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(A) The maximum electric field $E_{\max}$ that the dielectric can withstand is $20 \ MVm^{-1} = 20 	imes 10^6 \ Vm^{-1}$.The maximum potential differ

Vedclass · Accepted Answer

$4.6 \ \mu C$

Vedclass · Answer

(A) The maximum electric field $E_{\max}$ that the dielectric can withstand is $20 \ MVm^{-1} = 20 	imes 10^6 \ Vm^{-1}$.The maximum potential difference $V_{\max}$ across the capacitor is given by $V_{\max} = E_{\max} 	imes d$,where $d = 40 \ \mu m = 40 	imes 10^{-6} \ m$.$V_{\max} = 20 	imes 10^6 	imes 40 	imes 10^{-6} = 800 \ V$.The capacitance $C$ is given by $C = \frac{K \epsilon_0 A}{d}$,where $K = 4.0$,$A = (80 	imes 10^{-3} \ m)^2 = 6400 	imes 10^{-6} \ m^2$,and $\epsilon_0 = 9 	imes 10^{-12} \ Fm^{-1}$.$C = \frac{4.0 	imes 9 	imes 10^{-12} 	imes 6400 	imes 10^{-6}}{40 	imes 10^{-6}} = 4.0 	imes 9 	imes 10^{-12} 	imes 160 = 5760 	imes 10^{-12} \ F = 5.76 	imes 10^{-9} \ F$.The maximum charge $Q_{\max}$ is $Q_{\max} = C 	imes V_{\max} = 5.76 	imes 10^{-9} 	imes 800 = 4608 	imes 10^{-9} \ C = 4.6 \ \mu C$.

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