A 11.0-µF capacitor and a 50.0-µF capacitor are charged by being connected across separate 55.0-V batteries.

(a) Determine the resulting charge on each capacitor.

(b) The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge of each capacitor?

(c) What is the final potential difference across the 50.0-µF capacitor?

Are these connected in series or parallel?

a. Q1 = C*V = 11 * 55 = Micro Coulombs.

Q2 = 50 * 55 = Micro Coulombs.

b. The capacitors will discharge:
Q = C*V = 61 * 0 = 0.

c. 0 volts.

To solve this problem, you need to understand the concept of capacitance and how capacitors behave when connected in series or parallel.

The capacitance of a capacitor (C) is a measure of its ability to store electric charge. It is given in units of farads (F). When a capacitor is connected to a battery, it charges up to a certain voltage (V). The charge (Q) on a capacitor is related to its capacitance and voltage by the equation: Q = C * V.

Now let's solve each part of the problem step by step:

(a) To determine the resulting charge on each capacitor, we can use the equation Q = C * V, where Q is the charge, C is the capacitance, and V is the voltage. The given capacitance of the first capacitor is 11.0 µF, and the voltage is 55.0 V, so the charge on the first capacitor is Q1 = 11.0 µF * 55.0 V. Similarly, for the second capacitor with a capacitance of 50.0 µF and a voltage of 55.0 V, the charge is Q2 = 50.0 µF * 55.0 V.

(b) When the capacitors are disconnected from their batteries and connected to each other, they are connected in parallel. In this configuration, the total charge is conserved, and it is evenly distributed between the capacitors. So the final charge on each capacitor will be half of the total charge. Therefore, the final charge on the first capacitor is Q1_final = Q_total / 2 and on the second capacitor is Q2_final = Q_total / 2.

(c) The final potential difference across the 50.0 µF capacitor can be determined by using the equation: V_final = Q_final / C. We already found the final charge on the second capacitor (Q2_final), and we know the capacitance (C) of the second capacitor. Substituting these values, we can calculate the final potential difference across the 50.0 µF capacitor.

So, in summary, the capacitors are initially connected in parallel when charged, and their charges are determined by the given capacitance and voltage values. When the capacitors are then connected to each other, they are still in parallel, and the final charge on each capacitor is half of the total charge. The final potential difference across the 50.0 µF capacitor can be calculated using the final charge and the capacitor's capacitance.