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} $
$F’ = \frac{F}{8}$