Question

A charge $q$ is placed in the middle of a line joining the two equal and like point charges $Q$. This system will remain in equilibrum for which the value of $q$ is-

• A. $\frac{-Q}{3}$
• B. $\frac{-Q}{4}$
• C. $\frac{Q}{2}$
• D. $\frac{-Q}{2}$

Answer 1

The correct option is B $\frac{-Q}{4}$

If the system is in equilibrum then the net electric field at $B$ is zero due to other two charges.

$\frac{K{Q}^2}{l^2}+\frac{KqQ}{(l/2{)}^2}=0$

or $\frac{K{Q}^2}{l^2}=-\frac{4KqQ}{l^2}$
So, $q=\frac{-Q}{4}$