Question

A parallel plate capacitor is charged with a battery and then separated from it. Now if the distance between its two plates is increased what will be the changes in electric charge, potential difference and capacitance respectively.
A) Remains constant, decreases, decreases
B) Increases, decreases, decreases
C) Remains constant, increases, decreases
D) Decreases, remains constant, increases

Answer 1

Hint: We are considering a charged capacitor which is no more connected to source. The capacitor equation which relates the potential difference, electric charge and the capacitance which involves a distance parameter is used for approximating the changes.

Complete step by step answer:
Let us consider a charged capacitor which is no longer connected to an external power source. As a result, there are consequences to the capacitance and the related aspects. The capacitance of a parallel plate capacitor is given by –
\[\begin{align}
  & C=\dfrac{Q}{V} \\
\end{align}\]

But, we know that,
\[\begin{align}
 & V=Ed \\
 & \Rightarrow \text{ }C=\dfrac{Q}{Ed} \\
\end{align}\]
Where, Q is the charge on the plates, E is the electric field, d is the distance between the parallel plates and V is the potential difference on the plate.
From the above equation, we can understand that a capacitor which is no longer connected to an external source has a constant capacitance, with constant charge and potential difference in the capacitance.
Now, let us consider the situation of the capacitor when the ‘d’ the distance between the plates is increased.
In this case also, there exists no external source. The following can be inferred for this situation –
a) The electric charge on a plate remains constant when there is no external source. i.e., Q remains constant.
b) The distance ‘d’ is increased, this affects the potential difference as
\[V=Ed\]. As a result, the potential difference increases.
c) Again, as the potential difference increases, it affects the capacitance as –
\[\begin{align}
  & C=\dfrac{Q}{V} \\
 & \text{or} \\
 & C=\dfrac{Q}{Ed} \\
\end{align}\]
i.e., the capacitance decreases.
For a situation as given in the question – the electric charge remains constant, the potential difference increases and the capacitance decreases.

The correct answer is option C.

Note:
When the external source is still connected, the number of electric charges and hence the electric field keeps on increasing. In such a case, the potential difference remains constant and the capacitance increases due to higher number of electric charges.
There is a situation when the area of the parallel plate is changed. The area is related to the electric field.