Capacitor Disconnected from Battery, changing distance

[jeffc313]

1.A parallel-plate capacitor is fully charged and then disconnected from the battery. The plates
are then moved closer together, how does the charge on the plate change

it needs to be stated whether the charge on each plate increases, decreses or stays the same

Homework Equations

C=Q/V, C=epsilon_0 * A/d

The Attempt at a Solution

when the capacitor is disconnected, the charges cannot flow. So each one should stay the same, right? The solution for this exam problem that was on last years final says that it should increase. This doesn't make sense to me. Is the solution wrong, or am I wrong?

 

[Redbelly98]

1.A parallel-plate capacitor is fully charged and then disconnected from the battery. The plates
are then moved closer together, how does the charge on the plate change

it needs to be stated whether the charge on each plate increases, decreses or stays the same

Homework Equations

C=Q/V, C=epsilon_0 * A/d

The Attempt at a Solution

when the capacitor is disconnected, the charges cannot flow. So each one should stay the same, right? The solution for this exam problem that was on last years final says that it should increase. This doesn't make sense to me. Is the solution wrong, or am I wrong?

Welcome to Physics Forums.

I'm pretty certain you are correct, for exactly the reason you give.

Q = CV, so the charge would increase if the capacitor remained connected to the battery (V=constant, and C increases).

It sounds like the professor got confused about what the question was asking.