Explanation

(c) $F = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1\ }q_{2}}{r^{2}}$
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Taking the net force, we have
$F_{net} = 2Fcos\theta = 2\left\lbrack \frac{1}{4\pi\varepsilon_{0}}\frac{\mathcal{Q}qx}{\left( x^{2} + \frac{d^{2}}{4} \right)^{\frac{3}{2}}} \right\rbrack$
For maximum, $\frac{dF_{net}}{dx} = 0$
$\therefore\left( x^{2} + \frac{d^{2}}{4} \right)^{3/2} - \frac{3}{2}$ $x\left\lbrack x^{2} + \frac{d^{2}}{4} \right\rbrack^{1/2}(2x) = 0$
$\therefore\left( x^{2} + \frac{d}{4}^{2} \right)^{\frac{1}{2}}\left( x^{2} + \frac{d^{2}}{4} - 3x^{2} \right) = 0$
or$\ \ \ \ \ 2x^{2} = \frac{d^{2}}{4}$
$x = \frac{d}{2\sqrt{2}}$