Capacitors in paralell disconnected then connected to each other

[Staple Gun]

Homework Statement

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.

Homework Equations

Q=C/V

The Attempt at a Solution

This is as far as I got, but I'm guessing I went wrong somewhere.

Calculate equivalent capacitance
C_t = C₁ + C₂ = 7.85μF

Calculate total charge on the capacitors
Q_t = C_t/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.

 

[Doc Al]

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.

 

[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 Q_t = (C₁ + C₂)/V

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

Am I on the right track so far?

 

[nasu]

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 Q_t = (C₁ + C₂)/V

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

Am I on the right track so far?

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.

 

[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.

 

[Doc Al]

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

You are mixing up your formulas:
C = Q/V so: Q = CV (not C/V)

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?

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.

 

[Staple Gun]

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