Question

For the configuration of media of permitivities ${\epsilon }_o,\epsilon$ and ${\epsilon }_o$ between parallel plates each of area $A$, as show in figure, the equivalent capacitance is :

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

Answer 1

Answer: (D)

$\frac{{\epsilon }_0\epsilon A}{(2\epsilon +{\epsilon }_0)}$

Let the capacitance of the three capacitors be $C_1,C_2\&C_3$ respectively,
$C_1=\frac{{ϵ}_{0}A}{d},C_2=\frac{ϵA}{d}\&C_3=\frac{{ϵ}_{0}A}{d}$
Now, they are connected in series, So the equivalent capacitance is
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
$\Rightarrow \frac{1}{C}=\frac{d}{{ϵ}_{0}A}+\frac{d}{ϵA}+\frac{d}{{ϵ}_{0}A}$
$\frac{1}{C}=\frac{d}{A}[\frac{1}{ϵ}+\frac{2}{{ϵ}_0}]$
$\frac{1}{C}=\frac{d(2ϵ+{ϵ}_0)}{Aϵ{ϵ}_0}$
$\therefore C=\frac{{ϵ}_0ϵA}{d(2ϵ+{ϵ}_0)}$