# Finding the capacitance charge

1. Oct 5, 2007

### jhess12

1. The problem statement, all variables and given/known data

problem 41 on this link. I got the equivalent capacitance for part a. for the two capacitors with 15 and 3uF- $$C_{eq}$$=2.5uF
for capacitor with 6uf- i got the $$C_{eq}$$=8.5uF. final answer for a.=$$C_{eq}$$=5.96uF. this final anwer is correct according to the answers given to me.

2. Relevant equations

3. The attempt at a solution
capacitor with 15uF-$$Q_{1}$$
capacitor with 3uF-$$Q_{2}$$
capacitor with 6uF-$$Q_{3}$$
capacitor with 20uF-$$Q_{4}$$
for part b. i know that the Q for $$Q_{1}$$ and $$Q_{2}$$ are the same because they are in series. the Q i get for them is Q=[V/(1/C1+1/C2)]. If V=15V and
C1=15 and C2=3, then the answer i get is 37.5uC. but the answer in the book says $$Q_{1}$$ and $$Q_{2}$$ =26.3uC

for $$Q_{3}$$- Q=$$C_{3}$$V If $$C_{3}$$=6uF and V=15V, then Q should be 90uC. OR if Q=$$C_{eq}$$V, then Q=8.5uF(15)=127.5uC. but the book says it should be 63.2uC

for $$Q_{4}$$- Q=[V/(1/$$C_{eq}$$)] If $$C_{eq}$$=5.96uF and V=15V, then Q should be 89.5uC- this one i got correct.

I dont understand how i can get certain parts right, but others wrong, unless i am using the wrong equation. Please help, i need to know this for a test.

Last edited: Oct 5, 2007
2. Oct 5, 2007

### learningphysics

Start with C4... you did that right... the charge on C4 is just the charge on the Ceq...

What is the voltage across C4? Use V = Q/C...

So what is the voltage across C1 and C2? this is where you made your mistake... it's not 15V...

3. Oct 5, 2007

### jhess12

ok well, i figured that if Q4 is 89.5uC, then Q4=Q, then V3=Q/C=89.5/8.5=10.53V. if
V3C=Q, then 10.5(6uF)=63.2uC=Q3. this appears to be correct, even though i dont understand why. as for Q1 and Q2, i am still confused, i asked my professor to explain it, but he says i should know how to solve this so to figure it out on my own. I keep thinking that it should be Q=[V/(1/C1+1/C2)]. i can see that V=V1+V2, but beyond that....i just dont know

4. Oct 6, 2007

### learningphysics

No. what I meant what you know the charge on Q4 is 89.5uC. You know the capacitance of Q4 is 20uF. So the voltage across Q4 is 89.5uC/20uF = 4.475V.

15V - 4.475 = 10.525V

So the voltage across Q3 is 10.525V... so the charge across Q3 is 10.525*6uF = 63.15uC. etc...

5. Oct 6, 2007

### jhess12

i got it earlier from your earlier post, thanks so much