Question 25 Solution
Step 1: Energy Balance
The battery does work to charge the capacitor formed by the mercury drop spreading. This work goes into two places:
- Stored Electrostatic Energy ($U_E$)
- Surface Tension Energy ($U_S$)
Step 2: Formulate Terms
Let the area of the drop be $A$.
Capacitance $C = \frac{\epsilon_0 \epsilon_r A}{d}$.
Work done by battery (constant voltage $V$): $W_{battery} = V \cdot \Delta Q = V \cdot (V \Delta C) = V^2 C$.
Change in Electrostatic Energy: $\Delta U_E = \frac{1}{2} C V^2$.
Change in Surface Energy: The drop has a top surface and a bottom surface. Both expand by area $A$.
$\Delta U_S = \sigma(2A)$.
Step 3: Solve
$$V^2 C = \frac{1}{2} C V^2 + 2\sigma A$$ $$\frac{1}{2} C V^2 = 2\sigma A$$Substitute $C = \frac{\epsilon_0 \epsilon_r A}{d}$:
$$\frac{1}{2} \left(\frac{\epsilon_0 \epsilon_r A}{d}\right) V^2 = 2\sigma A$$Cancel $A$ and rearrange:
$$\frac{\epsilon_0 \epsilon_r V^2}{2d} = 2\sigma \implies V^2 = \frac{4\sigma d}{\epsilon_0 \epsilon_r}$$
Answer: $V = 2 \sqrt{ \frac{\sigma d}{\epsilon_0 \epsilon_r} }$