Question

A parallel plate capacitor of area $A$ placed at a separation d and capacitance $C$ is filled with three different dielectric materials having dielectric constant $K_1,K_2$ and $K_3$ respectively as shown in figure. If a dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K$ is given by

• A. $\frac{K_1{K}_2{K}_3}{K_1+K_2+K_3}$
• B. $\frac{2K_2(K_1+K_3)}{2K_2+K_1+K_3}$
• C. $\frac{2K_3(K_1+K_2)}{2K_3+K_1+K_2}$
• D. $\frac{2K_1(K_2+K_3)}{2K_1+K_2+K_3}$

Answer 1

The correct option is D $\frac{2K_1(K_2+K_3)}{2K_1+K_2+K_3}$

Let the capacitance of dielectric constant $K_1,K_2$ and $K_3$ be $C_1,C_2$ and $C_3$ respectively.

From the figure, it is clear that $C_2$ and $C_3$ are in parallel let their effective capacitance be $C{}^{\prime }$ also $C{}^{\prime }$ in series with $C_1$ respectively.

$C_1=K_1{C}_0=K_1\frac{{\epsilon }_{0}A}{\left(\frac{d}{2}\right)}=2K_1\left(\frac{A{\epsilon }_0}{d}\right)$

Similarly, $C_2$ and $C_3$ becomes,

$C_2=K_2{C}_0=K_2\frac{{\epsilon }_0\left(\frac{A}{2}\right)}{\left(\frac{d}{2}\right)}=K_2\left(\frac{A{\epsilon }_0}{d}\right)$

$C_3=K_3{C}_0=K_3\frac{{\epsilon }_0\left(\frac{A}{2}\right)}{\left(\frac{d}{2}\right)}=K_3\left(\frac{A{\epsilon }_0}{d}\right)$

$\therefore C^{{}^{\prime }}=C_2+C_3$

$C^{{}^{\prime }}=\left(\frac{A{\epsilon }_0}{d}\right)\left(K_2+K_3\right)$


$\therefore C_{eq}=\frac{C_1\times C^{{}^{\prime }}}{C_1+C^{{}^{\prime }}}=\frac{2K_1\left(\frac{A{\epsilon }_0}{d}\right)\times \left(K_2+K_3\right)\left(\frac{A{\epsilon }_0}{d}\right)}{2K_1\left(\frac{A{\epsilon }_0}{d}\right)+\left(K_2+K_3\right)\left(\frac{A{\epsilon }_0}{d}\right)}$

$C_{eq}=\left(\frac{A{\epsilon }_0}{d}\right)\left(\frac{2K_1(K_2+K_3)}{2K_1+K_2+K_3}\right).............\left(1\right)$

Let $C_{eq}=K\left(\frac{A{\epsilon }_0}{d}\right).........\left(2\right)$

On comparing eq (1) and (2) we get,

$K=\left(\frac{2K_1(K_2+K_3)}{2K_1+K_2+K_3}\right)$

Hence, option $\left(d\right)$ is correct.

Key concept: If dielectric divides the distance between the plates, the parts formed are in series and if dielectric divides the area of the plates, then parts formed are in parallel.