# Capacitors connected to each other

## Homework Statement

Two capacitors, one that has a capacitance of 4.30 µF and one that has a capacitance of 12.1 µF, are connected in parallel. The parallel combination is then connected across the terminals of a 9.0-V battery. Next, they are carefully disconneted so that tehy are not discharged. They are then reconnected to each other--the positive plate of each capacitor connected to the negative plate of the other. Find the potential difference across each capacitor after they are reconnected.

C1 = 4.30µF
C2= 12.1µF
V = 9v

C = Q/V

## The Attempt at a Solution

Before they are reconnected each capacitor is charged to the following:
C1(V) = Q1 --> (4.3µF)(9v) = 38.7µC
C2(v) = Q2 --> (12.1µF)(9v) =108.9µC

But I lose my way after the reconnection. I understand that the voltage will change are charge will rearrange but when it comes to the specifics im not entirely sure. Any help would be appreciated.

LowlyPion
Homework Helper
Welcome to PF.

What happens to the positive charge of the one capacitor and the negative charge of the other when they are connected?

Did I hear you say they cancel? Good.

So if they cancel what is the net charge across the 2 capacitors?

And since you are already armed with the equivalent capacitance ... and you know the charge ... hmm I wonder how to find the voltage?