Connected Capacitors: Charge & Potential Diffrences

[andrewdavis23]

Homework Statement

An isolated capacitor C₁ carries a charge Q₀. Its wires are then connected to those of a second capacitor C₂, previously uncharged. What charge will each carry now? What will be the potential difference across each?

Homework Equations

C=Q/V

The Attempt at a Solution

I figured the charge would eventually become evenly distributed (no measure of time was given). So, I figured...

Q₁=Q₂=1/2 Q₀

The potential difference (voltage) would be zero because the charges are equal. ?

 

[Mandeep Deka]

Are you sure?

Lets see what happens if we take your first answer as correct. You said (from a very obvious perspective) that charge gets equally distributed, but then if you apply Kirchhoff's loop rule (KVL) across the closed circuit, doesn't it get violated?? The total voltage difference across the loop turns out to be:

Q₀/2C₁-Q₀/2C₂

which is not equal to zero... which means that you answer isn't correct. Any any case you can't violate KVL!

Indeed you try to distribute the charge in such a way that KVL is satisfied, and since the sum of the charges present in the plates has to be Q₀... you will get the correct answer!