$(a)$ $A$ conductor $A$ with a cavity as shown in Figure $(a)$ 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 $Q+q$ [Figure $(b)$].
$(c)$ $A$ sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

(N/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 $q_{in}$ be the charge inside the conductor and $\epsilon_{0}$ be the permittivity of free space. According to Gauss's law, the electric flux $\phi$ is given by:
$\phi = \oint E \cdot ds = \frac{q_{in}}{\epsilon_{0}}$
Since $E = 0$ inside the conductor, we have $\frac{q_{in}}{\epsilon_{0}} = 0$, which implies $q_{in} = 0$.
Therefore, the net charge inside the material of the conductor is zero. Consequently, the entire charge $Q$ must reside on the outer surface of the conductor.
$(b)$ The outer surface of conductor $A$ initially has a charge $Q$. When another conductor $B$ with charge $q$ is inserted into the cavity and insulated from $A$, an equal and opposite charge $-q$ is induced on the inner surface of the cavity of $A$ due to electrostatic induction. To conserve the total charge of the isolated conductor $A$, a charge $+q$ must be induced on its outer surface. Thus, the total charge on the outer surface of conductor $A$ becomes $Q+q$.
$(c)$ $A$ sensitive instrument can be shielded from strong electrostatic fields in its environment by enclosing it fully inside a metallic box or cage. This phenomenon is known as electrostatic shielding, where the electric field inside a closed metallic conductor is always zero.