The gap between the plates of a parallel-plate capacitor is filled with an isotropic dielectric whos

Question

The gap between the plates of a parallel-plate capacitor is filled with an isotropic dielectric whose relative permittivity (dielectric constant) $K$ varies linearly from $K_1 = 2$ at the left plate to $K_2 = 6$ at the right plate, in a direction perpendicular to the plates. The area of each plate equals $S$, and the separation between the plates is $d$. The capacitance of the capacitor is given by the expression $C = \frac{\alpha \epsilon_0 S}{d \ln(\beta)}$, where $\alpha$ and $\beta$ are coprime positive integers. Find the value of $(\alpha + \beta)$.

Professor PC Thomas Classes · Answer

The dielectric constant as a function of distance $x$ from the left plate is $K(x) = K_1 + \frac{K_2 - K_1}{d}x$. Consider an elemental capacitor of thickness $dx$ at distance $x$. Its capacitance is $dC = \frac{K(x)\epsilon_0 S}{dx}$. Since the elemental dielectric layers are in series, the equivalent capacitance is found via: $\frac{1}{C} = \int_{0}^{d} \frac{dx}{K(x)\epsilon_0 S} = \frac{1}{\epsilon_0 S} \int_{0}^{d} \frac{dx}{K_1 + \frac{K_2 - K_1}{d}x}$. Integrating gives: $\frac{1}{C} = \frac{d}{\epsilon_0 S (K_2 - K_1)} \ln\left(\frac{K_2}{K_1}\right)$. $C = \frac{\epsilon_0 S (K_2 - K_1)}{d \ln(K_2/K_1)}$. Substituting $K_1 = 2$ and $K_2 = 6$: $C = \frac{4 \epsilon_0 S}{d \ln(3)}$. Comparing this with the given expression, we find $\alpha = 4$ and $\beta = 3$. Both are coprime integers. Therefore, $\alpha + \beta = 7$.