The capacitance of a capacitor made by a thin | vedclass

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(D) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{K \varepsilon_0 A}{d}$,where $A = l 	imes b$.Given:Capacitance $

Vedclass · Accepted Answer

$33.9$

Vedclass · Answer

(D) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{K \varepsilon_0 A}{d}$,where $A = l 	imes b$.Given:Capacitance $C = 2 	imes 10^{-6}\,F$Dielectric constant $K = 2.5$Thickness $d = 0.15 	imes 10^{-3}\,m$Width $b = 400 	imes 10^{-3}\,m = 0.4\,m$Permittivity of free space $\varepsilon_0 = 8.85 	imes 10^{-12}\,F/m$Substituting the values into the formula:$2 	imes 10^{-6} = \frac{2.5 	imes 8.85 	imes 10^{-12} 	imes (l 	imes 0.4)}{0.15 	imes 10^{-3}}$$2 	imes 10^{-6} = \frac{2.5 	imes 8.85 	imes 10^{-12} 	imes 0.4 	imes l}{0.15 	imes 10^{-3}}$$2 	imes 10^{-6} = \frac{8.85 	imes 10^{-12} 	imes l}{0.15 	imes 10^{-3}}$$l = \frac{2 	imes 10^{-6} 	imes 0.15 	imes 10^{-3}}{8.85 	imes 10^{-12}}$$l = \frac{0.3 	imes 10^{-9}}{8.85 	imes 10^{-12}} = \frac{300}{8.85} \approx 33.9\,m$.

The capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$,the dielectric constant of the paper is $2.5$,and the width of the paper is $400\,mm$,then the length of the foil will be.....$m$.

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