Question

A charge $Q$ is placed at a distance $L$ from the charge $-2q$ as shown in figure.

The system of three charges will be in equilibrium, if $Q$ is equal to

• A. $-4q$
• B. $+4q$
• C. $+8q$
• D. $-8q$

Answer 1

The correct option is C $+8q$

Since, the system of three charges is in equilibrium, so net forces on $-2q$ will be zero.
$\because$ direction of force $\left(F_{21}\right)$ on $-2q$ due to $+8q$ will be on the left side as shown in figure. So, the direction of force $F_{23}$ will be acted towards the $Q$, that means $Q$ attracting charge $-2q$.

$\therefore$ $Q$ charge will be positive in nature.

Since, $-2q$ is at equilibrium. So, from Coulomb's law,

$|F_{21}|=|F_{23}|$

$\Rightarrow \frac{k\left(8q\right)\left(2q\right)}{L^2}=\frac{k\left(2q\right)\left(Q\right)}{L^2}$

$\Rightarrow Q=8q$

$\therefore Q=+8q$

Hence, option $\left(c\right)$ is the correct answer.