Question 84 of 180

A parallel plate capacitor has plate area $40{cm}^{2}$ and plates separation $2mm$. The space between the plates is filled with a dielectric medium of a thickness $1mm$ and dielectric constant $5$. The capacitance of the system is:

A$24{\epsilon }_{0}F$B$\dfrac{3}{10}{\epsilon }_{0}F$C$\dfrac{10}{3}{\epsilon }_{0}F$D$10{\epsilon }_{0}F$

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Answer: (C) $\dfrac{10}{3}{\epsilon }_{0}F$

With insertion of dielectric, the system can be considered as combination of two capacitors in series.
Thus, equivalent capacitance of the system is
$\dfrac{1}{{C}_{eq}}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}$
$=\dfrac{1}{\dfrac{K{\epsilon }_{0}A}{t}}+\dfrac{1}{\dfrac{{\epsilon }_{0}A}{d-t}}$
$=\dfrac{t}{K{\epsilon }_{0}A}+\dfrac{d-t}{{\epsilon }_{0}A}$
$=\dfrac{1\times {10}^{-3}}{5{\epsilon }_{0}\times 40\times {10}^{-4}}+\dfrac{1\times {10}^{-3}}{{\epsilon }_{0}\times 40\times {10}^{-4}}$
$\Rightarrow \dfrac{1}{{C}_{eq}}=\dfrac{1}{20{\epsilon }_{0}}+\dfrac{1}{4{\epsilon }_{0}}$
$\Rightarrow {C}_{eq}=\dfrac{20\times 4{\epsilon }_{0}}{24}=\dfrac{10{\epsilon }_{0}}{3}F$.