Confusion with parallel plates

[Airsteve0]

Hi, I was wondering if someone could help me understand how the potential of a parallel plate capacitor can be altered. I understand that the potential difference across two plates is equal to the potential of the battery that charges them, but can this potential be altered by changing the plate separation distance? The reason I ask is because of the formula:

V=(Q*d)/(A*epsilon)​

where A is the area of the plates, d is the separation distance, epsilon is a constant and Q is the charge on the plates. According to this equation the voltage should change with distance but I suspect I am thinking about this incorrectly. As, well would this be different if there was no battery? Thanks you to anyone who can help clarify this problem I am having.

 

[Doc Al]

According to this equation the voltage should change with distance but I suspect I am thinking about this incorrectly.

The voltage would change with distance if the charge was held fixed. With a battery attached, the voltage is held fixed so the charge adjusts itself.

As, well would this be different if there was no battery?

Absolutely. Disconnect the battery and the charge cannot change. Now you'll see that the voltage does change with the distance between the plates.

 

[Airsteve0]

Thank you very much, that clears things up.