Physics - Electrostatics Question with Solution | TestHub

Let the two equal charges $Q$ be placed at positions $x=-a$ and $x=a$.

The charge $q$ is placed at the centre, $x=0$.

For the system to be in equilibrium, the net force on each charge must be zero.

The force on the central charge $q$ is zero due to symmetry, as it experiences equal and opposite forces from the two $Q$ charges.

For the charge $Q$ at $x=a$ to be in equilibrium, the sum of the forces exerted by the other charge $Q$ (at $x=-a$) and the charge $q$ (at $x=0$) must be zero.

The distance between the two $Q$ charges is $2a$, and the distance between $Q$ and $q$ is $a$.

Equating the net force on $Q$ to zero:

$\frac{1}{4\pi {ϵ}_0}\frac{Q^2}{(2a{)}^2}+\frac{1}{4\pi {ϵ}_0}\frac{Qq}{a^2}=0$

$\frac{Q^2}{4a^2}+\frac{Qq}{a^2}=0$

$\frac{Q}{4}+q=0$

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

Answer: $\frac{-Q}{4}$