Question

By the gauss’s theorem show that the electric field in a hollow spherical conductor is zero.

Answer 1

Hint: Gauss’s theorem is used to calculate the electric field of any closed surface area. We shall first mark a Gaussian surface and then find the closed integral of the electric field inside the gaussian equating it with the charge enclosed in the Gaussian surface divided by permittivity of free space.

Complete answer:
Since we have to find the electric field inside the hollow spherical conductor, thus the Gaussian surface would be a concentric hollow sphere whose radius, $r$ is less than the radius of the spherical conductor.
That is, if $R$is the radius of the given spherical conductor, then $rThe elemental area, $dA=4\pi {{r}^{2}}.dr$.
The Gauss theorem is mathematically expressed as:
$\oint{\vec{E}.d\vec{A}=\dfrac{{{q}_{enclosed}}}{{{\varepsilon }_{0}}}}$
Where,
$\vec{E}=$ electric field of Gaussian surface.
${{q}_{enclosed}}=$ charge enclosed inside the gaussian surface
${{\varepsilon }_{0}}=$ permittivity of free space or vacuum
We know that if we provide charge to a conductor, that charge resides only on the outer walls of the conductor and thus, there is zero charge inside the conductor.
Now, we have been given a hollow spherical conductor. In this case as well, the charge will reside on the outer wall of the conductor and there is zero charge anywhere inside the conductor.
$\Rightarrow {{q}_{enclosed}}=0$
Thus, applying Gauss theorem, we have
$\oint{\vec{E}.d\vec{A}=\dfrac{0}{{{\varepsilon }_{0}}}}$
$\Rightarrow \vec{E}=0$
Therefore, we get that the electric field in a hollow spherical conductor is zero using Gauss theorem.

Note:
The choice of the Gaussian surface plays a fundamental role in the determination of electric field using Gauss law. The Gaussian surface is an imaginary surface visualized to set boundaries for the application of Gauss law at or around a predefined area.