The capacitor is charged and then isolated, according to the question. It indicates that as plate separation increases or decreases, the charge on the isolated capacitor remains constant. As a result, because the capacitor is isolated, we may assume that the charge will remain constant. We now know that the charge is equal to the product of the capacitor’s capacitance and the capacitor’s potential difference.

Q=C×V

Now, we know that when the plate separation increases, the charge on the capacitor will remain constant. This indicates that as the plate difference increases, both capacitance and potential difference change.

The capacitance and potential difference are inversely proportional to each other, according to the charge formula, which indicates that when capacitance increases, the potential difference decreases and vice versa.

C= K x 1/V

Here, k is some constant. From the above formula we can say that the capacitance is inversely proportional to the potential difference.We have one more formula of capacitance,

C = ∈0 A/d

The area of the capacitor is represented by A, while the distance between plates is represented by d. Because the capacitance of the capacitor is inversely proportional to the plate separation, as the plate difference is increased, the capacitance of the capacitor will decrease. Because capacitance and potential difference are inversely related, the potential will rise whenever the capacitance falls. As a result, we may deduce that as plate separation increases, the charge will remain constant, the potential will increase, and the capacitance will decrease.