b
(b) Initially according to figure \((i)\) potential energy of \(Q\) is \({U_i} = \frac{{2kqQ}}{a}\) ......\((i)\)
According to figure \((ii)\) when charge \(Q\) is displaced by small distance \(x\) then it’s potential energy now
\({U_f} = kqQ\,\left[ {\frac{1}{{(a + x)}} + \frac{1}{{(a - x)}}} \right]\)\( = \frac{{2kqQa}}{{({a^2} - {x^2})}}\) .......\((ii)\)
Hence change in potential energy
\(\Delta U = {U_f} - {U_i} = 2kqQ\,\left[ {\frac{a}{{{a^2} - {x^2}}} - \frac{1}{a}} \right]\)\( = \frac{{2kqQ{x^2}}}{{({a^2} - {x^2})}}\)
Since \(x << a\) so \(\Delta U = \frac{{2kqQ{x^2}}}{{{a^2}}} \Rightarrow \Delta U \propto {x^2}\)