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

Answer 1

$\frac{m}{N}=\frac{x}{D}$
$d\left(\frac{1}{C}\right)=\frac{d x}{K_{m} \varepsilon_{0} A}=\frac{d x}{K \varepsilon_{0} A\left(1+\frac{m}{N}\right)}=\frac{d x}{K \varepsilon_{0} A\left(1+\frac{x}{D}\right)}$
$\frac{1}{C_{e q}}=\int d\left(\frac{1}{C}\right)=\int \limits_{0}^{D}\frac{d x}{K \left(\epsilon \right)_{0} A \left(\right. D + x \left.\right)}$
$\frac{1}{C_{e q}}=\frac{D}{K \epsilon _{0} A}ln2$
$C_{e q}=\frac{K \epsilon _{0} A}{D ln 2}.$ therefore $\alpha =1$