Capacitance of Parallel Plate Capacitor Filled with Stratified Dielectric Layers

To find the overall capacitance $C$ of the parallel plate capacitor, we note that the $N$ dielectric layers are arranged one after another between the plates, which means they are connected in series.

The capacitance of the $m^{\text{th}}$ dielectric layer, $C_m$, is given by: $C_m = \frac{K_m \epsilon_0 A}{\delta}$

Given that the thickness of each layer is $\delta = \frac{d}{N}$ and its dielectric constant is $K_m = K\left(1 + \frac{m}{N}\right)$, we substitute these into the expression for $C_m$: $C_m = \frac{K\left(1 + \frac{m}{N}\right) \epsilon_0 A}{\frac{d}{N}} = \frac{N K \epsilon_0 A \left(1 + \frac{m}{N}\right)}{d}$

Since the capacitors are connected in series, the reciprocal of the equivalent capacitance $C$ is the sum of the reciprocals of the individual capacitances: $\frac{1}{C} = \sum_{m=1}^N \frac{1}{C_m}$

Substituting $C_m$ into the sum gives: $\frac{1}{C} = \sum_{m=1}^N \frac{d}{N K \epsilon_0 A \left(1 + \frac{m}{N}\right)} = \frac{d}{K \epsilon_0 A} \sum_{m=1}^N \frac{1}{N \left(1 + \frac{m}{N}\right)}$

For a very large value of $N$ ($N > 10^3$), the summation can be approximated by a definite integral using the Riemann sum definition where $x = \frac{m}{N}$ and $dx = \frac{1}{N}$: $\lim_{N \to \infty} \sum_{m=1}^N \frac{1}{N \left(1 + \frac{m}{N}\right)} = \int_{0}^{1} \frac{dx}{1 + x}$

Evaluating this integral: $\int_{0}^{1} \frac{dx}{1 + x} = \Big[ \ln(1 + x) \Big]_{0}^{1} = \ln(2) - \ln(1) = \ln 2$

Thus, the reciprocal of the equivalent capacitance becomes: $\frac{1}{C} = \frac{d \ln 2}{K \epsilon_0 A}$

Inverting this expression yields the capacitance $C$: $C = \frac{K \epsilon_0 A}{d \ln 2}$

Comparing this result with the given expression $C = \alpha \left(\frac{K \epsilon_0 A}{d \ln 2}\right)$, we get: $\alpha = 1$