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(C) The system can be treated as two capacitors in series: one with a dielectric of thickness $t = 1\,mm$ and another with air of thickness $(d - t) =

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$\frac{10}{3} \varepsilon_0\,F$

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(C) The system can be treated as two capacitors in series: one with a dielectric of thickness $t = 1\,mm$ and another with air of thickness $(d - t) = 1\,mm$.The equivalent capacitance $C_{eq}$ for capacitors in series is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$Where $C_1 = \frac{K \varepsilon_0 A}{t}$ and $C_2 = \frac{\varepsilon_0 A}{d - t}$.Substituting the values:$A = 40\,cm^2 = 40 	imes 10^{-4}\,m^2$$d = 2\,mm = 2 	imes 10^{-3}\,m$$t = 1\,mm = 1 	imes 10^{-3}\,m$$K = 5$$\frac{1}{C_{eq}} = \frac{t}{K \varepsilon_0 A} + \frac{d - t}{\varepsilon_0 A}$$\frac{1}{C_{eq}} = \frac{1 	imes 10^{-3}}{5 \varepsilon_0 	imes 40 	imes 10^{-4}} + \frac{1 	imes 10^{-3}}{\varepsilon_0 	imes 40 	imes 10^{-4}}$$\frac{1}{C_{eq}} = \frac{1}{20 \varepsilon_0} + \frac{1}{4 \varepsilon_0} = \frac{1 + 5}{20 \varepsilon_0} = \frac{6}{20 \varepsilon_0} = \frac{3}{10 \varepsilon_0}$Therefore,$C_{eq} = \frac{10}{3} \varepsilon_0\,F$.

$A$ parallel plate capacitor has a plate area of $40\,cm^2$ and a plate separation of $2\,mm$. 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:

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