A parallel plate capacitor has plate area 40 | vedclass

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(C) The system can be viewed as two capacitors in series: one with the dielectric and one with the air gap.$C_1 = \frac{K \varepsilon_0 A}{t} = \frac{

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

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(C) The system can be viewed as two capacitors in series: one with the dielectric and one with the air gap.$C_1 = \frac{K \varepsilon_0 A}{t} = \frac{5 \varepsilon_0 	imes 40 	imes 10^{-4}}{1 	imes 10^{-3}} = 20 \varepsilon_0 \ F$$C_2 = \frac{\varepsilon_0 A}{d-t} = \frac{\varepsilon_0 	imes 40 	imes 10^{-4}}{1 	imes 10^{-3}} = 4 \varepsilon_0 \ F$Since they are in series,the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{20 \varepsilon_0} + \frac{1}{4 \varepsilon_0}$$\frac{1}{C_{eq}} = \frac{1 + 5}{20 \varepsilon_0} = \frac{6}{20 \varepsilon_0} = \frac{3}{10 \varepsilon_0}$$C_{eq} = \frac{10}{3} \varepsilon_0 \ F$

$A$ parallel plate capacitor has plate area $40 \ cm^2$ 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)

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