Capacitance of a parallel plate capacitor with plates area A and separation d which is filled with two wedges of dielectric medium of dielectric constants K₁ and K₂ is $\mathrm{C}=\mathrm{N}.\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\log 2$. If K₂ = 8 and K₁ = 4, then find the value of N.

For an elemental length dx capacity of lower and upper elements are

${\mathrm{dC}}_1=\frac{{\mathrm{K}}_1{\mathrm{\epsilon }}_0\mathrm{bdx}}{\mathrm{y}};{\mathrm{dC}}_2=\frac{{\mathrm{K}}_2{\mathrm{\epsilon }}_0\mathrm{bdx}}{\mathrm{d}-\mathrm{y}}$

b = plate width As two elements are in series,

$\frac{1}{\mathrm{dC}}=\frac{1}{{\mathrm{dC}}_1}+\frac{1}{{\mathrm{dC}}_2}$

Using 

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}\Rightarrow \mathrm{dC}=\frac{{\mathrm{K}}_1{\mathrm{K}}_2{\mathrm{\epsilon }}_0\mathrm{bdx}}{{\mathrm{K}}_1\mathrm{d}+\left({\mathrm{K}}_2-{\mathrm{K}}_1\right)\mathrm{y}}$

Integrating $\mathrm{C}=\frac{{\mathrm{K}}_1{\mathrm{K}}_2{\mathrm{\epsilon }}_0\mathrm{A}}{\left({\mathrm{K}}_2-{\mathrm{K}}_1\right)\mathrm{d}}\log \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)$

As K₂ = 8, K₁ = 4, we get $\mathrm{C}=\frac{8\times 4{\mathrm{\epsilon }}_0\mathrm{A}}{(8-4)\mathrm{d}}\log \left(\frac{8}{4}\right)=8\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\log 2$

So, N = 8.