Question 16

The total capacitance $C$ is the sum of the capacitances of the left and right chambers ($C = C_L + C_R$) as they are in parallel.

In each chamber, the liquid and air act as two capacitors in series. For a chamber with liquid height $h$:

$C_{\text{chamber}} = \frac{\epsilon_0 A}{\frac{h}{\epsilon_r} + \frac{2-h}{1}} = \frac{\epsilon_0 (1)}{\frac{h}{15} + 2 - h} = \frac{15\epsilon_0}{30 - 14h}$

At $t = 0$, $h_1 = 2$ and $h_2 = 0$:

$C(0) = \frac{15\epsilon_0}{30 - 28} + \frac{15\epsilon_0}{30 - 0} = 7.5\epsilon_0 + 0.5\epsilon_0 = 8\epsilon_0$

At $t = 500 \text{ s}$, $h_1 = 1.25 \text{ m}$ and $h_2 = 0.75 \text{ m}$:

$C_L(500) = \frac{15\epsilon_0}{30 - 14(1.25)} = \frac{15\epsilon_0}{12.5} = 1.2\epsilon_0$

$C_R(500) = \frac{15\epsilon_0}{30 - 14(0.75)} = \frac{15\epsilon_0}{19.5} = \frac{10}{13}\epsilon_0 \approx 0.769\epsilon_0$

$C(500) = (1.2 + 10/13)\epsilon_0 = \frac{128}{65}\epsilon_0 \approx 1.969\epsilon_0$

The difference is $C(0) - C(500) = 8\epsilon_0 - 1.969\epsilon_0 = (8 - 1.969)\epsilon_0$.

Comparing with $(8-n)\epsilon_0$, we get $n = 128/65 \approx 1.97$.