Q: Two identical metals balls with charge +2Q and -Q are separated by some distance and exert a force F on each other. They are joined by a conducting wire, which is then removed. The force between them will not be

(a) F

(b) F/2

(c) F/4

(d) F/8

Ans: (d)

Sol: $\large F = \frac{1}{4\pi \epsilon_0} \frac{2Q \times Q}{r^2}$

When they are joined by a conducting wire , New charge on each is

$\large Q’ = \frac{2Q + (-Q)}{2} = \frac{Q}{2}$

$\large F’ = \frac{1}{4\pi \epsilon_0} \frac{Q/2 \times Q/2}{r^2}$

$\large \frac{F’}{F} = \frac{1}{8} $

$F’ = \frac{F}{8}$