Question

Find the percentage increase in the capacitance of a parallel plate capacitor after it is half-filled with a dielectric of dielectric constant $5$ , if it is known that its capacitance with air as the medium is $C$ .

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

Answer 1

Answer: (B)

When it is half filled with dielectric of $K=5$ , its capacitance becomes $C_s$ which is series combination of
$C_1=\frac{K{ϵ}_{0}A}{d/1}=\frac{{ϵ}_{0}A}{d/2K}=\frac{{ϵ}_{0}A}{d/10}$ and $C_2=\frac{{ϵ}_{0}A}{d/2}$
$\therefore \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{d/10}{{\left(ϵ\right)}_{0}A}+\frac{d/2}{{\left(ϵ\right)}_{0}A}=\frac{(3d/5)}{{\left(ϵ\right)}_{0}A}$
$\Rightarrow C_s=\frac{5}{3}\frac{{\left(ϵ\right)}_{0}A}{d}=\frac{5C}{3}(using(i))$
Percentage increase in capacitance $=\frac{C_s-C}{C}\times 100$
$=\frac{\frac{5}{3}C-C}{C}\times =\frac{2}{3}\times 100=66.6\%$