Question

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

Answer 1

Hint: The capacitor is isolated after charging. Capacitors capacitance is inversely proportional to the Potential difference of the capacitor. Moreover, the capacitance of the capacitor is inversely proportional to the distance of plates of the capacitor.
Formula used:
$\text{Q=C}\times \text{V}$
$\text{C=}{\in }_0{\displaystyle \frac{\text{A}}{\text{d}}}$

Complete answer:
In the question it is given that the capacitor is charged and then isolated. It means 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, we know that the charge is equal to the product of capacitance of the capacitor and potential difference across the capacitor.

$\text{Q=C}\times \text{V}$

Here, Q represents the charge of the capacitor
C represents the Capacitance of the capacitor and,
V represents the Potential difference across the capacitor.

Now, as we know that the charge on the capacitor will remain constant on increasing the plate separation. This means that both Capacitance and potential difference varies on increasing the plate difference.
From the formula of the charge, we can say that the capacitance and potential difference are inversely proportional to each other, which means that if capacitance will increase the potential difference will decrease and vice versa.
$\Rightarrow \text{C=k}{\displaystyle \frac{1}{\text{V}}}$
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,
$\text{C=}{\in }_0{\displaystyle \frac{\text{A}}{\text{d}}}$
Here, A represents the area and d represents the distance between plates of the capacitor.
Now, on increasing the plate difference, the capacitance of the capacitor will decrease because it is inversely proportional to the plate separation.
Now, as we know that the capacitance and potential difference and inversely proportional to each other the potential will increase whenever the capacitance will decrease.
Hence, we can conclude that the charge will remain constant, the potential will increase and the capacitance will decrease on increasing the plate separation.

So, the correct answer is “Option D”.

Note:
The charge of the capacitance will remain constant in this case because the capacitor is isolated after charging. Moreover, the capacitance of the capacitor and the potential difference of the capacitor will not remain constant; instead they will vary and will be inversely proportional to each other.