A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the battery disconnected, the amount of charge on the plates remains constant.) What happens to the potential difference between the plates as they are being separated?

There will be an increase in potential difference.

Explanation:

As we know that the potential difference depends upon the capacitance.

ΔV = Q/C

When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.

In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.