Determine Capacitance Difference Between Metal Plates Over Time

To determine the value of $n$, we break the problem into two parts: determining the heights of the liquid in the two chambers at $t = 500\text{ s}$, and evaluating the total capacitance of the system at $t = 0$ and $t = 500\text{ s}$.

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Step 1: Determining the liquid heights at $t = 500\text{ s}$

The container has dimensions $2\text{ m} \times 2\text{ m} \times 1\text{ m}$ (height $\times$ total length $\times$ breadth). Since it is partitioned into two equal chambers:

• Area of cross-section of each chamber, $A = 1\text{ m} \times 1\text{ m} = 1\text{ m}^2$.
• Total height of container, $H = 2\text{ m}$.
• Hole area, $a = \sqrt{10}\text{ cm}^2 = \sqrt{10} \times 10^{-4}\text{ m}^2$.

Let $h_1$ and $h_2$ be the heights of the liquid in the left and right chambers at time $t$, respectively. By conservation of volume: $A h_1 + A h_2 = A H \implies h_1 + h_2 = 2 \implies h_2 = 2 - h_1$

The pressure difference at the hole near the bottom is $\Delta P = \rho g (h_1 - h_2)$. Using Torricelli's Law, the speed of efflux is: $v = \sqrt{2g(h_1 - h_2)}$

The rate of decrease of height $h_1$ is given by: $-A \frac{dh_1}{dt} = a v = a \sqrt{2g(h_1 - h_2)}$

Substituting $h_2 = 2 - h_1$: $h_1 - h_2 = 2h_1 - 2 = 2(h_1 - 1)$ $-\frac{dh_1}{dt} = \frac{a}{A} \sqrt{4g(h_1 - 1)} = \frac{2a}{A} \sqrt{g(h_1 - 1)}$

Substituting the given numerical values ($g = 10\text{ ms}^{-2}$, $a = \sqrt{10} \times 10^{-4}\text{ m}^2$, $A = 1\text{ m}^2$): $-\frac{dh_1}{dt} = 2 \left(\sqrt{10} \times 10^{-4}\right) \sqrt{10(h_1 - 1)} = 2 \times 10^{-3} \sqrt{h_1 - 1}$

Separating the variables and integrating from $t = 0$ to $t = 500\text{ s}$: $\int_{2}^{h_1} \frac{-dh_1}{\sqrt{h_1 - 1}} = 2 \times 10^{-3} \int_{0}^{500} dt$ $\left[ -2\sqrt{h_1 - 1} \right]_{2}^{h_1} = 2 \times 10^{-3} \times 500 = 1$ $-2\sqrt{h_1 - 1} + 2\sqrt{2 - 1} = 1$ $2 - 2\sqrt{h_1 - 1} = 1 \implies \sqrt{h_1 - 1} = 0.5 \implies h_1 - 1 = 0.25$

Thus, at $t = 500\text{ s}$: $h_1 = 1.25\text{ m} = \frac{5}{4}\text{ m}$ $h_2 = 2 - 1.25 = 0.75\text{ m} = \frac{3}{4}\text{ m}$

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Step 2: Capacitance Calculation

Each chamber acts as two capacitors in series: a liquid column of height $h$ with dielectric constant $\epsilon_r = 15$ and an air column of height $(2 - h)$ with dielectric constant $\epsilon_r = 1$.

For a chamber with liquid height $h$:

• Capacitance of the liquid dielectric part: $C_{\text{liq}} = \frac{\epsilon_r \epsilon_0 A}{h} = \frac{15 \epsilon_0}{h}$
• Capacitance of the air part: $C_{\text{air}} = \frac{\epsilon_0 A}{2 - h} = \frac{\epsilon_0}{2 - h}$

Since these two regions are stacked vertically between the plates, their equivalent capacitance $C_{\text{chamber}}$ is: $\frac{1}{C_{\text{chamber}}} = \frac{1}{C_{\text{liq}}} + \frac{1}{C_{\text{air}}} = \frac{h}{15\epsilon_0} + \frac{2-h}{\epsilon_0} = \frac{1}{\epsilon_0} \left( 2 - \frac{14}{15}h \right)$ $C_{\text{chamber}}(h) = \frac{\epsilon_0}{2 - \frac{14}{15}h}$

The left and right chambers are in parallel across plates $\text{M}_1$ and $\text{M}_2$, so the total capacitance is: $C_{\text{total}} = C_L(h_1) + C_R(h_2)$

At $t = 0$:

Here $h_1 = 2\text{ m}$ and $h_2 = 0\text{ m}$: $C_L(2) = \frac{\epsilon_0}{2 - \frac{28}{15}} = \frac{15}{2}\epsilon_0 = 7.5\epsilon_0$ $C_R(0) = \frac{\epsilon_0}{2 - 0} = 0.5\epsilon_0$ $C(0) = 7.5\epsilon_0 + 0.5\epsilon_0 = 8\epsilon_0$

At $t = 500\text{ s}$:

Here $h_1 = \frac{5}{4}\text{ m}$ and $h_2 = \frac{3}{4}\text{ m}$: $C_L\left(\frac{5}{4}\right) = \frac{\epsilon_0}{2 - \frac{14}{15} \times \frac{5}{4}} = \frac{\epsilon_0}{2 - \frac{7}{6}} = \frac{6}{5}\epsilon_0 = 1.2\epsilon_0$ $C_R\left(\frac{3}{4}\right) = \frac{\epsilon_0}{2 - \frac{14}{15} \times \frac{3}{4}} = \frac{\epsilon_0}{2 - \frac{7}{10}} = \frac{10}{13}\epsilon_0$

The total capacitance at $t = 500\text{ s}$ is: $C(500) = \left( \frac{6}{5} + \frac{10}{13} \right)\epsilon_0 = \frac{128}{65}\epsilon_0$

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Step 3: Finding $n$

The difference in capacitance between $t = 0$ and $t = 500\text{ s}$ is: $\Delta C = C(0) - C(500) = 8\epsilon_0 - \frac{128}{65}\epsilon_0 = \left( 8 - \frac{128}{65} \right)\epsilon_0$

Comparing this with the given expression $(8 - n)\epsilon_0$: $n = \frac{128}{65} \approx 1.97$