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 electric potential across the 1.00-nF capacitor?

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Quizplus · Accepted Answer

When capacitors are connected in parallel, the charge is redistributed between them, but the total charge is conserved. 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 $Q_{\text{total}} = Q_1 + Q_2 = 30.0 \, \text{nC} + 80.0 \, \text{nC} = 110.0 \, \text{nC}$. The total capacitance is $C_{\text{total}} = C_1 + C_2 = 1.00 \, \text{nF} + 2.00 \, \text{nF} = 3.00 \, \text{nF}$. The final voltage across both capacitors is $V = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{110.0 \, \text{nC}}{3.00 \, \text{nF}} = 36.7 \, \text{V}$.

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