Two equal point charges are fixed at x = –a and x = a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to :

x²

$U_t=\frac{2 K Q q}{a}$

and $U_t=K Q q\left[\frac{1}{a+x}+\frac{1}{a-x}\right]$

Here, $K =\frac{1}{4 \pi \varepsilon_0}$

$\Delta U=U_f-U_t$

or  $|\Delta U|=\frac{2 K Q q x^2}{a^3}$ for x << a

∴ $\Delta U \propto x^2$