Question

Which of the following arrangement shown in the diagram shows the analogue of stable equilibrium of a charge $(+q)$ placed at mid of line joining two fixed charges $(+Q)$ and displaced axially?
$[$Ball is in equilibrium in each case$]$

• A. $A$
• B. $B$
• C. $C$
• D. All the above

Answer 1

The correct option is A $A$

For the given situations of charges,
Here, $F_{32}>F_{31}$
$\Rightarrow {\overrightarrow{F}}_{net}={\overrightarrow{F}}_{32}-{\overrightarrow{F}}_{31}$
$($towards the equilibrium position$)$
Therefore, it is a stable equilibrium.

For $A$:
When the ball is displaced from its equilibrium position, it will try to come at its original position.
Thus, it is in stable equilibrium.

For $B$:
When the ball is displaced from its equilibrium position, it will never come back to its original position.
Thus, it is in an unstable equilibrium.

For $C$:
When the ball is displaced from its equilibrium position, it has no tendency to go anywhere.
Thus, it is in neutral equilibrium.

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