Question

A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be:

• A. $400\%$
• B. $66.6\%$
• C. $33.3\%$
• D. $200\%$

Answer 1

The correct option is B $66.6\%$

$C=\frac{{ϵ}_{0}A}{d}$ ....(i)
When it is half filled with dielectric of K = 5, its capacitance becomes $C_s$ which is series combination of

$C_1=\frac{K{ϵ}_{0}A}{\frac{d}{2}}=\frac{{ϵ}_{0}A}{\frac{d}{2K}}=\frac{{ϵ}_{0}A}{\frac{d}{10}}and{C}_2=\frac{{ϵ}_{0}A}{\frac{d}{2}}$
$\therefore \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{\frac{d}{10}}{{ϵ}_{0}A}+\frac{\frac{d}{2}}{{ϵ}_{0}A}=\frac{\frac{3d}{5}}{{ϵ}_{0}A}$
$\Rightarrow C_s=\frac{5}{3}\frac{{ϵ}_{0}A}{d}=\frac{5C}{3}$
(using (i))
Percentage increase in capacitance = $\frac{C_s-C}{C}\times 100$
= $\frac{\frac{5}{3}C-C}{C}\times 100=\frac{2}{3}\times 100=66.6\%$