# Parallel capacitors

1. Jul 18, 2009

### wcelectric

1. The problem statement, all variables and given/known data

Two capacitors are connected in parallel to a battery with voltage of 28V. Let $$C_1 = 9.0 \mu F$$ and $$C_2 = 4.0 \mu F$$.
Suppose the charged capacitors are disconnected from the source and from each other, and then reconnected to each other with plates of opposite signs together. By how much does the energy of the system decrease?

2. Relevant equations
$$u=\frac{1}{2}CV^2$$

$$C=\frac{Q}{V}$$

3. The attempt at a solution

I have $$Q_1 = 252 \mu C$$ and $$Q_2 = 112 \mu C$$. Beyond that I'm stuck.

2. Jul 18, 2009

### Redbelly98

Staff Emeritus
That's a reasonable start.

Next, they say that the plate with +252 μC plate is connected to the -112 μC plate on the other capacitor. How much net charge would there be, being shared between these two plates?

3. Jul 18, 2009

### wcelectric

So would I just add them?

I am confused about the plates of opposite signs being together. It's still a parallel circuit?

4. Jul 18, 2009

### Redbelly98

Staff Emeritus
Yes.
Yes.

5. Jul 18, 2009

### wcelectric

Ok, so the net charge is 140 μC. But I don't understand why.

Last edited: Jul 18, 2009
6. Jul 18, 2009

### diazona

I'd use the formula
$$U = \frac{1}{2}\frac{Q^2}{C}$$

7. Jul 18, 2009

### wcelectric

Oops. I mixed up C with Coulombs.

8. Jul 19, 2009

### Redbelly98

Staff Emeritus
It might help to think of an atom, say hydrogen for example. It contains two charged particles:

(1) a proton with charge 1.6 × 10-19 C
(2) an electron with an opposite charge of -1.6 × 10-19 C

To get the net charge of the hydrogen atom, we just add up all the charges in the atom, which is zero net charge in this example.

Similarly, whenever you combine charges, you simply add up the separate charges to get a net total charge.

That's right. You have Q, and just need to figure out what C is for this parallel capacitor combination.