A parallel plate capacitor has a capacity 80 | vedclass

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Question

(A) Initial capacitance with air is $C = 80 	imes 10^{-6} \,F$.When the dielectric slab of constant $k = 20$ is inserted, the new capacitance becomes

Vedclass · Accepted Answer

$45.6 	imes 10^{-3} \,C$

Vedclass · Answer

(A) Initial capacitance with air is $C = 80 	imes 10^{-6} \,F$.When the dielectric slab of constant $k = 20$ is inserted, the new capacitance becomes $C' = kC = 20 	imes 80 	imes 10^{-6} = 1600 	imes 10^{-6} \,F$.The capacitor is connected to a battery of $V = 30 \,V$, so the charge on the capacitor is $q' = C'V = (1600 	imes 10^{-6}) 	imes 30 = 48000 	imes 10^{-6} \,C = 48 	imes 10^{-3} \,C$.When the dielectric slab is removed, the capacitance returns to $C = 80 	imes 10^{-6} \,F$. The new charge on the capacitor is $q = CV = (80 	imes 10^{-6}) 	imes 30 = 2400 	imes 10^{-6} \,C = 2.4 	imes 10^{-3} \,C$.The charge that passes through the wire is $\Delta q = q' - q = 48 	imes 10^{-3} - 2.4 	imes 10^{-3} = 45.6 	imes 10^{-3} \,C$.

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