1. Physical setup

   • Two rigid conducting plates are separated by height $h$.
   • The side surface is insulating.
   • Small light, soft, conducting-coated balls of radius $r$ are placed on the lower plate.
   • A potential difference $V_0$ is applied across the plates, so the electric field between plates is approximately uniform: $E = \frac{V_0}{h}$

2. How current is produced

   Each ball acts like a charge carrier:

   • When touching the lower plate, it acquires charge.
   • It is then pulled upward by the electric field.
   • On touching the upper plate, it transfers charge there.
   • Then it may return and repeat the process.

   Thus, average current is $I \sim (\text{charge carried per trip}) \times (\text{number of trips per second})$

3. Charge acquired by a ball

   A conducting sphere of radius $r$ at potential $V$ has charge of order $q \sim 4\pi \varepsilon_0 r V$

   Here the potential scale involved is of order $V_0$, so $q \propto V_0$

4. Force on the charged ball

   Electric force on the ball is $F = qE$

   Since $q \propto V_0$ and $E = V_0/h \propto V_0$, $F \propto V_0 \cdot V_0 = V_0^2$

5. Speed / time of flight

   The balls are light and soft, so we consider their motion under this electric force. For a given mass $m$ and travel distance $h$, acceleration is $a = \frac{F}{m} \propto V_0^2$

   Time to cross distance $h$ under constant acceleration satisfies $h \sim \frac{1}{2}at^2$ so $t \propto \frac{1}{\sqrt{a}} \propto \frac{1}{V_0}$

   Therefore, number of trips per second is $f \propto \frac{1}{t} \propto V_0$

6. Average current

   Hence, $I \propto q f \propto V_0 \cdot V_0 = V_0^2$

7. Option check

   • A: proportional to $V_0^2$ ✅
   • B: proportional to $V_0$ ❌
   • C: zero ❌
   • D: proportional to $V_0^{1/2}$ ❌

Therefore, the correct answer is A.