Q.A positive charge $Q$ is distributed uniformly around a circular ring of radius $R$ lying in a plane. A small point test charge $q$ is placed at the centre O of the ring. The axis of the ring is the straight line through O perpendicular to the plane of the ring. Consider small displacements of $q$ either in the plane of the ring or along the axis. Which of the following statements is/are correct?

Concept

For a uniformly charged ring (charge $Q>0$, radius $R$), the on-axis potential is

which is a maximum at $z=0$ (it decreases as $|z|$ grows). Since $\nabla^2 V = 0$ in the empty region at the centre, the sum of the curvatures of $V$ is zero: an axial maximum forces an in-plane minimum of $V$.

In-plane behaviour

• $V$ has a minimum at O in the plane, so $U = qV$ for $q>0$ has a minimum ⇒ restoring force ⇒ stable. Physically, displacing $q>0$ toward the ring brings it nearer to one arc whose stronger repulsion pushes it back. So (a) is correct and (d) is wrong (it is stable, not unstable, in the plane).
• For $q<0$, $U=qV$ has a maximum at O in the plane ⇒ unstable; once displaced, the nearer arc attracts it, pulling it further out until it strikes the ring. So (b) is correct. …

Method: Testing Stability of Equilibrium at an Electrostatic Null Point

Use this whenever a test charge sits at a field-free point of symmetry (a null point) and you must decide whether small displacements grow or shrink.

Steps

Step 1: Confirm E = 0 at the point by symmetry.

For example, at the centre of a uniformly charged ring, every element's field contribution is cancelled by its diametrically opposite partner.

Step 2: Displace the test charge slightly along ONE direction at a time and check whether the resulting local field pushes it back (restoring ⇒ stable) or further away (⇒ unstable).

For a small displacement, the field grows roughly linearly from zero:

Combine this with $\vec{F} = q\vec{E}$ — the SIGN of the test charge then decides whether that field direction attracts it back or repels it further.

Step 3: Use Earnshaw's theorem as a consistency check. …