Question

A charge $q$ is placed at the centre of line joining two equal point charges each equal to $+Q$. The system of $3$ charges will be in equilibrium if $q$ is equal to

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

Answer 1

Answer: (A)

The situation is as shown in the figure.

Let two equal charges Q each placed at points A and B at a distance r apart. C is the centre of AB where charge q is placed.
For equilibrium, net force on charge Q = 0
$\therefore \frac{1}{\pi {\epsilon }_0}\frac{1}{4\pi {\epsilon }_0}{\frac{Qq}{r/2}}^2=0$ $\therefore \frac{1}{\pi {\epsilon }_0}\frac{QQ}{r^2}+\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{(r/2{)}^2}=0$
$\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{r^2}=-\frac{1}{4\pi {\epsilon }_0}\frac{4Qq}{r^2}$ or $Q=-4q$ or $q=-\frac{Q}{4}$