What Happens to Charge When Capacitor Plate Separation Doubles?

AI Thread Summary
When the separation between the plates of a parallel plate capacitor is doubled while maintaining a constant potential difference, the charge on the plates is halved. This conclusion is based on the relationship between capacitance, charge, and plate separation, where capacitance is inversely proportional to the distance between the plates. The discussion reveals an error in the answer key, which incorrectly states that the charge would double. The correct understanding aligns with the formula for capacitance, confirming that increasing the distance decreases the charge. Overall, the charge on the capacitor plates decreases when the plate separation is increased under constant voltage conditions.
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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=εoA/2d

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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Hi physgrl! :smile:

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

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