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 $K_m=K\left(1+\frac{m}{N}\right).$ For a very large $N\left(>10^3\right)$, the capacitance $C$ is $\alpha \left(\frac{k{\epsilon }_{0}A}{D\ln 2}\right)$.The value of $\alpha$ will be - [ ${ϵ}_0$ is the permittivity of free space]

Answer 1

Answer: 1

$\frac{m}{N}=\frac{x}{D}$
$d\left(\frac{1}{C}\right)=\frac{dx}{K_m{\epsilon }_{0}A}=\frac{dx}{K{\epsilon }_{0}A\left(1+\frac{m}{N}\right)}=\frac{dx}{K{\epsilon }_{0}A\left(1+\frac{x}{D}\right)}$
$\frac{1}{C_{eq}}=\int d\left(\frac{1}{C}\right)=\underset{0}{\overset{D}{\int }}\frac{dx}{K{\left(ϵ\right)}_{0}A(D+x)}$
$\frac{1}{C_{eq}}=\frac{D}{K{ϵ}_{0}A}ln2$
$C_{eq}=\frac{K{ϵ}_{0}A}{Dln2}.$ therefore $\alpha =1$