Three point charges of equal value q are placed at the vertices of al equilateral triangle. The resulting lines of force should be sketch as in
 

When Q is at position x, then
$\begin{array}{l}\mathrm{U}=\frac{\mathrm{qQ}}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{(\mathrm{a}+\mathrm{x})}+\frac{1}{(\mathrm{a}-\mathrm{x})}\right]\\ =\frac{2\mathrm{aqQ}}{4{\mathrm{\pi \epsilon }}_0\left({\mathrm{a}}^2-{\mathrm{x}}^2\right)}\end{array}$
$\therefore \frac{\mathrm{dU}}{\mathrm{dx}}=\frac{2\mathrm{aqQ}}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{2\mathrm{x}}{{\left({\mathrm{a}}^2-{\mathrm{x}}^2\right)}^2}\right]=\frac{2\mathrm{aqQ}}{4{\mathrm{\pi \epsilon }}_0}\times \frac{2\mathrm{x}}{{\mathrm{a}}^4}$
$\left(\because {\mathrm{a}}^2-{\mathrm{x}}^2={\mathrm{a}}^2\text{ when }\mathrm{x}\text{ is small }\right)$
or   $\mathrm{dU}\propto \mathrm{xdx}\text{ or }\mathrm{U}\propto {\mathrm{x}}^2$.