Question

(a) A conductor $A$ with a cavity as shown in Fig. (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 $Q+q$.

(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Answer 1

(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 ${\in }_0$ is the permittivity of free space.

According to Gauss's law,

Flux, $\varphi =\overline{E},\overline{ds}=\frac{q}{{\in }_0}$

Here, $E=0$

$\frac{q}{{\in }_0}=0$

$\because {\in }_0\ne 0$

$\therefore q=0$

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.