Question
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 
is the surface charge
density near the hole.
, where is the unit vector in the outward
normal direction, and is the surface charge
density near the hole.
Answer :
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.
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.
. 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.
No comments:
Post a Comment