(a) A conductor A with a cavity as shown in Fig, is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is `(Q+q)` fig.
(c) A sensitive instrument is to be shielded from the strong electrostatic field in its environment. Suggest a possible way.

Explanation

a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.
Let q is the charge inside the conductor and `epsilon_(0)` is the permittivity of free space
According to Gauss’s law,
`phi=bar(E).bards=(q)/(epsilon_(0))`
flux,
Here, E=0
`(q)/(epsilon^(0))=0`
`therefore epsilon_(0)ne0`
`therefore q-=0`
Therefore, charge inside the conductor is zero. The entire charge Q appears on the outer surface of the conductor.
(b) The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount −q will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is Q + q.
(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.