Physics - Electrostatics Question with Solution | TestHub

Initial capacitance

$C=\frac{{\epsilon }_{0}A}{d}$



When it is half filled by a dielectric of dielectric constant$K$, then

${\text{C}}_1=\frac{\mathit{\text{K}}{\mathit{\epsilon }}_0\mathit{\text{A}}}{\mathit{\text{d}}/2}=2\mathit{\text{K}}\frac{{\epsilon }_0\mathit{\text{A}}}{\mathit{\text{d}}}$

and${\mathit{\text{C}}}_2=\frac{{\mathit{\epsilon }}_0\mathit{\text{A}}}{\mathit{\text{d}}/2}=\frac{{2\epsilon }_0\mathit{\text{A}}}{\mathit{\text{d}}}$

$\therefore \frac{1}{C\text{'}\text{'}}=\frac{1}{C_1}+\frac{1}{C_2}$

$=\frac{d}{2{\epsilon }_{0}A}\left(\frac{1}{K}+1\right)=\frac{d}{2{\epsilon }_{0}A}\left(\frac{1}{5}+1\right)$

$=\frac{6}{10} \frac{d}{{\epsilon }_{0}A}$

$\therefore C^{\text{'}\text{'}}=\frac{5{\epsilon }_{0}A}{3d}$

Hence, increase in capacitance

$= \frac{\frac{5}{3}\frac{{\epsilon }_{0}A}{d}-\frac{{\epsilon }_{0}A}{d}}{\frac{{\epsilon }_{0}A}{d}}=\frac{5}{3}-1=\frac{2}{3}=66.6\%$