A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric…
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
\(400 \%\)
\(66.6 \%\)
\(33.3 \%\)
\(200 \%\)
Solution
Initial capacitance \(=\frac{\epsilon \mathrm{A}}{\mathrm{d}}\)
When it is half filled by a dielectric of dielectric constant \(k\) then
$\mathrm{C}_{1}=\frac{\mathrm{k} \mathrm{e}, \mathrm{A}}{\frac{\mathrm{d}}{2}}=2 \mathrm{k} \frac{\mathrm{\epsilon} \mathrm{A}}{\mathrm{d}}$
$\text { and } \mathrm{C}_{2}=\frac{\epsilon, \mathrm{A}}{\frac{\mathrm{d}}{2}}=\frac{2 \varepsilon, \mathrm{A}}{\mathrm{d}}$
$\begin{array}{l}
\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}} \\
=\frac{\mathrm{d}}{2 \mathrm{e} \mathrm{A}}\left(\frac{1}{\mathrm{k}}+1\right)
\end{array}$
$\begin{array}{l}
=\frac{\mathrm{d}}{2 \epsilon, \mathrm{A}}\left(\frac{1}{5}+1\right) \\
=\frac{\mathrm{d}}{2 \epsilon, \mathrm{A}}\left(\frac{6}{5}\right) \\
=\frac{3 \mathrm{~d}}{5 \epsilon, \mathrm{A}} \\
\therefore \mathrm{C}=\frac{5 \epsilon \mathrm{A}}{3 \mathrm{~d}}
\end{array}$
Hence, % increase in capacitance is
$=\left(\frac{\frac{\left(\frac{5}{3}\right) \epsilon, \mathrm{A}}{\mathrm{d}}-\frac{\epsilon, \mathrm{A}}{\mathrm{d}}}{\frac{\epsilon \mathrm{A}}{\mathrm{d}}}\right) \times 100=\frac{2}{3} \times 100=66.6 \%$
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