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Capacitance and charge

by physgrl
Tags: capacitance, charge
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physgrl
#1
Jan31-12, 08:02 AM
P: 138
1. The problem statement, all variables and given/known data

If the separation between the plates of a parallel plate capacitor is doubled and the potential difference is held constant by a battery, the magnitude of the charge on the plates will:

*a. double

b. quadruple

c. be cut in half

d. not change

2. Relevant equations

CV=Q
C=εoA/2d

3. The attempt at a solution

CV=Q
C proportional to Q
εoA/2d proportional to Q
1/d proportional to Q

now if the separation d is doubled the charge Q should be cut in half instead of doubled (as the answer key says) right? so have I made a mistake? Is it supposed to be done differently?
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#2
Jan31-12, 08:34 AM
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P: 6,188
Hi physgrl!

You are right and the answer key is wrong.

Btw, I have a different formula for the capacitance of 2 parallel plates.
It's ##C={\epsilon A \over d}##, as you can see on the wiki page: http://en.wikipedia.org/wiki/Capacitance

But that does not change your line of reasoning, which is correct.
physgrl
#3
Jan31-12, 08:56 AM
P: 138
ohh yeah the 1/2 was a mistake...thanks


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