Electric Charges Homework: Potential Difference & Capacitance

[science.girl]

Homework Statement

Two parallel conducting plates, separated by a distance d, are connected to a battery of emf \epsilon. Which of the following is correct if the plate separation is doubled while the battery remains connected?
a. The electric charge on the plates is doubled.
b. The electric charge on the plates is halved.
c. The potential difference between the plates is doubled.
d. The potential difference between the plates is halved
e. The capacitance is unchanged.

Homework Equations

\DeltaV = \epsilon - Ir

The Attempt at a Solution

I'd actually appreciate it if someone could explain this question to me. I'm having some difficulty with these concepts. What exactly is the question asking? I can probably take it from there.

 

[Dick]

The question is asking which option is true. That was easy. But the function of a battery is to maintain a constant potential difference between the two plates. What does that tell you about options c) and d)? Now what does the potential difference between the plates have to do with the E field between the plates and their separation? Finally what does the charge on a plate have to do with the E field between the plates? Think Gauss' law.

 

[science.girl]

The question is asking which option is true. That was easy. But the function of a battery is to maintain a constant potential difference between the two plates. What does that tell you about options c) and d)? Now what does the potential difference between the plates have to do with the E field between the plates and their separation? Finally what does the charge on a plate have to do with the E field between the plates? Think Gauss' law.

Thank you very much! I understand now