A parallel plate capacitor with area 200,cm^2 | vedclass

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(C) Given:Area of parallel plate capacitor,$A = 200\,cm^2 = 200 	imes 10^{-4}\,m^2 = 2 	imes 10^{-2}\,m^2$Separation between the plates,$d = 1.5\,cm

Vedclass · Accepted Answer

$250$

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(C) Given:Area of parallel plate capacitor,$A = 200\,cm^2 = 200 	imes 10^{-4}\,m^2 = 2 	imes 10^{-2}\,m^2$Separation between the plates,$d = 1.5\,cm = 1.5 	imes 10^{-2}\,m$Force of attraction between the plates,$F = 25 	imes 10^{-6}\,N$Permittivity of free space,$\epsilon_0 = 8.85 	imes 10^{-12}\,C^2/Nm^2$The force of attraction between the plates of a capacitor is given by the formula:$F = \frac{Q^2}{2A\epsilon_0}$Since $Q = CV = \frac{\epsilon_0 A V}{d}$,we substitute $Q$ into the force equation:$F = \frac{(\frac{\epsilon_0 A V}{d})^2}{2A\epsilon_0} = \frac{\epsilon_0^2 A^2 V^2}{d^2 \cdot 2A\epsilon_0} = \frac{\epsilon_0 A V^2}{2d^2}$Rearranging for $V^2$:$V^2 = \frac{2F d^2}{\epsilon_0 A}$Substituting the values:$V^2 = \frac{2 	imes (25 	imes 10^{-6}) 	imes (1.5 	imes 10^{-2})^2}{(8.85 	imes 10^{-12}) 	imes (200 	imes 10^{-4})}$$V^2 = \frac{50 	imes 10^{-6} 	imes 2.25 	imes 10^{-4}}{8.85 	imes 10^{-12} 	imes 2 	imes 10^{-2}}$$V^2 = \frac{112.5 	imes 10^{-10}}{17.7 	imes 10^{-14}} \approx 6.356 	imes 10^4 \approx 63560$$V = \sqrt{63560} \approx 252.1\,V$Thus,the value of $V$ is approximately $250\,V$.

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