- #1

Leeoku

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## Homework Statement

I got question 10 and need help on 11

[PLAIN]http://lulzimg.com/i23/abf9f8.jpg

## Homework Equations

C = Q/V

## 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.. that's 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

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