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, where is the unit vector in the outward normal direction, and 
is the surface charge density near the hole. Asked by Shagufta Last Modified
Answer 
Apoorva Purohit
Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.
Let E is the electric field just outside the conductor, q is the electric charge, 
is the charge density, and 
is the permittivity of free space.
Charge 
According to Gauss’s law,
Therefore, the electric field just outside the conductor is
. This field is a superposition of field due to the cavity 
and the field due to the rest of the charged conductor. These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.
Therefore, the field due to the rest of the conductor is.
Hence, proved.
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