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(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 $C' = K 	imes C = 20 	imes 80 	imes 10^{-6} = 1600 	imes 10^{-6} \ F$. The charge on the capacitor with the dielectric is $q_1 = C' V = (1600 	imes 10^{-6}) 	imes 30 = 48000 	imes 10^{-6} \ C = 48 	imes 10^{-3} \ C$. After removing the dielectric,the capacitance returns to $C = 80 	imes 10^{-6} \ F$. The new charge on the capacitor is $q_2 = C V = (80 	imes 10^{-6}) 	imes 30 = 2400 	imes 10^{-6} \ C = 2.4 	imes 10^{-3} \ C$. The charge that flows through the wire is $\Delta q = q_1 - q_2 = 48 	imes 10^{-3} - 2.4 	imes 10^{-3} = 45.6 	imes 10^{-3} \ C$.

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