A 1.00-nF capacitor initially charged to 30.0 V and a 2.00-nF capacitor charged to 40.0 V are connected to each other with the negative plate of each connected to the negative plate of the other.What is the final charge on the 2.00-nF capacitor?

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Quizplus · Accepted Answer

When two capacitors are connected in parallel, the total charge is conserved, and the final voltage across both capacitors is the same. The initial charge on the 1.00-nF capacitor is $Q_1 = C_1V_1 = (1.00\, \text{nF})(30.0\, \text{V}) = 30.0\, \text{nC}$, and the initial charge on the 2.00-nF capacitor is $Q_2 = C_2V_2 = (2.00\, \text{nF})(40.0\, \text{V}) = 80.0\, \text{nC}$. The total initial charge is $30.0\, \text{nC} + 80.0\, \text{nC} = 110.0\, \text{nC}$. The final charge on the 2.00-nF capacitor, with the same final voltage $V_f$ for both capacitors, is $Q_{2f} = C_2V_f$. Since the total charge is conserved, $V_f = \frac{110.0\, \text{nC}}{1.00\, \text{nF} + 2.00\, \text{nF}} = 36.7\, \text{V}$, and thus $Q_{2f} = (2.00\, \text{nF})(36.7\, \text{V}) = 73.4\, \text{nC}$, which rounds to $73.3\, \text{nC}$, matching choice B.

A 100-NF Capacitor Initially Charged to 30