Question

Two equal positive point charges $Q$ are fixed at points $B(a,0)$ and $A(-a,0)$. Another test charge $q_0$ is placed at $O(0,0)$. Then select the most appropriate statement regarding the equilibrium position of $q_0$ for displacement along $x-$ axis.

• A. It is not an equilibrium position
• B. It is a stable equilibrium position
• C. It is a neutral equilibrium position
• D. It is an unstable equilibrium position

Answer 1

The correct option is B It is a stable equilibrium position

Assume that the charge $q_o$ is positive.


The force exerted by charge at $A$ on charge at $O$ is $F_2$ along $AB$ and the force exerted by charge at $B$ on charge at $O$ is $F_1$ along $BA$
So,
$|F_1|=|F_2|=\frac{kQ{q}_0}{a^2}$
At position $O$, the $F_{\text{net}}=0$ for $q_0$.
Thus, it is an equilibrium position because if it is displaced slightly along -ve $x-$ axis, repulsive force $F_2$ increases as separation between charges decreases. Thus, restoring force will tend to bring it back to position $O$.
Similarly if displaced towards +ve $x-$ axis, repulsive force $F_1$ increases and will tend to restore its position
$\therefore$ position of charge $q_0$ at $O(0,0)$ is stable equilibrium.

$\begin{array}{c}\text{Why this question ?}\\ \text{Even if the charge is assumed to be negative the answer would have remain same.}\\ \text{Instead of repulsive force, attractive forces will be the restoring force.}\end{array}$