Question

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_1$ and dielectric constant $K_1$ and the other has thickness $d_2$ and dielectric constant $K_2$ as shown in figure. This arrangement can be thought as a dielectric slab of thickness $d(=d_1+d_2)$ and effective dielectric constant $K$. The $K$ is

• A. $\frac{K_1{d}_1+K_2{d}_2}{d_1+d_2}$
• B. $\frac{K_1{d}_1+K_2{d}_2}{K_1+K_2}$
• C. $\frac{K_1{K}_2\left(d_1+d_2\right)}{K_2{d}_1+K_1{d}_2}$
• D. $\frac{2K_1{K}_2}{K_1+K_2}$

Answer 1

Answer: (C)

The capacities of two individual condensers are
$C_1=\frac{K_1{\epsilon }_{0}A}{d_1}$ and $C_2=\frac{K_2{\epsilon }_{0}A}{d_2}$
The arrangement is equivalent to two capacitors joined in series
$\therefore$ Equivalent capacitance, $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$,
$=\frac{d_1}{K_1{\epsilon }_{0}A}+\frac{d_2}{K_2{\epsilon }_{0}A}$
$=\frac{1}{{\epsilon }_{0}A}\left[\frac{d_1}{K_1}+\frac{d_2}{K_2}\right]=\frac{1}{{\epsilon }_{0}A}\left[\frac{K_2{d}_1+K_1{d}_2}{K_1{K}_2}\right]$
or $C_{eq}={\epsilon }_{0}A\left(\frac{K_1{K}_2}{K_2{d}_1+K_1{d}_2}\right)\dots \left(i\right)$
Also $C_{eq}=\frac{K{\epsilon }_{0}A}{d_1+d_2}\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
${\epsilon }_{0}A\left(\frac{K_1{K}_2}{d_2{K}_1+d_1{K}_2}\right)={\epsilon }_{0}A\left(\frac{K}{d_1+d_2}\right)$
$\therefore K=\frac{K_1{K}_2\left(d_1+d_2\right)}{d_2{K}_1+d_1{K}_2}$