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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 charge density near the hole.

1 Answer

Consider the conductor with the hole filled up.
Then the field just outside is $(\frac{\sigma}{\in_0}) \hat {n}$ and is zero inside.
View this field as a superposition of the field due to the filled up hole plus the field due to the rest of the charged conductor.
Inside the conductor, these fields are equal and opposite.
Outside they are equal both in magnitude and direction.
Hence, the field due to the rest of the conductor is $\large\frac{\sigma}{2 \in _0} $$\hat n$
answered Jun 4, 2014 by meena.p
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