1. Electric Field Analysis:

The electric field on the axis of a uniformly charged ring of radius $R$ at height $z$ is:

$$ E(z) = \frac{Q z}{4\pi\epsilon_0 (R^2 + z^2)^{3/2}} $$

This field starts at 0 (at center), increases to a maximum, and then decreases. The maximum electric field occurs at $z = \frac{R}{\sqrt{2}}$.

2. Stability Requirement:

For stable equilibrium (oscillatory motion), the restoring force must oppose displacement.

• If the bead moves up ($z \uparrow$), the upward electric force must decrease so gravity pulls it back.
• If the bead moves down ($z \downarrow$), the upward electric force must increase to push it back.

This implies we must be in the region where $E(z)$ decreases with height. This corresponds to the region $z > \frac{R}{\sqrt{2}}$. The critical limit case corresponds to the point of maximum field, $z = \frac{R}{\sqrt{2}}$.

3. Calculating the Charge:

We solve for the charge $q$ required to maintain equilibrium at this critical point of maximum field efficiency ($z = R/\sqrt{2}$):

First, calculate $E_{max}$:

$$ E_{max} = \frac{Q (R/\sqrt{2})}{4\pi\epsilon_0 (R^2 + R^2/2)^{3/2}} $$ $$ E_{max} = \frac{Q R}{\sqrt{2} \cdot 4\pi\epsilon_0 (3R^2/2)^{3/2}} = \frac{Q}{4\pi\epsilon_0 R^2} \left( \frac{2}{3\sqrt{3}} \right) $$

Now, apply the equilibrium condition $q E_{max} = mg$:

$$ q \left[ \frac{2 Q}{4\pi\epsilon_0 R^2 3\sqrt{3}} \right] = mg $$

Solving for $q$:

$$ q = \frac{mg (4\pi\epsilon_0 R^2) 3\sqrt{3}}{2 Q} $$ $$ q = \frac{6\sqrt{3} \pi \epsilon_0 R^2 m g}{Q} $$