A capacitor has a capacitance of 15 mu F when | vedclass

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(B) Initial charge on the first capacitor: $Q_1 = C_1 V = (15\ \mu F) 	imes 100\ V = 1500\ \mu C$.Initial charge on the second capacitor: $Q_2 = C_2

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(B) Initial charge on the first capacitor: $Q_1 = C_1 V = (15\ \mu F) 	imes 100\ V = 1500\ \mu C$.Initial charge on the second capacitor: $Q_2 = C_2 V = (1\ \mu F) 	imes 100\ V = 100\ \mu C$.When the dielectric is removed from the first capacitor,its new capacitance becomes $C_1' = \frac{15\ \mu F}{15} = 1\ \mu F$.When connected in parallel,the total charge $Q_{total} = Q_1 + Q_2 = 1500\ \mu C + 100\ \mu C = 1600\ \mu C$.The equivalent capacitance is $C_{eq} = C_1' + C_2 = 1\ \mu F + 1\ \mu F = 2\ \mu F$.The final voltage $V_{final} = \frac{Q_{total}}{C_{eq}} = \frac{1600\ \mu C}{2\ \mu F} = 800\ V$.

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