internal resistance drops 2 V under this load.

Question 13

At a junction, currents of 3 A and 2 A flow in along two wires, and a single wire carries current out. The outgoing current is:
(a) 5 A (b) 1 A (c) 6 A (d) 2.5 A

Accepted Answer

(a) 5 A.

By the junction rule (charge conservation), current in = current out, so the outgoing current is 3 + 2 = 5 A.

Question 14

In a balanced Wheatstone bridge, P = 10 Ω, Q = 20 Ω and S = 40 Ω. The unknown resistance R is:
(a) 20 Ω (b) 80 Ω (c) 10 Ω (d) 5 Ω

Accepted Answer

(a) 20 Ω.

Balance condition P/Q = R/S gives R = P·S/Q = 10 × 40 / 20 = 20 Ω (no current then flows through the galvanometer).

Question 15

The electric potential at 1 m from a point charge of 1 μC is (k = 9 × 10⁹):
(a) 9000 V (b) 9 × 10⁹ V (c) 9 × 10⁻³ V (d) 900 V

Accepted Answer

(a) 9000 V.

V = kQ/r = 9 × 10⁹ × 10⁻⁶ / 1 = 9 × 10³ = 9000 V. Potential is a scalar and falls as 1/r (one power less than the field).

Practice — Electrostatics and Current Electricity

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