JEE Advanced 2019 — Paper 1 — Question 17

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 $\delta=\frac{\mathrm{d}}{\mathrm{N}}$. The dielectric constant of the $m^{\text {th }}$ layer is $K_{m}=K\left(1+\frac{m}{N}\right)$. For a very large $N\left(>10^{3}\right)$, the capacitance $C$ is $\alpha\left(\frac{K \varepsilon_{0} A}{d \ln 2}\right)$. The value of $\alpha$ will be $\_\_\_\_$ . [ $\epsilon_{0}$ is the permittivity of free space]