Question

A charge $Q$ is placed at each of the two opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the resultant force on each charge $q$ is zero, then

• A. $q=\sqrt{2}Q$
• B. $q=-\sqrt{2}Q$
• C. $q=2\sqrt{2}Q$
• D. $q=-2\sqrt{2}Q$

Answer 1

The correct option is D $q=-2\sqrt{2}Q$


From the configuration of charges shown in figure, the charge $q$ will experience repulsive force from the charge $\left(q\right)$ at the diagonally opposite end.
Thus, to achieve $F_{\text{net}}=0$, the adjacent charge must be of opposite nature.
Also, due to symmetry the attractive force $\left(F\right)$ will be equal from both charges $\left(Q\right)$.

The force experienced by charge $q$ due to one of the charges $Q$ is given by
$F=\frac{kQq}{a^2}$
The component of the force $F$, along the diagonal = $F\cos {45}^{\circ }$
$=\frac{F}{\sqrt{2}}$

Force exerted by charge $q$ on other charge $q$ is
$F_1=\frac{k\left(q\right)\left(q\right)}{(\sqrt{2}a{)}^2}$
$\Rightarrow F_1=\frac{k{q}^2}{2a^2}$
Applying equilibrium condition on charge $q$,
$F\sqrt{2}=F_1$
$\Rightarrow \sqrt{2}\left(\frac{kQq}{a^2}\right)=\frac{k{q}^2}{2a^2}$
$\Rightarrow \sqrt{2}Q=\frac{q}{2}$
$\therefore q=2\sqrt{2}Q$
Since $q$ and $Q$ are of opposite nature,
$\Rightarrow q=-2\sqrt{2}Q$
Hence, option (d) is correct.