A charge Q is fixed at each of two opposite corners of a square. A charge q is placed at each of the other two corners. If the resultant electric force on Q is zero, then Q and q are related as

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$Q = 2 \sqrt{2} q$

$Q = - \sqrt{2} q$

$Q = \sqrt{2} q$

$Q = - 2 \sqrt{2} q$

answer is B.

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Detailed Solution

Here $\frac{Q^{2}}{(\sqrt{2} a)^{2}} = - \frac{2 Qqcos 45^{\circ}}{a^{2}}$
or
$\begin{aligned} \frac{Q^{2}}{2 a^{2}} = - \frac{2 Qq}{\sqrt{2} a^{2}} \\ Q = - 2 \sqrt{2} q \end{aligned}$

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