Two equal point charges are fixed at $x = -a$ and $x = \pm a$ 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 $\mathbf{X}$ along the x-axis, is approximately proportional to
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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
$U_{f} = kqQ \left[\frac{1}{(a+x)} + \frac{1}{(a-x)}\right]$ $= \frac{2kqQa}{\left(a^{2} - x^{2}\right)}$ .......(ii)
Hence change in potential energy
$\Delta U = U_f - U_i = 2kqQ \left[ \frac{a}{a^2 - x^2} - \frac{1}{a} \right]$ $= \frac{2kqQx^{2}}{\left(a^{2} - x^{2}\right)}$
Since x << a so $\Delta U = \frac{2 k q Q x^{2}}{a^{2}} \Rightarrow \Delta U \propto x^{2}$
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