Capacitors

Concept:The system consists of two parallel-plate capacitors (the chambers) in parallel, each with two dielectrics in series: liquid and air. The capacitance changes as liquid flows from left to right.Explanation:Given: Total height $H = 2\ \mathrm{m}$, each chamber floor area $A = 1\ \mathrm{m}^2$ (since length = 1 m, breadth = 1 m).For a chamber with liquid height $h$ and dielectric constant $\varepsilon_r = 15$, the equivalent capacitance is:$$C(h) = \frac{\varepsilon_0 A}{H - h\left(1 - \frac{1}{\varepsilon_r}\right)} = \frac{\varepsilon_0}{2 - h\left(1 - \frac{1}{15}\right)}.$$At $t=0$: left chamber $h_1 = 2\ \mathrm{m}$, right chamber $h_2 = 0$.$C_{\text{left},0} = \frac{\varepsilon_0}{2 - 2 \times \frac{14}{15}} = 7.5\varepsilon_0$.$C_{\text{right},0} = \frac{\varepsilon_0}{2 - 0} = 0.5\varepsilon_0$.Total $C_{t=0} = 8\varepsilon_0$.At $t = 500\ \mathrm{s}$: left chamber height $h_1 = 1.25\ \mathrm{m}$, so right chamber height $h_2 = 2 - 1.25 = 0.75\ \mathrm{m}$ (by conservation of liquid volume).$C_{\text{left},500} = \frac{\varepsilon_0}{2 - 1.25 \times \frac{14}{15}} = \frac{\varepsilon_0}{2 - \frac{7}{6}} = 1.2\varepsilon_0 = \frac{6}{5}\varepsilon_0$.$C_{\text{right},500} = \frac{\varepsilon_0}{2 - 0.75 \times \frac{14}{15}} = \frac{\varepsilon_0}{2 - \frac{7}{10}} = \frac{10}{13}\varepsilon_0$.Total $C_{t=500} = \frac{6}{5}\varepsilon_0 + \frac{10}{13}\varepsilon_0 = \frac{78+50}{65}\varepsilon_0 = \frac{128}{65}\varepsilon_0$.The capacitance difference is $\Delta C = C_{t=0} - C_{t=500}$. Given $\Delta C = (8 - n)\varepsilon_0$, we have:$8\varepsilon_0 - \frac{128}{65}\varepsilon_0 = (8 - n)\varepsilon_0 \implies n = \frac{128}{65} \approx 2$.Answer:$n = 2$.