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(A) Initially,when the switch $S$ is open,the two capacitors on the left are in parallel,and this combination is in series with the capacitor on the r

Vedclass · Accepted Answer

$3.3 	imes 10^{-4} \ C$

Vedclass · Answer

(A) Initially,when the switch $S$ is open,the two capacitors on the left are in parallel,and this combination is in series with the capacitor on the right.Let $C = 5.0 \ \mu F = 5 	imes 10^{-6} \ F$.The equivalent capacitance is $C_{eq1} = \frac{(C + C) 	imes C}{(C + C) + C} = \frac{2C^2}{3C} = \frac{2}{3}C$.$C_{eq1} = \frac{2}{3} 	imes 5 	imes 10^{-6} \ F = \frac{10}{3} 	imes 10^{-6} \ F$.The total charge stored is $Q_i = C_{eq1} 	imes V = (\frac{10}{3} 	imes 10^{-6}) 	imes 50 = \frac{500}{3} 	imes 10^{-6} \ C \approx 1.67 	imes 10^{-4} \ C$.When the switch $S$ is closed,the capacitor on the right is short-circuited. Only the two parallel capacitors remain in the circuit.The equivalent capacitance is $C_{eq2} = C + C = 2C = 10 	imes 10^{-6} \ F$.The new total charge stored is $Q_f = C_{eq2} 	imes V = 10 	imes 10^{-6} 	imes 50 = 5.0 	imes 10^{-4} \ C$.The charge that flows through $AB$ is the difference between the final and initial charges:$\Delta Q = Q_f - Q_i = 5.0 	imes 10^{-4} - 1.67 	imes 10^{-4} = 3.33 	imes 10^{-4} \ C$.Thus,the charge flowing through $AB$ is $3.3 	imes 10^{-4} \ C$.

Each capacitor shown in the figure has a capacitance of $5.0 \ \mu F$. The battery has an $emf$ of $50 \ V$. If the switch $S$ is closed,what amount of charge will flow through $AB$?

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