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Questions  >>  CBSE XII  >>  Math  >>  Model Papers

A hollow charged conductor has a tiny hole cut into its surface . show that the electric field in the hole is $ \bigg( \large\frac{\sigma}{2 \in_0}\bigg)$$\hat n$, where $\hat n$ is the unit vector in the outward normal direction and $\sigma$ is the surface change density near the hole.

1 Answer

Answer :
We can calculate the electric field due to hollow charge sphere by using guass theorem
Let the surface charge density near the hole = $ \sigma$
Let $\hat n$ be the normal vector which is directed outwards .
Let P be the point on the hole.
According of guass theorem
$\int E.ds =\large\frac{q}{\in_o}$
Where q is the charge near the hole.
$\sigma= \large\frac{q}{ds}$
$=> q= \sigma ds$
$E ds \cos \theta= \large\frac{\sigma ds}{\in_0}$
But the angle between the electric field and area vector is $0^{\circ}$
$E ds= \large\frac{\sigma ds}{\in_0}$
$E= \large\frac{\sigma }{\in_0}$
$E= \large\frac{\sigma }{\in_0}$$\hat n$
The electric field is due to the filled up hole and the field due to the rest of the change conductor.
Hence there is no electric field inside the conductor .
But outside the conductor, the electric fields are equal and are in the same direction .
Hence the electric field at P due to each part $\large\frac{1}{2}$$E=\frac{\sigma}{2 \in_0}\hat n$
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