- #1
TheEvenfall
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Hello PF,
The capacitor C_1 is connected to a battery as shown. After being charged, it is connected to the capacitor C_2 and the second battery as shown. When all the charges come to rest, what will be the charge on capacitor C_2?
http://imgur.com/4ciJRrn
C = Q/V
The initial charge on C_1 is Q_1 = C_1 V_1
After it is connected to the other circuit and in steady state, I used the loop rule to get the equation (q_1/C_1) + (q_2/C_2) = V_2, where q_1 is the new charge on C_1 and q_2 is the charge on C_2.
Now the problem is relating q_1, q_2 and Q_1. According to my professor's solution, -Q_1 = -q_1 + q_2.
He got that equation by doing what's shown here: http://imgur.com/WvKJ6rQ, and I have no idea where he got that from, and this is the point that I'm stuck on and why I'm posting this.
Using that equation we can find q_2 easily, but I don't understand how he got it.
Thank you.
Homework Statement
The capacitor C_1 is connected to a battery as shown. After being charged, it is connected to the capacitor C_2 and the second battery as shown. When all the charges come to rest, what will be the charge on capacitor C_2?
http://imgur.com/4ciJRrn
Homework Equations
C = Q/V
The Attempt at a Solution
The initial charge on C_1 is Q_1 = C_1 V_1
After it is connected to the other circuit and in steady state, I used the loop rule to get the equation (q_1/C_1) + (q_2/C_2) = V_2, where q_1 is the new charge on C_1 and q_2 is the charge on C_2.
Now the problem is relating q_1, q_2 and Q_1. According to my professor's solution, -Q_1 = -q_1 + q_2.
He got that equation by doing what's shown here: http://imgur.com/WvKJ6rQ, and I have no idea where he got that from, and this is the point that I'm stuck on and why I'm posting this.
Using that equation we can find q_2 easily, but I don't understand how he got it.
Thank you.