# I have a problem with some capacitors connected

1. Sep 10, 2014

### anachin6000

I've found a problem about 2 identic capacitors which have the same voltage. One of them is sinked in oil and after that the 2 capacitors are connected in series. The problem ask for the for the energy variation. I've done the problem but I don't understand something:

When the capacitors are conected they must change charge between them because of the voltage difference. They alredy had the same charges and I know that two capacitors connected in series will have the same charge. So how is possible to change charge between them?

2. Sep 10, 2014

### skeptic2

What effect might the oil have on the capacitor?

3. Sep 10, 2014

### anachin6000

It fills the space between the capacitor's plates and change it's capacity, so as a result it's voltage will be changed.

4. Sep 10, 2014

### skeptic2

Why will the oil change the capacitor's capacity and will it increase or decrease it?

5. Sep 11, 2014

### anachin6000

The oil is conssidered a dielectic that have the permitivity bigger than the air so the capacity will increase.

But it's not relevant to know that for my question. I just want to know what happens with the charge.

6. Sep 11, 2014

### Staff: Mentor

If there is no external path leading from one end of a capacitor to its other end, then initial charge on its plates remains unchanged.

7. Sep 11, 2014

### anachin6000

When I said about charge Iwas refering to the topic question not to that capacitor.

8. Sep 12, 2014

### sophiecentaur

It is totally relevant to your original question.
Adding oil to one of the capacitors is changing its Capacity. When you do this and the capacitors are connected in parallel, the charges will no longer be equal (Q=CV) so the voltage will change (if disconnected from the supply) or some balancing current will flow if they are connected to a supply.

There is an extra factor that you need to consider and that is the energy expended in either case. There has to be a finite resistance in connecting wires or in the supply and you get a discrepancy in the before and after energy situation. This will be resolved if you allow the energy to be dissipated in the resistances.