Casey A. said:
Homework Statement
Capacitors of 7.6 µF and 2.1 µF are charged as a parallel combination across a 333V battery. The capacitors are disconnected from the battery and each other. They are then connected positive plate to negative plate and negative plate to positive plate.
Find the resulting charge on the first capacitor.
Answer in units of µC.
Homework Equations
C = Q/V
The Attempt at a Solution
I'll preface this by saying that I looked at other threads about this, but it's still just not working out for me so I think I need to actually converse about it.
Knowing that parallel capacitors have the same voltage:
q1 = (7.6e-6)(300V) = .00228C
q2 = (2.1e-6)(300V) = 6.3e-4C
Before we go any further, is that supposed to be 300 V, or 333 V?
Then:
Qtotal = q1 + q2 = .00291C
To check:
Qtotal = CtotalV
(7.6e-6C + 2.1e-6c)(300V) = .00291C
It's just that from here, I don't know where to go at all. When the charges are connected one end to another, are they in a series or are they parallel?
Hints aren't working very well for me. I don't know if that's because I'm behind in physics or what, but nothing that's being thrown against the wall is sticking.
After correcting any possible minor mistakes mistakes so far (e.g., 300 V vs. 333 V), here is my advice as to where to go from there:
(a) When a given parallel capacitor is charged to some amount
q, it really has +
q on its positive plate and -
q on its negative plate. So we say that it is charged to "
q" even though it really has net zero charge with +
q on one plate and -
q on the other. So for the next step, work out what these values are for each capacitor and then label each capacitor plate with +
q on one and -
q on the other. (Substitute the actual numbers in for the
qs.) Actually draw it out and label them on paper.
(b) Now connect the capacitors together. [Edit: after flipping one of them upside down.] Actually draw this result out on your paper too. It looks like one big capacitor now doesn't it? Find the charge on each of the "big combination"'s capacitor plates. You should notice that on a given plate, it has positive and negative charge. So some of the charge cancels out [on each "big" plate]. But now knowing the total charge, and the capacitance (treat this big, combo capacitor as the parallel of the smaller capacitors), find the voltage.
(c) Now you can separate the capacitors from each other. And you know the voltage and capacitance of each of the smaller capacitors, so you can find the charge on each, individual capacitor.