Question

A parallel plate capacitor of capacitance $C$
has spacing d between two plates having area $A$. The region between the plates is filled with N dielectric layers, parallel to its plates, each with thickness $\delta =\frac{d}{N}$. The dielectric constant of the $m^{th}$ layer is $K_m=K(1+\frac{m}{N})$. For a very large $N(>{10}^3)$, the capacitance $C$ is $\alpha \left(\frac{K{\in }_{0}A}{dln2}\right)$. The value of $\alpha$ will be . [${\in }_0$ is the permittivity of free space]

Answer 1

Since the plates are equally distributed,
$\frac{x}{m}=\frac{d}{N}$
$d\left(\frac{1}{C}\right)$ $=\frac{dx}{K_m{\epsilon }_{0}A}=$ $\frac{dx}{K{\epsilon }_{0}A(1+\frac{m}{N})}=\frac{dx}{K{\epsilon }_{0}A(1+\frac{x}{d})}$
$\frac{1}{C_{eq}}=\int d\left(\frac{1}{C}\right)={\int }_0^d\frac{dx}{K{\epsilon }_{0}A(1+\frac{x}{d})}$
$\frac{1}{C_{eq}}=\frac{d}{K{\epsilon }_{0}A}ln2$
$C_{eq}=\frac{K{\epsilon }_{0}A}{dln2}$.
Therefore $\alpha =1$.