Parallel Capacitors: Find Energy Decrease

[wcelectric]

Homework Statement

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?

Homework Equations

u=\frac{1}{2}CV^2

C=\frac{Q}{V}

The Attempt at a Solution

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

 

[Redbelly98]

The Attempt at a Solution

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

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?

 

[wcelectric]

So would I just add them?

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

 

[Redbelly98]

So would I just add them?

Yes.

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

Yes.

 

[wcelectric]

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

 

[diazona]

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

 

[wcelectric]

Oops. I mixed up C with Coulombs.

 

[Redbelly98]

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

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.

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

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