Problem Statement Summary

A grounded metallic ball of radius $r$ is placed on the center of a uniformly charged thin insulating disc of radius $R$, carrying a total charge $Q$. We are given that $R \gg r$. We need to find the force of electrostatic interaction between the ball and the disc.

Analysis and Derivation

1. Electric Potential and Field of the Disc

Since the radius of the disc $R$ is much larger than the radius of the ball $r$ ($R \gg r$), and the ball is placed on the center, we can approximate the potential and field near the center of the disc.

The surface charge density of the disc is:

$$ \sigma = \frac{Q}{\pi R^2} $$

The potential $V(z)$ on the axis of a uniformly charged disc at a distance $z$ from the center is given by:

$$ V(z) = \frac{\sigma}{2\epsilon_0} \left( \sqrt{R^2 + z^2} – z \right) $$

For $z \ll R$, we can expand this expression:

$$ V(z) \approx \frac{\sigma}{2\epsilon_0} (R – z) = \frac{\sigma R}{2\epsilon_0} – \frac{\sigma}{2\epsilon_0}z $$

This potential corresponds to a constant potential $V_0 = \frac{\sigma R}{2\epsilon_0}$ superimposed with a uniform electric field $E_0 = \frac{\sigma}{2\epsilon_0}$ directed along the z-axis (away from the disc).

2. Induced Charge on the Grounded Ball

The ball is grounded, meaning its potential is zero ($V_{ball} = 0$). It is placed on the center of the disc. Let’s assume the ball rests on the surface, so its center is at height $z=r$.

The external potential at the position of the ball’s center ($z=r$) is approximately:

$$ V_{ext} \approx \frac{\sigma R}{2\epsilon_0} $$

(Since $r \ll R$, the $E_0 z$ term is negligible compared to the $V_0$ term for calculating the total induced monopole charge).

To keep the grounded sphere at zero potential in the presence of this external potential $V_{ext}$, a charge $q_{ind}$ must be induced on it. For a sphere of radius $r$, the potential at its surface due to its own charge is $q_{ind}/(4\pi\epsilon_0 r)$.

Using the superposition principle ($V_{total} = V_{ext} + V_{self} = 0$):

$$ V_{ext} + \frac{q_{ind}}{4\pi\epsilon_0 r} = 0 $$ $$ \frac{\sigma R}{2\epsilon_0} + \frac{q_{ind}}{4\pi\epsilon_0 r} = 0 $$

Solving for the induced charge $q_{ind}$:

$$ q_{ind} = -4\pi\epsilon_0 r \left( \frac{\sigma R}{2\epsilon_0} \right) = -2\pi \sigma R r $$

3. Force Calculation

The force exerted on the ball is due to the interaction of the induced charge with the local electric field created by the disc. The local electric field $E$ near the center of the disc acts on the induced charge $q_{ind}$.

The electric field $E$ near the center of the disc (behaving like an infinite sheet due to $R \gg r$) is:

$$ E = \frac{\sigma}{2\epsilon_0} $$

The force $F$ is simply the product of the induced charge and the external field:

$$ F = q_{ind} E $$ $$ F = (-2\pi \sigma R r) \left( \frac{\sigma}{2\epsilon_0} \right) $$ $$ F = -\frac{\pi \sigma^2 R r}{\epsilon_0} $$

The negative sign indicates attraction. We need the magnitude. Now, substitute $\sigma = \frac{Q}{\pi R^2}$:

$$ F = \frac{\pi R r}{\epsilon_0} \left( \frac{Q}{\pi R^2} \right)^2 $$ $$ F = \frac{\pi R r}{\epsilon_0} \frac{Q^2}{\pi^2 R^4} $$ $$ F = \frac{Q^2 r}{\pi \epsilon_0 R^3} $$

Thus, the magnitude of the force of electrostatic interaction is:

$$ |F| = \frac{Q^2 r}{\pi \epsilon_0 R^3} $$