# Homework Help: Finding the cafpacitance, I thought it was in series

1. Oct 9, 2005

### mr_coffee

Hello everyone...I'm runnning int problems on this problem:
he two metal objects in Fig. 25-26 have net charges of +62 pC and -62 pC, which result in a 14 V potential difference between them.
image is here: http://www.webassign.net/hrw/hrw7_25-26.gif
(a) What is the capacitance of the system?
4.429 pF
(b) If the charges are changed to +226 pC and -226 pC, what does the capacitance become?
wrong check mark pF
(c) What does the potential difference become?
V

If the charges are the same, i thought that ment it would be in series, which is Q = CVtotal
So i put in:
C = 226pC/14V = 16.14 pF
I also tried 0, both wrong. ANy ideas? Thanks

2. Oct 9, 2005

### Physics Monkey

This isn't a question about series capacitors. The two metal objects form a single capacitor. Is the capacitance of a parallel plate capacitor dependent on the charge on the plates? Is the capacitance of a general capacitor dependent on the charge?

3. Oct 9, 2005

### mr_coffee

The capcitance of a parallel plate capacitor is dependent on the charge on the plates because the formula is. Q total = V(Ceq). Now is the capacitence of a general capacitor depedent on the charge? I"m assuming yes, the genral formula is Q = CV...

4. Oct 9, 2005

### Physics Monkey

The capacitance of a parallel plate capacitor is $$C = \epsilon_0 A /d$$ with no mention of charge. The capacitance of a system is dependent only on the geometry which is partly why it's a useful concept. In the equation $$Q = CV$$, any change in Q is accompanied by a change in V so that C is always the same (so long as the geometry is the same). There lies the answer to your question. Once you know the capacitance of your system, it's always the same unless you change the geometry.

5. Oct 9, 2005

### mr_coffee

ohhh i c what your saying, thank you very much, i got the problem now! :)