Question

A parallel-plate capacitor is constructed using three different dielectric materials as shown in the figure. What is the capacitance across P and Q?

• A. $(\frac{K_1}{2}+\frac{K_2{K}_3}{K_2+K_3})\frac{{\epsilon }_{0}A}{t}$
• B. $(K_1+\frac{K_2{K}_3}{K_2+K_3})\frac{{\epsilon }_{0}A}{t}$
• C. $(K_1+\frac{2K_2{K}_3}{K_2+K_3})\frac{{\epsilon }_{0}A}{t}$
• D. $(K_1+\frac{K_2{K}_3}{2(K_2+K_3)})\frac{{\epsilon }_{0}A}{t}$

Answer 1

The correct option is A $(\frac{K_1}{2}+\frac{K_2{K}_3}{K_2+K_3})\frac{{\epsilon }_{0}A}{t}$

On redrawing the circuit we get,
$C_1=K_1\frac{{\epsilon }_{0}A/2}{t}=K_1\frac{{\epsilon }_{0}A}{2t}$

$C_2=K_2\frac{{\epsilon }_{0}A/2}{t/2}=K_2\frac{{\epsilon }_{0}A}{t}$

$C_3=K_3\frac{{\epsilon }_{0}A/2}{t/2}=K_3\frac{{\epsilon }_{0}A}{t}$

Now as $C_2$,$C_3$ are in series and this combination is parallel with $C_1$. Therefore the effective capacitance will be

$C_{eq}=\frac{{\epsilon }_{0}A}{t}[\frac{K_2\times K_3}{K_2+K_3}+\frac{K_1}{2}]$