# Capacitor questions

1. Mar 25, 2008

### jcpwn2004

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

You have a single capacitor connected to a battery. However, the two plates on the capacitor do not have the same area. Does each plate still have the same charge? Explain

2. Relevant equations

C=Q/V=Eo(A/d)

3. The attempt at a solution

I really have no clue, I'm pretty sure the answer is yes but have no way to explain it.

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

http://img149.imageshack.us/img149/3907/physicsef8.jpg [Broken]

2. Relevant equations

Q=CV Ceq = C1 + C2

3. The attempt at a solution

I just need help with part B for finding the charges and voltage across for C1 & C2.

Well I got the Ceq to be 12uF and then to find the charges I use Q1 = C1V but that comes out to like .01 and I know that's not right...

Last edited by a moderator: May 3, 2017
2. Mar 25, 2008

### Snazzy

I can't say that I know the answer for question 1. My gut instinct is the same as yours.

Also, your equivalent capacitance is off. You might want to re-check that number.

After you do that, start rebuilding your circuit from the equivalent capacitor. Capacitors in series will have the same charge, but different voltage differences. Capacitors in parallel will have different charges, but same voltage differences.

3. Mar 26, 2008

### jcpwn2004

Yah I wrote that equation wrong, I already solved for the equivalent capacitor and got the right answer in the back of the book. I dont get how I go from there though and start solving for other stuff. Like the Ceq is 4 micro farads, what do I do to start finding out other stuff?

4. Mar 26, 2008

### Snazzy

{1/12 + 1/4 + 1/6)^-1 isn't 4.

Like I said, you start rebuilding the circuit and work your way from there with the hints I've given you.

5. Mar 26, 2008

### jcpwn2004

Lol wow I'm an idiot I was looking at the wrong problem. Ok here's what I have the Ceq is = to 2uF. Q3 = .0002 C Q4 = .0002 C V3 = 33.3V & V4 = 50V. I don't understand how to get the charge for the capacitors in series...

6. Mar 26, 2008

### Snazzy

The charge for any number of capacitors in series is the same. Remember that for series capacitors, the charge on each capacitor is equal, and for parallel capacitors, the charges add up.

7. Mar 26, 2008

### jcpwn2004

I still don't get it haha thanks for the help though. Right now I'm looking at Q1 = C1xV. so Q1 = (8uF)x(100V)....which is .0008 which isn't right :(.

8. Mar 26, 2008

### Snazzy

Don't start off with C1 first, mate. First of all, you can combine all four capacitors into one giant capacitor, right? You can then find the charge in the giant capacitor. You can then break this giant capacitor into three smaller capacitors, each one of them having that charge you calculated earlier since they are in series. You can then find the potential difference across the first, second, and third capacitors that you obtained by breaking the giant capacitor, yes?

Now the first capacitor in that circuit can be further broken into two smaller capacitors in parallel with each other. You know that for parallel components, the voltage drop is equal for both capacitors but the charges are different and from this you can find the charge for C1 and C2.

Last edited: Mar 26, 2008
9. Mar 26, 2008

### jcpwn2004

wow i finally figured it out, thanks so much man I love you!!!@(!@!@!