How to Calculate Final Charge on Capacitors Connected in Series?

[member 392791]

Homework Statement

If a 3.0-μF capacitor charged to 40 V and a 5.0-μF capacitor charged to 18 V are connected
to each other, with the positive plate of each connected to the negative plate of the
other, what is the final charge on the 3.0-μF capacitor?

a . 11 μC
b. 15 μC
c. 19 μC
d. 26 μC
e. 79 μC

Homework Equations

The Attempt at a Solution

The positive to negative terminal from the plates means they are connected in series, right? So if there are 210 μC available (40x3) + (5x18), shouldn't the charge be the same since charge is the same for plates connected in series? Thus maybe the average or something, yet that is not one of the possibilities. The answer is a, but not sure why

 

[CWatters]

(40x3) + (5x18)

That would be the total charge if they were connected +ve to +ve.

Better to think of the capacitors as connected in parallel but with opposite charge.

 

[member 392791]

That's the same thing as being connected in series. Ok, so then do I just subtract the charges (120uC - 90uC). Then there is 30uC charge available, and since they are in series, they should have equal amounts, 15uC? Yet that isn't the answer, which is 11uC.

 

[lewando]

... they should have equal amounts, 15uC?

Looking at the 2 capacitors from the parallel perspective, the voltage across them needs to be the same. Using Q_3uF=C_3uFV and Q_5uF=C_5uFV, the charge must be different.

 

[member 392791]

why would I look at them from a parallel perspective? If the negative and positive plates of each capacitor are connected, doesn't that demand that they be in series?

 

[lewando]

why would I look at them from a parallel perspective?

Why would you not?

If the negative and positive plates of each capacitor are connected, doesn't that demand that they be in series?

I am not familiar with that rule. Fact is, you can look at the two connected caps from either perspective. If you are interested in how current might flow in a loop, look at them in series. If you are interested in the voltage across 2 points (the two junctions), consider them in parallel.

 

[CWatters]

In general (not this problem)...

Series capacitors have equal charge (because same current flows through both)
Parallel capacitors have equal voltage (because they are connected between same nodes).

In this problem the voltage must end up the same but the charge will be different so it makes more sense to think of the capacitors as being in parallel.Lets say Q1 and Q2 are the initial charges and Q1' and Q2' are the final charges. We know these must rearrange so the final voltage is the same so

V = Q1'/C1 = Q2'/C2 .....(1)

The total charge is conserved so

Q1 - Q2 = Q1' + Q2' ...(2)

It's -ve on the left because the charge on one capacitor is the opposite polarity.

Two equations and two unknowns.

 

[CWatters]

In general (not this problem)...

Oops I meant "not just this problem".