Question

A parallel capacitor plate is charged and then isolated. The effects of increasing of plate separation on charge, potential, capacitance respectively are
(A) constant, decreases, decreases
(B) increases, decreases, decreases
(C) constant, decrease, increases
(D) constant, increases, decreases.

Answer 1

The capacitor is isolated after charging. The capacitance of the capacitor is inversely proportional to the distance of plates of the capacitor and the potential difference of the capacitor.

Formula Used: The following formulas are used to solve this question.
$\Rightarrow Q = C \times V$ where $Q$ is the charge on the capacitor, $C$ is the capacitance of the capacitor and $V$ is the potential difference of the capacitor.
$\Rightarrow C = { \in _0}\dfrac{A}{d}$ where $A$ is the area of the capacitor and $d$ is the distance separating the two plates of the capacitor.

Complete step by step answer
It is given in the question, that the capacitor is charged and then isolated.
This implies that the charge on the isolated capacitor will not change on increasing or decreasing plate separation. Hence, we can conclude that the charge will remain the same because the capacitor is isolated.
Now, the capacitance of a capacitor is the product of charge on the capacitor and the potential difference of the capacitor.
$\therefore$ Capacitance of the capacitor $C$ is given by,
$\Rightarrow C = Q \times V$, where $Q$ is the charge on the capacitor and $V$ is the potential difference of the capacitor.
We know that the charge on the capacitor will remain constant on increasing the plate separation. It may be thus said, the capacitance and potential difference changes on increasing the plate separation in between the capacitor plates.
The capacitance and potential difference are inversely proportional to each other. This implies that, if the capacitance will increase, the potential difference will decrease and vice versa.
$\therefore C\alpha \dfrac{1}{V}$
Now, we know that, capacitance of the capacitor is given by $C.$
$\therefore$ $C = { \in _0}\dfrac{A}{d}$ where $A$ is the area of the capacitor and $d$ is the distance separating the two plates of the capacitor.
From here, it is evident that the capacitance of the capacitor is inversely proportional to $d$, the distance separating the two plates of the capacitor.
Since, capacitance is inversely proportional to distance of separation as well as the potential difference of the capacitor, we can say that, whenever distances increase, capacitance decreases and potential difference increases.
$\therefore$ The charge remains constant, the potential difference increases and the capacitance decreases on increasing the plate separation.

So, the correct answer is Option D.

Note
The charge on the capacitor remains constant because the capacitor is isolated after charging. So there will not be any loss of the charge of the capacitor.