Question

A charge q is placed at the centre of the line joining two exactly equal positive charges Q. The systems of these charges will be in equilibrium, if q is equal to

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

Answer 1

Answer: (A)

$-\frac{Q}{4}$

At equilibrium, net force will be zero, hence
$\therefore \frac{1}{4\pi {\epsilon }_0}\times \frac{Qq}{(r/2{)}^2}+\frac{1}{4\pi {\epsilon }_0}\times \frac{Q^2}{r^2}=0$

$4q=-Q$ or $q=-\frac{Q}{4}$.