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

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(B) Initial charge on capacitor $A$: $Q_1 = C_{A, 	ext{initial}} 	imes V = 15\ \mu F 	imes 100\ V = 1500\ \mu C = 15 	imes 10^{-4}\ C$.Initial cha

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(B) Initial charge on capacitor $A$: $Q_1 = C_{A, 	ext{initial}} 	imes V = 15\ \mu F 	imes 100\ V = 1500\ \mu C = 15 	imes 10^{-4}\ C$.Initial charge on capacitor $B$: $Q_2 = C_B 	imes V = 1\ \mu F 	imes 100\ V = 100\ \mu C = 10^{-4}\ C$.After removing the dielectric from capacitor $A$, its new capacitance is $C_A' = \frac{C_{A, 	ext{initial}}}{K} = \frac{15\ \mu F}{15} = 1\ \mu F$.When connected in parallel, the total charge $Q_{	ext{total}} = Q_1 + Q_2 = 1500\ \mu C + 100\ \mu C = 1600\ \mu C$.The equivalent capacitance in parallel is $C_{	ext{eq}} = C_A' + C_B = 1\ \mu F + 1\ \mu F = 2\ \mu F$.The common potential is $V_{	ext{common}} = \frac{Q_{	ext{total}}}{C_{	ext{eq}}} = \frac{1600\ \mu C}{2\ \mu F} = 800\ V$.

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