Question

A spherical capacitor has an outer radius of ‘b’ and inner radius ‘a’ with the air as the dielectric medium. If the ‘a’ is kept constant and the outer radius ‘b’ is increased, what happens to the capacitance?
A) Increases
B) Decreases
C) May increase or decrease
D) None of these.

Answer 1

Hint: We need to find the relation between the outer and inner radii of the spherical capacitor to its capacitance. We are to assume that the spherical capacitor is not connected to an external source. Using these we can find the change in capacitance.

Complete answer:
We are to find the capacitance of a spherical capacitor for understanding the relations between the inner radius and the outer radius of the capacitor in which the dielectric medium is kept.
The capacitance for any shape can be expressed as –
\[C=\dfrac{{{\varepsilon }_{0}}A}{d}\]
The area of the spherical shell is given as –
\[A=4\pi ab\]
The distance between the plates can be given as –
\[d=b-a\]
We can substitute these values in the formula of capacitance to get the capacitance of a spherical capacitor. The capacitance can be given as –

\[\begin{align}
  & C=\dfrac{{{\varepsilon }_{0}}A}{d} \\
 & \Rightarrow C=\dfrac{4\pi {{\varepsilon }_{0}}ab}{(b-a)} \\
\end{align}\]
Now, we have the relation between the outer radius ‘b’ and the inner radius ‘a’ of a spherical shell with the dielectric medium as air with the capacitance of the spherical capacitor. We can use this knowledge to understand the change in capacitance when one of these quantities change.
We are given that the outer radius of the given capacitor is increasing. That means, the distance between the parallel plates is increasing. Also, the area of the spherical shell is also increasing, which is less prominent than the former.
We understand for \[b>a\],the capacitance decreases with the increase in the outer radius.

The correct answer is option B.

Note:
The capacitance of any shape is dependent on whether the capacitor is supplied to an external source. In this case, we consider the capacitor not being connected to an external source, otherwise, the capacitance of the shell would have increased.