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A parallel plate capacitor has plate area $40\,cm ^2$ and plates separation $2\,mm$. The space between the plates is filled with a dielectric medium of a thickness $1\,mm$ and dielectric constant $5$ . The capacitance of the system is :
$24 \varepsilon_0\; F$
$\frac{3}{10} \varepsilon_0\; F$
$\frac{10}{3} \varepsilon_0\; F$
$10 \varepsilon_0 \; F$
Solution
This can be seen as two capacitors in series combination so
$\frac{1}{C_{e q}}=\frac{1}{C_1}+\frac{1}{C_2}$
$=\frac{1}{\frac{K \in_0 A}{t}}+\frac{1}{\frac{\epsilon_0 A}{d-t}}$
$=\frac{ t }{ K \in_0 A }+\frac{ d – t }{\epsilon_0 A }$
$=\frac{1 \times 10^{-3}}{5 \epsilon_0 \times 40 \times 10^{-4}}+\frac{1 \times 10^{-3}}{\epsilon_0 40 \times 10^{-4}}$
$\frac{1}{ C _{ eq }}=\frac{1}{20 \epsilon_0}+\frac{1}{4 \epsilon_0}$
$C _{ eq }=\frac{20 \times 4 \epsilon_0}{24}=\frac{10 \epsilon_0}{3} F$
