A parallel plate capacitor has plate area $40 \mathrm{~cm}^2$ and plate separation 2 mm . The space between…
A parallel plate capacitor has plate area $40 \mathrm{~cm}^2$ and plate separation 2 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5 . The capacitance of the system is ( $\varepsilon_0=$ permittivity of vacuum)
$24 \varepsilon_0 \mathrm{~F}$
$\frac{3}{10} \varepsilon_0 \mathrm{~F}$
$\frac{10}{3} \varepsilon_0 \mathrm{~F}$
$10 \varepsilon_0 \mathrm{~F}$
Solution
As the two capacitors are in series combination,
$\begin{aligned}
& \frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}=\frac{1}{\frac{\mathrm{~A} \varepsilon_0 k}{t}}+\frac{1}{\frac{A \varepsilon_0}{d-t}}=\frac{t}{\mathrm{~A} \varepsilon_0 \mathrm{k}}+\frac{d-t}{\mathrm{~A} \varepsilon_0} \\
& =\frac{1 \times 10^{-3}}{40 \times 10^{-4} \times 5 \varepsilon_0}+\frac{\left(2 \times 10^{-3}-1 \times 10^{-3}\right)}{40 \varepsilon_0 \times 10^{-4}} \\
& \therefore \quad \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{20 \varepsilon_0}+\frac{1}{4 \varepsilon_0} \\
& \therefore \quad \mathrm{C}_{\mathrm{eq}}=\frac{10}{3} \varepsilon_0 \quad \mathrm{~F}
\end{aligned}$