Question

Three slabs of same dimension having dielectric constants $K_1,K_2$ and $K_3$ completely fill the space between the plates of a parallel plate capacitor as shown in figure. If $C$ is the initial capacitance of the capacitor, the new capacitance after the introduction of slabs will be:

• A. $\left(\frac{K_1{K}_2}{K_1+K_2}\right)C$
• B. $\left(\frac{K_1+K_2+K_3}{3}\right)C$
• C. $(K_1+K_2)C$
• D. $\left(\frac{K_1+K_2}{2}\right)C$

Answer 1

The correct option is B $\left(\frac{K_1+K_2+K_3}{3}\right)C$

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

All three capacitor $C_1$ $C_2$ and $C_3$ are across the same potential difference, so they are connected in parallel combination.

$C_{eq}=C_1+C_2+C_3$

Initially, $C=\frac{{ϵ}_{0}A}{d}$

$C_1=K_1\frac{{\epsilon }_0\left(\frac{A}{3}\right)}{d}=K_1\left(\frac{{\epsilon }_{0}A}{3d}\right)$

Similarly,

$C_2=K_2\left(\frac{{\epsilon }_{0}A}{3d}\right)$

$C_3=K_3\left(\frac{{\epsilon }_{0}A}{3d}\right)$

$C_{eq}=K_1\left(\frac{{\epsilon }_{0}A}{3d}\right)+K_2\left(\frac{{\epsilon }_{0}A}{3d}\right)+K_3\left(\frac{{\epsilon }_{0}A}{3d}\right)$

$C_{eq}=\left(\frac{K_1+K_2+K_3}{3}\right)\left(\frac{{\epsilon }_{0}A}{d}\right)$

$C_{eqn}=\left(\frac{K_1+K_2+K_3}{3}\right)C$

Hence, option $\left(b\right)$ is the correct answer.