Capacitors connected to each other

[magevivi]

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

Homework Equations

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 I am not entirely sure. Any help would be appreciated.

 

[LowlyPion]

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?

 

[nlightner]

I would approach this problem by using the equation Q=CV to determine the charge on each capacitor before and after the reconnection. Initially, each capacitor has a charge of Q1=38.7 µC and Q2=108.9 µC. When they are reconnected, the total charge on the two capacitors remains the same, but it is now distributed differently.

Since the positive plate of C1 is connected to the negative plate of C2, they will have equal and opposite charges. This means that the charge on the positive plate of C1 will decrease by some amount, while the charge on the negative plate of C2 will increase by the same amount. Similarly, the charge on the negative plate of C1 will decrease, and the charge on the positive plate of C2 will increase.

To determine the potential difference across each capacitor after the reconnection, we can use the equation Q=CV again. The total charge on the two capacitors is still the same, so we can set up the following equation:

Q1 + Q2 = C1(V1) + C2(V2)

Since the capacitors are connected in parallel, they have the same voltage across them (V1 = V2 = V). We can also substitute in the values for C1, C2, Q1, and Q2 that we calculated earlier. This gives us:

38.7µC + 108.9µC = (4.30µF)(V) + (12.1µF)(V)

Solving for V, we get V = 6.75V. This means that after the reconnection, each capacitor will have a potential difference of 6.75V across it. This is lower than the initial voltage of 9V, which makes sense since some of the charge has been redistributed between the two capacitors.

In summary, when two capacitors are connected in parallel and then reconnected with the positive plate of one connected to the negative plate of the other, the charges on each capacitor will redistribute and the potential difference across each capacitor will be equal.