Solution:
If we assume that the hole has been filled, then Gauss’s formula may be used to calculate the electric field intensity at a location near to the surface of the conductor, where
Flux,
here, is the surface charge density
ds is the area of the Gaussian surface
Let be the electric field due to the hole and is the electric field due to the rest of the conductor. and are opposite in direction and equal in magnitude since the total electric field inside the conductor is zero.
The electric field is due to the superposition of and .
From equation (1) we know
Therefore,
Therefore, the electric field in the hole