Answer to Question #86448 in Electric Circuits for Steven Rearden

Question

A 2 microfarad capactior is charged to 50 V and then connected in parallel (positive plate to positive plate) with a 4 microfarad capacitor charged to 100 V.
a. What are the final charges on the capactiors?
b. What is the potential difference across each?
c. Repeat a and b if the positive plate of one capacitor is connected to the negative plate of the other.

Answer 1

Before the connection the first and the second capacitors had a charge

$$Q_1=C_1V_1,$$ $$Q_2=C_2V_2.$$

a) After they are connected in parallel, the charge (since both positive and both negative charges repel) will be

$$Q_{a)}=Q_1+Q_2=500\space\mu\text{C}.$$

b) We know also that the total capacitance for series connection is given by

$$C_{||}=C_1+C_2,$$

and (remember about the charges in part a)

$$V_{b)}=\frac{Q_{a)}}{C_{||}}=83.33\text{ V}.$$

c) In this case charges of different signs will attract to each other and will cancel, and the total charge on two capacitors will be

$$Q_{c)}=Q_2-Q_1=300\space\mu\text{C}.$$

And this total charge defines the voltage:

$$V_{c)}=\frac{Q_{c)}}{C_{||}}=50 \text{ V}.$$

Individual charges:

$$Q_{1c)}=C_1V_{c)}=100\mu\text{C},$$ $$Q_{2c)}=C_2V_{c)}=200\mu\text{C}.$$