Question

For configuration of media of permittivity ${\epsilon }_0,\epsilon ,{\epsilon }_0$ between parallel plates each of area $A$ as shown in the figure; the equivalent capacitance is

• A. $\frac{{\epsilon }_{0}A}{d}$
• B. $\frac{\epsilon {\epsilon }_{0}A}{d}$
• C. $\frac{\epsilon {\epsilon }_{0}A}{d(\epsilon +{\epsilon }_0)}$
• D. $\frac{\epsilon {\epsilon }_{0}A}{(2\epsilon +{\epsilon }_0)d}$

Answer 1

The correct option is D $\frac{\epsilon {\epsilon }_{0}A}{(2\epsilon +{\epsilon }_0)d}$

From question,

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
and
$C_1=\frac{{\epsilon }_{0}A}{d};C_2=\frac{\epsilon A}{d};C_3=\frac{{\epsilon }_{0}A}{d}$

On solving we get
$C_{eq}=\frac{A\left({\epsilon }_0\epsilon \right)}{d(2ϵ+{\epsilon }_0)}$