# Capacitor charged then Connected to another Capacitor.

1. Apr 11, 2013

### Jonnyto

1. The problem statement, all variables and given/known data
One capacitor is charged until its stored energy is 4.0 J, the charging battery then being removed. A second uncharged capacitor is then connected to it in parallel. (a) If the charge distributes equally, what is now the total energy stored in the electric fields?(b) Where did the excess energy go?

2. Relevant equations
u=1/2*ε*E2 Where E is the electric field and u is the energy density(U/Volume)

3. The attempt at a solution
Okay so I'm somewhat lost. Since the charge is spread equally, and they are in parallel, so I assume they have the same capacitance. So for that I assumed that the total energy stored in each is half of that? However my reasoning is not correct since energy disappears? How can I approach the problem correctly?

2. Apr 11, 2013

### Dick

The charge will redistribute between the two capacitors. Each of two capacitors will have half the charge of the original capacitor. How will that affect the electric field in each capacitor?

3. Apr 11, 2013

### Jonnyto

Would that cut the electric field in half?

4. Apr 11, 2013

### Dick

Yes, it would. What will that do to the total stored energy?

5. Apr 11, 2013

### Jonnyto

1/4 the original energy on each.

6. Apr 11, 2013

### Dick

Right again. So now you've got half the energy you started with. Where did it go?

7. Apr 11, 2013

### Jonnyto

Hmm that's what I'm unsure of now. Was this energy lost pushing the charge to the other capacitor?

8. Apr 11, 2013

### Dick

Well, it has to flow through a wire. What kind of property of wire might make that not free?

9. Apr 11, 2013

### haruspex

Yes, but how exactly?
In the real world, the connection will have some inductance and some resistance, however small. What will that lead to?