Question

Three charges $+Q$, $q$, $+Q$ are placed respectively, at distance, $0,\frac{d}{2}$ and $d$ from the origin, on the $x-$axis. If the net force experienced by $+Q$, placed at $x=0$, is zero, then the value of $q$ is:

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

Answer 1

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


Force due to charge $+Q$,

$F_a=\frac{\text{KQ.Q}}{d^2}$

Force due to charge $q$,

$F_b=\frac{\text{KQq}}{{\left(\frac{d}{2}\right)}^2}$

For equilibrium,
$\overrightarrow{F_a}+\overrightarrow{F_b}=0$

$\Rightarrow \frac{k{Q}^2}{d^2}+\frac{kQq}{(d/2{)}^2}=0\therefore q=-\frac{Q}{4}$

Hence, option $\left(A\right)$ is correct.