Q 1.29: A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is , where is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.
Let the surface area of the sphere be S.
And assume that the hole is filled. For a point B, just above the hole, considering a gaussian surface passing through B, we have
Now, Since the electric field is always perpendicular to the surface of the conductor.
Using Superposition principle, ,
where is due to the hole and is due to the rest of the conductor. (Both pointing outwards, i.e away from the centre)
Again, for a point A, just below the hole, Electric field will be zero because of electrostatic shielding.
Using the superposition principle, this will be due to pointing inwards(towards the centre) and due to (Pointing away from the centre)
Using this relation, we get:
Since this is pointing outwards,
is the electric field in the hole.
(Trick: 1. Assume the hole to be filled.
2. Consider 2 points just above and below the hole.
3. Electric fields at these points will be due to the hole and rest of the conductor. Use superposition principle.)