Explanation

(b) Potential energy of charges $q_{1}\ and\ q_{2},r$ distance apart
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U = \ \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r}$
For $r = 0.1$m,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{12 \times 10^{- 6} \times 5 \times 10^{- 6}}{0.1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{9 \times 10^{9} \times 60 \times 10^{- 12}}{0.1} = 5.4\ J$
For $r = 0.06m,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{2} = \frac{9 \times 10^{9} \times 60 \times 10^{- 12}}{0.06}\ = 9\ J\ $
$\therefore$ Work done = (9 – 5.4) J = 3.6 J