What Happens to Charge When Capacitor Plate Separation Doubles?

[physgrl]

Homework Statement

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

Homework Equations

CV=Q
C=ε_oA/2d

The Attempt at a Solution

CV=Q
C proportional to Q
ε_oA/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?

 

[I like Serena]

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]

ohh yeah the 1/2 was a mistake...thanks