A parallel plate capacitor with plate area $S$ and separation $d$ is initially filled entirely with

Question

A parallel plate capacitor with plate area $S$ and separation $d$ is initially filled entirely with air, yielding a capacitance $C_0$. Two different configurations for filling exactly half the volume of the gap with a uniform dielectric of relative permittivity $K$ are evaluated.\ Configuration 1: The dielectric fills the region between the plates from $x = 0$ to $x = d/2$ (layers parallel to plates).\ Configuration 2: The dielectric fills the region covering half the area of the plates, from $y = 0$ to $y = L/2$ across the entire depth $d$ (layers perpendicular to plates).\ If the equivalent capacitance in Configuration 1 is $C_1$ and in Configuration 2 is $C_2$, what is the ratio $C_2 / C_1$?

Professor PC Thomas Classes · Answer

In Configuration 1, the system acts as two capacitors in series: one with dielectric (thickness $d/2$) and one with air (thickness $d/2$). Their capacitances are $C_d = \frac{K \epsilon_0 S}{d/2} = 2K C_0$ and $C_a = \frac{\epsilon_0 S}{d/2} = 2 C_0$. The equivalent series capacitance is $C_1 = \frac{C_d C_a}{C_d + C_a} = \frac{(2K C_0)(2 C_0)}{2K C_0 + 2 C_0} = C_0 \frac{2K}{K+1}$.\ In Configuration 2, the system acts as two capacitors in parallel, each with area $S/2$. Their capacitances are $C_d' = \frac{K \epsilon_0 (S/2)}{d} = K \frac{C_0}{2}$ and $C_a' = \frac{\epsilon_0 (S/2)}{d} = \frac{C_0}{2}$. The equivalent parallel capacitance is $C_2 = C_d' + C_a' = C_0 \frac{K+1}{2}$. The requested ratio is $C_2 / C_1 = \frac{C_0 (K+1)/2}{C_0 2K/(K+1)} = \frac{(K+1)^2}{4K}$.