Explanation

(c) Suppose third charge is similar to $Q$ and it is $q$
So net force on it
$F_{net} = 2F\cos\theta$
[image:1_image349.wmf][image:1_image349.wmf][image:1_image350.wmf][image:1_image350.wmf]
Where $F = \frac{1}{4\pi\varepsilon_{0}}.\frac{Qq}{\left( x^{2} + \frac{d^{2}}{4} \right)}$ and $\cos\theta = \frac{x}{\sqrt{x^{2} + \frac{d^{2}}{4}}}$
$\therefore F_{net} = 2 \times \frac{1}{4\pi\varepsilon_{0}}.\ \frac{Qq}{\left( x^{2} + \frac{d^{2}}{4} \right)} \times \frac{x}{\left( x^{2} + \frac{d^{2}}{4} \right)^{1/2}}$
$= \frac{2Qqx}{4\pi\varepsilon_{0}\left( x^{2} + \frac{d^{2}}{4} \right)^{3/2}}$
For $F_{net}$ to be maximum $\frac{dF_{net}}{dx} = 0$
$i.e.\frac{d}{dx}\left\lbrack \frac{2Qqx}{4\pi\varepsilon_{0}\left( x^{2} + \frac{d^{2}}{4} \right)^{3/2}} \right\rbrack = 0$
or $\left\lbrack \left( x^{2} + \frac{d^{2}}{4} \right)^{- 3/2} - 3x^{2}\left( x^{2} + \frac{d^{2}}{4} \right)^{- 5/2} \right\rbrack = 0$
$i.e.\ \ x = \pm \frac{d}{2\sqrt{2}}$