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(B) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula: $C

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$\frac{20 \epsilon_0}{9}$

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(B) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula: $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.Given:Area $A = 50 \ cm^2 = 50 	imes 10^{-4} \ m^2 = 5 	imes 10^{-3} \ m^2$.Separation $d = 3 \ mm = 3 	imes 10^{-3} \ m$.Thickness of dielectric $t = 1 \ mm = 1 	imes 10^{-3} \ m$.Dielectric constant $K = 4$.Substituting the values:$C = \frac{\epsilon_0 (5 	imes 10^{-3})}{3 	imes 10^{-3} - 1 	imes 10^{-3} + \frac{1 	imes 10^{-3}}{4}}$$C = \frac{\epsilon_0 (5 	imes 10^{-3})}{2 	imes 10^{-3} + 0.25 	imes 10^{-3}}$$C = \frac{\epsilon_0 (5 	imes 10^{-3})}{2.25 	imes 10^{-3}}$$C = \frac{5 \epsilon_0}{2.25} = \frac{5 \epsilon_0}{9/4} = \frac{20 \epsilon_0}{9}$.

$A$ parallel plate capacitor has a plate area of $50 \ cm^2$ and a plate separation of $3 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and a dielectric constant of $4$. Calculate the capacitance. ($\epsilon_0$ is the permittivity of free space)

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