# Capacitors circuit question

## Homework Statement

A battery of 15V supplies two capacitors via a two-way switch. Initially the switch is to the left hand side, calculate the charge on the 1mf capacitor when the switch is switched to the other side (as shown in diagram).

## Homework Equations

Q=VC, Total Capacitance in Parallel = C1 + C2 + C3 etc., 1/Total capacitance in series = 1/c1 + 1/c2 + 1/c3 etc.

http://img15.imageshack.us/img15/7147/capacitors.png [Broken]

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ideasrule
Homework Helper
There's only one connected capacitor when the switch is on the left side.

Initially the 1mf capacitor has 15mc which I understand but once the switch is flipped it loses 5mC to have only 10mC quantity of charge. I'm also having trouble of thinking the problem through, once the switch is flipped the capacitors are in series? as the battery can be neglected due to broken circuit?

Anyways an mathematical explanation as well as a qualitative one would be most appreciated.
Thanks!

Yeah when the switch is on the left hand side only one capacitor, the 1mf capacitor, so the process I used to find the charge initially was Q=VC, 15x(1x10^-3) = 15mC.

ideasrule
Homework Helper
Who cares about whether the capacitors are in series or not. You know that total charge can't change, and that the voltage across both capacitors have to be equal. You can write 2 equations with these facts and solve for Q1 and Q2.

You're absolutely correct, I for some reason kept thinking the battery would continue to function but it can be neglected and the 1mf capacitor would begin to discharge to reach an equilibrium with the 0.5mf capacitor.

But like you said the initial charge cannot change and so is conserved and then ratio's of the capacitance values will give the ratio to which the charge is deposited.

I was overcomplicating the problem and not thinking about my 'definitions' correctly.

Thanks again.