Question

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left( \dfrac{\sigma }{2{{\varepsilon }_{0}}} \right)\widehat{n}$, where $\widehat{n}$is the unit vector in the outward normal direction and $\sigma$ is the surface charge density near the hole.

Answer 1

Since, the conductor is charged and hollow, then we can say by the property of a conductor that the charge distribution inside the volume of conductor will be zero and also electric field inside the cavity will be zero. So, we will be calculating the Electric field at the periphery of the hole.

Complete answer:
Let us first define some terms that we are going to use in our
equations later on.
Let ‘E ‘be the electric field just outside the conductor ‘q’ be the electric charge, $\sigma$ the charge density and ${{\varepsilon }_{0}}$, the permittivity of free space. Then we can say that, for a very small area ,
$\Rightarrow q=\sigma .ds$
Now, applying Gauss Law on this very small piece of area, we can write:
The net electric flux through this area is:
$\begin{aligned} & \Rightarrow \phi =\overrightarrow{E}.\overrightarrow{ds} \\\ & \Rightarrow \overrightarrow{E}.\overrightarrow{ds}=\dfrac{q}{{{\varepsilon }_{0}}} \\\ & \Rightarrow \overrightarrow{E}.\overrightarrow{ds}=\dfrac{\sigma .ds}{{{\varepsilon }_{0}}}\widehat{n} \\\ & \Rightarrow \overrightarrow{E}=\dfrac{\sigma }{{{\varepsilon }_{0}}}\widehat{n} \\\ \end{aligned}$
Therefore, the electric field just outside the conductor is $\dfrac{\sigma }{{{\varepsilon }_{0}}}\widehat{n}$ . This field is basically the superposition of two fields. The field due to cavity and the field due to the rest of the charged conductor. These fields are equal in magnitude and opposite in direction inside the conductor, that is why the net field inside the conductor is zero. But equal in direction and magnitude outside the conductor.
Thus, let this field be E’ , then we can write:
$\begin{aligned} & \Rightarrow E'+E'=\dfrac{\sigma }{{{\varepsilon }_{0}}}\widehat{n} \\\ & \therefore E'=\dfrac{\sigma }{2{{\varepsilon }_{0}}}\widehat{n} \\\ \end{aligned}$
Hence, the electric field in the hole is $\left( \dfrac{\sigma }{2{{\varepsilon }_{0}}} \right)\widehat{n}$, has been proved.

Note:
The tricky part of this question was to analyze the direction and magnitude of electric fields due to the hole and conductor inside and outside the conductor. Once this relation was established, the rest of the problem was pretty much easy to solve and get the required relation.