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$(a)$ 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 $A \text { is } Q+q$ [Figure $(b)$]
$(c)\;A$ sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Solution
$(a)$ 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 is $\epsilon_{0}$ the permittivity of free space. According to Gauss's law, Flux,
$\phi=E . d s=\frac{q}{\epsilon_{0}}$
Here, $E =0 \Rightarrow \frac{q}{\epsilon_{0}}=0 \Rightarrow q=0$
$\left[\text { as } \epsilon_{0} \neq 0\right]$
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.
