A parallel plate capacitor has rectangular pl | vedclass

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(B) Given: Area $A = 400 \, cm^2 = 400 	imes 10^{-4} \, m^2 = 4 	imes 10^{-2} \, m^2$. Distance $d = 2 \, mm = 2 	imes 10^{-3} \, m$. Potential dif

Vedclass · Accepted Answer

$3.54 	imes 10^{-8} \, C$

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(B) Given: Area $A = 400 \, cm^2 = 400 	imes 10^{-4} \, m^2 = 4 	imes 10^{-2} \, m^2$. Distance $d = 2 \, mm = 2 	imes 10^{-3} \, m$. Potential difference $V = 200 \, V$. Permittivity of free space $\epsilon_0 = 8.854 	imes 10^{-12} \, F/m$. The capacitance $C$ is given by $C = \frac{\epsilon_0 A}{d}$. Substituting the values: $C = \frac{8.854 	imes 10^{-12} 	imes 4 	imes 10^{-2}}{2 	imes 10^{-3}} = 17.708 	imes 10^{-11} \, F$. The charge $Q$ is given by $Q = CV$. $Q = (17.708 	imes 10^{-11}) 	imes 200 = 3541.6 	imes 10^{-11} \, C = 3.5416 	imes 10^{-8} \, C$. Thus,the charge on the plates is approximately $3.54 	imes 10^{-8} \, C$.

$A$ parallel plate capacitor has rectangular plates of area $400 \, cm^2$ separated by a distance of $2 \, mm$ with air as the medium. If a potential difference of $200 \, V$ is applied across the capacitor,what is the charge on the plates?

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