# Charge on capacitor

1. Jul 11, 2011

### Leeoku

1. The problem statement, all variables and given/known data
I got question 10 and need help on 11
[PLAIN]http://lulzimg.com/i23/abf9f8.jpg [Broken]

2. Relevant equations
C = Q/V

3. The attempt at a solution
At first i thought it was a simple c = q/v and solve for Q but that wasnt right. Is that because the circuit splits off in two parts and a different number of charge is passed through each end based on the capacitors there?

This makes me think, if we use C = q/v, we would use the answer from question 10 to find the total charge available. Charge must be conserved so we take a percentage C_a is in comparison to the rest.. thats wrong too

Edit: I think i got the right answer but I need to double check on reasoning.
Using the idea that charge is convserved, i read in book that charges on capacitors in series are equal. Thus, it is a percentage split between (C_a+C_b) and (C_c+C_d). So what i did was find the percentage the capactiance in series for top and bottom were relative to the total capacitance of the 4. Using C = q/v with total capacitance, i found the total charge, then multiplied by the percentage.
C1+c2 was 2.05e-6
C3+c4 was 1.367e-6
Total was 3.417e-6

Thus c1+c2 was about 60% of total.
C(total) = q/v
q = 2.955e-4
q * 0.6 = 1.77e-4 C

Last edited by a moderator: May 5, 2017
2. Jul 11, 2011

### ehild

Ca and Cb connected in series, and the total voltage across them is V=86.5 V. What is the relation among the charges on series connected capacitors?

ehild

3. Jul 11, 2011

### Leeoku

i just edited while u posted, im guessing thats the right reasoning?

4. Jul 11, 2011

### ehild

Yes, it is correct, but the calculation would be simpler if you used Q=Cr * V with Cr the resultant capacitance of Ca and Cb. Series capacitors have the same charge and it is the same as the charge on their resultant.

ehild