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(C) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ in a gap of separation $d$ is gi

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$\frac{10}{3}\epsilon_0 	ext{ F}$

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(C) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ in a gap of separation $d$ is given by the formula:$C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$Given:Area $A = 40 	ext{ cm}^2 = 40 	imes 10^{-4} 	ext{ m}^2 = 4 	imes 10^{-3} 	ext{ m}^2$Separation $d = 2 	ext{ mm} = 2 	imes 10^{-3} 	ext{ m}$Thickness $t = 1 	ext{ mm} = 1 	imes 10^{-3} 	ext{ m}$Dielectric constant $K = 5$Substituting the values:$C = \frac{\epsilon_0 (4 	imes 10^{-3})}{2 	imes 10^{-3} - 1 	imes 10^{-3} + \frac{1 	imes 10^{-3}}{5}}$$C = \frac{\epsilon_0 (4 	imes 10^{-3})}{1 	imes 10^{-3} + 0.2 	imes 10^{-3}}$$C = \frac{4 	imes 10^{-3} \epsilon_0}{1.2 	imes 10^{-3}}$$C = \frac{4}{1.2} \epsilon_0 = \frac{40}{12} \epsilon_0 = \frac{10}{3} \epsilon_0 	ext{ F}$

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

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