Question

The spheres shown in the figure are connected by a conductor. The capacitance of this setup will be

• A. $4\pi {ϵ}_{0}b$
• B. $4\pi {ϵ}_0\frac{ab}{(b-a)}$
• C. $4\pi {ϵ}_{0}a$
• D. $4\pi {ϵ}_0\frac{b^2}{(b-a)}$

Answer 1

The correct option is A $4\pi {ϵ}_{0}b$

For a while, let's assume that there is no connection between the inner and outer spheres. The two spheres on given some charges $q_1\&q_2$.


Now, $V_a=\frac{k{q}_1}{a}+\frac{k{q}_2}{b}$

$V_b=\frac{k{q}_1}{b}+\frac{k{q}_2}{b}$

So, $V_a>V_b$

This directly implies that any charges given to inner sphere will move to the outer sphere if the two are provided with a connection.
Therefore, the setup is equivalent to an isolated spherical conductor of radius $b$.

And we know for such a case

$C=4\pi {ϵ}_{0}R=4\pi {ϵ}_{0}b$

Hence, option $\left(c\right)$ is correct.

Key concept. Principle of Van de Graaff generator and capacitance of an isolated metallic sphere.