(a) A conductor A with a cavity as shown in Fig. 1.36(a) 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. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way

Refer figure given in Question

Ans (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 be the charge inside the conductor and is ε₀ the permittivity of free space.

According to Gauss’s law, Flux, φ = E. ds = q/ε₀

Here, E = 0 => q/ε₀ = 0   => q = 0 [as ε₀ ≠ 0 ]

Therefore, charge inside the conductor is zero.

The entire charge Q appears on the outer surface of the conductor.

Ans (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.

Ans (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.