Average Current Dependence on Voltage in Charging Cylindrical Chamber

To determine the dependence of the average current $I_{\text{avg}}$ on the potential $V_0$ in the steady state, we analyze the charge acquired by the balls and their motion between the plates.

1. Electric Field in the Chamber

The bottom plate is maintained at potential $+V_0$ and the top plate at $-V_0$. The potential difference across the height $h$ of the cylindrical chamber is: $\Delta V = V_0 - (-V_0) = 2V_0$

Assuming the electric field $E$ inside the chamber is uniform (similar to a parallel plate capacitor): $E = \frac{\Delta V}{h} = \frac{2V_0}{h}$ Thus, the electric field is directly proportional to $V_0$: $E \propto V_0$

2. Charge Acquired by a Ball

When a conducting spherical ball of radius $r$ comes into contact with a conducting plate placed in a uniform electric field $E$, it acquires a charge $q$ given by: $q = \left(\frac{\pi^2}{6}\right) 4 \pi \varepsilon_0 r^2 E$ Since $E \propto V_0$, the charge $q$ acquired by each ball is directly proportional to $V_0$: $q \propto V_0$

3. Acceleration and Time of Flight

The electrostatic force $F$ acting on a ball of mass $m$ carrying charge $q$ in the field $E$ is: $F = q E$ Substituting the proportionalities $q \propto V_0$ and $E \propto V_0$: $F \propto V_0^2$

Ignoring gravity, the acceleration $a$ of the ball is: $a = \frac{F}{m} \propto V_0^2$

Since the coefficient of restitution is zero, the ball comes to rest relative to the plate upon impact, recharges to the plate's potential, and then accelerates from rest ($u = 0$) towards the opposite plate.

The time $t$ taken by the ball to traverse the distance $h$ between the plates is obtained from: $h = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2h}{a}}$ Since $a \propto V_0^2$, the time of flight scales as: $t \propto \frac{1}{V_0}$

4. Average Current

The average current $I_{\text{avg}}$ in the circuit is proportional to the rate of charge transferred by the balls between the plates: $I_{\text{avg}} \propto \frac{q}{t}$

Substituting $q \propto V_0$ and $t \propto \frac{1}{V_0}$: $I_{\text{avg}} \propto \frac{V_0}{1/V_0} = V_0^2$

Thus, the average current registered by the ammeter in the steady state is proportional to $V_0^2$.

Correct Option: (D)