Question

A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of three charges will be in equilibrium if $q$ is equal to:

• A. $-\frac{Q}{2}$
• B. $-\frac{Q}{4}$
• C. $+\frac{Q}{4}$
• D. $+\frac{Q}{2}$

Answer 1

Answer: (B)

$-\frac{Q}{4}$

For $Q$ to be in equilibrium, net force acting on $Q$ must be $0$.

So, $F=\frac{kQq}{(\frac{r}{2}{)}^2}+\frac{k{Q}^2}{r^2}=0$

$\Rightarrow \frac{4q}{r^2}+\frac{Q}{r^2}=0$

$\Rightarrow q=-\frac{Q}{4}$