Question

A positively charged particle having some mass is resting in equilibrium at a height H above the center of a fixed, uniformly and positively charged ring of radius R. The force of gravity (mg) acts downwards. The equilibrium of the particle at the given position (H) for small vertical displacement will be: (Assume g is uniform)

• A. stable if $H<\frac{R}{2}$
• B. stable if $H=\frac{R}{\sqrt{2}}$
• C. unstable if $H<\frac{R}{\sqrt{2}}$
• D. stable if $H=\frac{R}{2}$

Answer 1

The correct option is C unstable if $H<\frac{R}{\sqrt{2}}$

For the uniformly charged ring , Potential at a distance H from the axis is $E=\frac{KQH}{\sqrt{R^2+H^2}}$
This function has its maximum value at $H=\frac{R}{\sqrt{2}}$
At $H=\frac{R}{\sqrt{2}}$, force is maximum and hence area under the E-H Curve i.e potential Energy is also maximum.
(i) If $H<\frac{R}{\sqrt{2}}$
In this case, if we displace the particle slightly downwards, E will decrease, so upwards $qE$ force will also decrease. So the particle will move downwards i.e. away from equilibrium $\Rightarrow$ unstable equilibrium.
(ii) If $H>\frac{R}{\sqrt{2}}$
In this case, if we displace the particle slightly upwards, $E$ decreases, so $qE$ force also decreases. Thus the particle will move downwards i.e. towards equilibrium $\Rightarrow$ stable equilibrium.