Question

The difference in the capacitance (in F) between the metal plates at \(t=0\) and that at \(t=500\,\text{s}\) is \((8-n)\epsilon_0\), where \(\epsilon_0\) is the permittivity of free space. The value of \(n\) is:

Hint:

Chambers side-by-side act as parallel capacitors ($C_{total} = C_1 + C_2$).
Layers of different dielectrics within a chamber act as capacitors in series ($1/C = 1/C_1 + 1/C_2$).
The formula $C = \frac{\epsilon_0 A}{d_1/\epsilon_{r1} + d_2/\epsilon_{r2}}$ is a very useful shortcut for layered capacitors.

Answer 1

Step 1: Understanding the Question:
The container is partitioned into two chambers, which act as two capacitors connected in parallel.
In each chamber, the liquid (dielectric $\epsilon_r$) and the air above it (dielectric $1$) form a series combination because the electric field passes through both layers.
The area of each chamber is $1\text{ m}^2$. Total height is $2\text{ m}$.

Step 2: Key Formula or Approach:
Capacitance of a chamber with liquid height $h$:
$C_{liq} = \frac{\epsilon_0 \epsilon_r A}{h}, C_{air} = \frac{\epsilon_0 (1) A}{2-h}$.
$C_{chamber} = \frac{C_{liq} C_{air}}{C_{liq} + C_{air}} = \frac{\epsilon_0 A}{(2-h) + \frac{h}{\epsilon_r}} = \frac{\epsilon_0}{2 - h + \frac{h}{15}} = \frac{\epsilon_0}{2 - \frac{14h}{15}}$.
Total Capacitance $C = C_{left} + C_{right}$.

Step 3: Detailed Explanation:

• At $t = 0$:
$h_1 = 2\text{ m}, h_2 = 0\text{ m}$.
$C_{left} = \frac{\epsilon_0}{2 - \frac{14 \times 2}{15}} = \frac{\epsilon_0}{2 - 28/15} = \frac{\epsilon_0}{2/15} = 7.5 \epsilon_0$.
$C_{right} = \frac{\epsilon_0}{2 - 0} = 0.5 \epsilon_0$.
$C_{total}(0) = 7.5 \epsilon_0 + 0.5 \epsilon_0 = 8 \epsilon_0$.

• At $t = 500\text{ s$:}
$h_1 = 1.25\text{ m}$ (from Q.15), so $h_2 = 2 - 1.25 = 0.75\text{ m}$.
$C_{left} = \frac{\epsilon_0}{2 - \frac{14 \times 1.25}{15}} = \frac{\epsilon_0}{2 - \frac{14 \times 5/4}{15}} = \frac{\epsilon_0}{2 - 7/6} = \frac{\epsilon_0}{5/6} = 1.2 \epsilon_0$.
$C_{right} = \frac{\epsilon_0}{2 - \frac{14 \times 0.75}{15}} = \frac{\epsilon_0}{2 - \frac{14 \times 3/4}{15}} = \frac{\epsilon_0}{2 - 42/60} = \frac{\epsilon_0}{2 - 0.7} = \frac{\epsilon_0}{1.3} \approx 0.7692 \epsilon_0$.
$C_{total}(500) = 1.2 \epsilon_0 + 0.7692 \epsilon_0 = 1.9692 \epsilon_0$.

• Finding $n$:
Difference $= C(0) - C(500) = 8 \epsilon_0 - 1.9692 \epsilon_0 = (8 - 1.9692) \epsilon_0$.
Comparing with $(8-n)\epsilon_0$, we get $n = 1.9692$.
Rounding off to two decimal places: $n = 1.97$.

Step 4: Final Answer:
The total capacitance of the system decreases as the liquid levels equalize because the chamber with higher level provides much more capacitance than the increase in the other.
The value of $n$ is $1.97$.