Question

Two positive point charges of 12μc and 8μc are placed 10cm apart in air. The work done to bring them 4 cm closer is

(A) Zero

(B) 4.8J

(C) 3.5J

(D) – 5.8J

Answer 1

The correct option (D) – 5.8 J

Explanation:

initial potential Energy = [(kQ1Q2)/r1]

PE1 = [{k(12)(8)(10–12)}/(0.1)]

They are brought 4 cm closer means distance is 6 cm

∴ PE2 = [{k(12)(8)(10–12)}/(0.6)]

∴ workdone = ∆ Energy

= PE1 – PE2

= K × 12 × 8 ×10–12 [(1/0.1) – (1/0.6)]

= 96 × 10–12 × 9 × 109 [– 6.66]

= – 5760 × 10–3

= – 5.76J