Answer:

Explanation:
The capacitor of a parallel-plate capacitor is given by:

where
A is the area of each plate
d is the separation between the plates
is the vacuum permittivity
The energy stored in a capacitor instead is given by

where
Q is the charge stored in each plate
Substituting the expression we found for C inside the last formula,

And re-arranging it

Now if we substitute

We find the charge stored on the capacitor:

Answer:
Increases
Explanation:
The expression for the capacitance is as follows as;

Here, C is the capacitance,
is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.
It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.
The expression for the energy stored in the parallel plate capacitor is as follows;

Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.
Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.
Therefore, the option (c) is correct.
Answer:c
Explanation: i guessed but got it right bc im a bad bleep lol