# Capacitors in paralell disconnected then connected to each other

1. Oct 7, 2008

### Staple Gun

1. The problem statement, all variables and given/known data
Capacitors C1 = 5.95 μF and C2 = 1.90 μF are charged as a parallel combination across a 277 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on capacitor C1.

2. Relevant equations
Q=C/V

3. The attempt at a solution
This is as far as I got, but i'm guessing I went wrong somewhere.

Calculate equivalent capacitance
Ct = C1 + C2 = 7.85μF

Calculate total charge on the capacitors
Qt = Ct/V = 0.002174 C

Charge on C1 is half of the total, or 0.001087 C

Calculate voltage on C1
V1 = Q1/C1 = 182.7 volts

I am stuck from here though, even if what I've done so far is right, I have no idea where to go.

Last edited: Oct 7, 2008
2. Oct 7, 2008

### Staff: Mentor

Calculate the original charge on each capacitor.

When those capacitors are reconnected, what happens to the charge on them? How does the charge rearrange itself? Hint: Consider the total charge and the fact that they are reconnected in parallel.

3. Oct 7, 2008

### Staple Gun

Thanks for the reply, I am still a bit confused though. I thought the original charge on each capacitor would be half of the total charge of the equivalent capacitor.

Therefore Qt = (C1 + C2)/V

And the charge of each capacitor would be 1/2 Qt

Am I on the right track so far?

4. Oct 7, 2008

### nasu

This is your mistake. The capacitors have different capacitances! They will have different charges. (Q=C*V). They "have" the same voltage, when in parallel.

5. Oct 7, 2008

### Staple Gun

Ok, so the charge on capacitor C1 would be...

Q1 = 5.95μF/ 277 V = 2.14*10-8C
and likewise the charge on C2 would be 6.859*10-9C

I really don't have a clue what happens when they are reconnected though.

My guess is that by using Q=CV I can take the sum of the charges, and then using the capacitance of both I can figure out the new electric field across the capacitors, and somehow find the new charge?

Thanks for the help so far! This is for an online class so I never really learned the concepts behind capacitors and i'm struggling to wrap my head around it, but this is helping a lot.

6. Oct 7, 2008

### Staff: Mentor

You are mixing up your formulas:
C = Q/V so: Q = CV (not C/V)
Two hints:
(1) When reconnected, the charges rearrange but the total charge can't change.
(2) When reconnected, the voltage across each must be the same.

7. Oct 7, 2008

### Staple Gun

EDIT: I figured it out, thanks so much for the help!!

Last edited: Oct 7, 2008