Question

A 1.00-\muF capacitor charged to \(50.0 \mathrm{~V}\) and a 2.00 - \(\mu \mathrm{F}\) capacitor charged to \(20.0 \mathrm{~V}\) are connected, with the positive plate of each connected to the negative plate of the other. What is the final charge on the \(1.00-\mu \mathrm{F}\) capacitor?

Short answer

Answer: The final charge on the 1.00-μF capacitor is 2.00 x 10^-5 C.

Step-by-step solution

Write the expressions for initial charges of the capacitors

We are given the capacitance and voltage of both capacitors. We can use the formula, \(Q = CV\), to find the initial charges on each capacitor. Let's denote the initial charges as \(Q_1\) for the 1.00-μF capacitor and \(Q_2\) for the 2.00-μF capacitor: \(Q_1 = C_1 V_1 = (1.00 \times 10^{-6} \,\mathrm{F})(50.0\,\mathrm{V})\) \(Q_2 = C_2 V_2 = (2.00 \times 10^{-6} \,\mathrm{F})(20.0\,\mathrm{V})\)

Apply charge conservation principle

When the capacitors are connected, charge conservation must be maintained. This means that the total initial charge must be equal to the total final charge. We can express this as: \(Q_1 + Q_2 = Q_{1f} + Q_{2f}\)

Apply energy conservation principle

Similarly, we need to conserve the energy stored in the capacitors before and after connecting them. The stored energy in a capacitor can be calculated using the formula, \(U = \frac{1}{2}C V^2\). Let \(U_1\) and \(U_2\) represent the initial energies, and \(U_{1f}\) and \(U_{2f}\) represent the final energies: \(U_1 + U_2 = U_{1f} + U_{2f}\) \(\frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2 = \frac{1}{2}C_1 V_{1f}^2 + \frac{1}{2}C_2 V_{2f}^2\)

Find the final voltages

Next, we need to find the final voltages across the capacitors. We can do this by dividing the total final charge by the capacitance of each capacitor: \(V_{1f} = \frac{Q_{1f}}{C_1}\) \(V_{2f} = \frac{Q_{2f}}{C_2}\)

Substitute and solve for the final charge

Now we can substitute the equations for the final voltages into the energy conservation equation: \(\frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2 = \frac{1}{2}C_1 \left(\frac{Q_{1f}}{C_1}\right)^2 + \frac{1}{2}C_2 \left(\frac{Q_{2f}}{C_2}\right)^2\) Solve the above equation for \(Q_{1f}\), knowing that \(Q_{1f} + Q_{2f} = Q_1 + Q_2\): \(Q_{1f} = \frac{C_1 C_2 (V_1 + V_2)}{C_1 + C_2} \) Finally, substitute the given values into the equation for \(Q_{1f}\): \(Q_{1f} = \frac{(1.00 \times 10^{-6}\,\mathrm{F})(2.00 \times 10^{-6}\,\mathrm{F})(50.0\,\mathrm{V} + 20.0\,\mathrm{V})}{(1.00 \times 10^{-6}\,\mathrm{F}) + (2.00 \times 10^{-6}\,\mathrm{F})}\) Calculate the final charge on the 1.00-μF capacitor: \(Q_{1f} = 2.00 \times 10^{-5}\,\mathrm{C}\) So, the final charge on the 1.00-μF capacitor is \(2.00 \times 10^{-5}\,\mathrm{C}\).