Question

A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the 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

The correct option is B

$-\frac{Q}{4}$

For the equilibrium of charge $Q$, the force applied by both the charges i.e. another $Q$ and $q$ should be equal in magnitude and opposite in direction. This is only possible when $q$ has negative charge.

If $d$ is the distance between both pairs of $q$ and $Q$, then we can use coloumbs law:
$F_{Qq}=-F_{QQ}$
$\frac{1}{4\pi {\epsilon }_0}.\frac{qQ}{d^2}=\frac{1}{4\pi {\epsilon }_0}.\frac{Q^2}{(2d{)}^2}$

$⟹q=-\frac{Q}{4}$