# What is the initial charge on capacitor C1?

1. Homework Statement

A 7.50 F capacitor, C1, is initially connected to a 90.0 V battery and is given a charge Q. The capacitor is disconnected from the battery and then connected to an initially uncharged capacitor, C2. Assume the two capacitors are connected such that the positive plates of each are connected and the negative plates of each are connected. The voltage across both capacitors C1 and C2 drops to 12 V after they are connected.

a. What is the initial charge on capacitor C1?
b. What are the charges on capacitors C1 and C2?
c. What is the capacitance of capacitor C2?
2. Homework Equations

q=CV

3. The Attempt at a Solution
a. q=CV
c1=7.5
V=90
q=675 C ok, this one was easy

but I'm stuck in part b of the question. Will the charges be the same or not?
This is what I'm trying to do:
q1 + q2 = (7.5 + C_2)12
675 + q2 = 90 + 12C_2

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dynamicsolo
Homework Helper
If the two capacitors are only connected to each other, the voltages across each will be the same. Since V = Q/C , the charges will not be the same unless both capacitances are equal. (Given the context of the problem, I suspect they are not.) With the voltages being equal, you will be able to find a proportionality between the charges on the two capacitors.

BTW, is the capacitance of C1 supposed to be milli-farads or micro-farads or something? There's a character in the first line of the problem that doesn't show on this board. I ask because I doubt the charge on C1 is 675 Coulombs...

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Yes, now I figured out, according to the context, that they aren't equal, but I think there are many combinations for q2 and c2 that will make V=12... so I think I should leave them in variables?

dynamicsolo
Homework Helper
Yes, now I figured out, according to the context, that they aren't equal, but I think there are many combinations for q2 and c2 that will make V=12... so I think I should leave them in variables?
You now know that Q1/C1 = Q2/C2 and that Q1 + Q2 = 675 (micro?)Coulombs . (Charge is conserved, so whatever charge was on C1 when it was transferred from the battery has to now be shared with C2. There is a unique solution.)

Oh right... After doing the calculations:
Q1/7.5=12
Q1=90microC
Q2=585microC
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