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(D) The maximum electric field $E$ that the dielectric can withstand is its dielectric strength,$E = 3.6 	imes 10^{7} \, Vm^{-1}$.The capacitance of

Vedclass · Accepted Answer

$2.33$

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(D) The maximum electric field $E$ that the dielectric can withstand is its dielectric strength,$E = 3.6 	imes 10^{7} \, Vm^{-1}$.The capacitance of a parallel plate capacitor with a dielectric is $C = \frac{K \varepsilon_{0} A}{d}$.The maximum charge $q$ is given by $q = CV = C(Ed) = \left( \frac{K \varepsilon_{0} A}{d} ight) Ed = K \varepsilon_{0} A E$.Rearranging for the dielectric constant $K$,we get $K = \frac{q}{\varepsilon_{0} A E}$.Given $q = 7 	imes 10^{-6} \, C$,$A = 30 \pi 	imes 10^{-4} \, m^{2}$,$E = 3.6 	imes 10^{7} \, Vm^{-1}$,and $\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \, Nm^{2}C^{-2}$,which implies $\varepsilon_{0} = \frac{1}{36 \pi 	imes 10^{9}}$.Substituting these values: $K = \frac{7 	imes 10^{-6}}{\left( \frac{1}{36 \pi 	imes 10^{9}} ight) 	imes (30 \pi 	imes 10^{-4}) 	imes (3.6 	imes 10^{7})}$.$K = \frac{7 	imes 10^{-6} 	imes 36 \pi 	imes 10^{9}}{30 \pi 	imes 10^{-4} 	imes 3.6 	imes 10^{7}} = \frac{7 	imes 36 	imes 10^{3}}{30 	imes 3.6 	imes 10^{3}} = \frac{252}{108} = 2.33$.

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