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(B) The capacitance of a parallel plate capacitor with a dielectric medium is given by the formula: $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$.Here

Vedclass · Accepted Answer

$4.52$

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(B) The capacitance of a parallel plate capacitor with a dielectric medium is given by the formula: $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$.Here,$C = 2 	imes 10^{-4} \, \mu F = 2 	imes 10^{-4} 	imes 10^{-6} \, F = 2 	imes 10^{-10} \, F$.The area $A = 0.01 \, m^2$,distance $d = 2 \, mm = 2 	imes 10^{-3} \, m$,and permittivity of free space $\varepsilon_0 = 8.85 	imes 10^{-12} \, F/m$.Rearranging the formula to solve for the dielectric constant $\varepsilon_r$:$\varepsilon_r = \frac{C \cdot d}{\varepsilon_0 \cdot A}$Substituting the values:$\varepsilon_r = \frac{(2 	imes 10^{-10}) 	imes (2 	imes 10^{-3})}{(8.85 	imes 10^{-12}) 	imes (0.01)}$$\varepsilon_r = \frac{4 	imes 10^{-13}}{8.85 	imes 10^{-14}}$$\varepsilon_r = \frac{40}{8.85} \approx 4.52$.

$A$ parallel plate capacitor has a potential of $20 \, kV$ and a capacitance of $2 \times 10^{-4} \, \mu F$. If the area of the plates is $0.01 \, m^2$ and the distance between the plates is $2 \, mm$,find the dielectric constant of the medium.

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