Question

A 5.00-nF capacitor charged to \(60.0 \mathrm{~V}\) and a 7.00 -nF capacitor charged to \(40.0 \mathrm{~V}\) are connected negative plate to negative plate. What is the final charge on the 7.00 -nF capacitor?

Short answer

Answer: The final charge on the 7.00-nF capacitor is 338.33 nC.

Step-by-step solution

Calculate the initial charges on each capacitor

To find the initial charges on each capacitor, use the formula: \(Q=CV\), where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the voltage. For the 5.00-nF capacitor: \(Q_1=C_1V_1=(5.00\,\mathrm{nF})(60.0\,\mathrm{V}) = 300.0\,\mathrm{nC}\) For the 7.00-nF capacitor: \(Q_2=C_2V_2=(7.00\,\mathrm{nF})(40.0\,\mathrm{V}) = 280.0\,\mathrm{nC}\)

Calculate the total initial charge in the system

Now, add the initial charges of both capacitors. \(Q_{\text{initial}}=Q_1 + Q_2 = 300.0\,\mathrm{nC} + 280.0\,\mathrm{nC} = 580.0\,\mathrm{nC}\)

Calculate the equivalent capacitance of the connected capacitors

Since the capacitors are connected in parallel, the equivalent capacitance (\(C_{\text{eq}}\)) is simply the sum of their individual capacitances. \(C_{\text{eq}}=C_1 + C_2 = 5.00\,\mathrm{nF} + 7.00\,\mathrm{nF} = 12.00\,\mathrm{nF}\)

Calculate final voltage across capacitors

Since the total charge must remain constant after connecting the capacitors in parallel and they have the same voltage after connecting, we can calculate the final voltage using the equivalent capacitance and total initial charge. \(V_{\text{final}}=\frac{Q_{\text{initial}}}{C_{\text{eq}}} = \frac{580.0\,\mathrm{nC}}{12.00\,\mathrm{nF}} = 48.33\,\mathrm{V}\)

Calculate the final charge on the 7.00-nF capacitor

The final charge on the 7.00-nF capacitor (\(Q_2^{\text{final}}\)) can be calculated using the final voltage and the capacitance of this capacitor. \(Q_2^{\text{final}}=C_2V_{\text{final}} = (7.00\,\mathrm{nF})(48.33\,\mathrm{V}) = 338.33\,\mathrm{nC}\) So, the final charge on the 7.00-nF capacitor is 338.33 nC.