(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 instruments is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Explanation

(a) As we know, electric field inside the charge conductor is zero. `vecE_(in)=0`
Select a Gaussian surface which encloses the cavity in the conductor as shown in figure (a). From Gauss.s law,
`oint vecE. dvecS = q/epsilon_(0)`
`therefore vecE =0` (Inside)
`therefore` Charge enclosed q=0
Hence, charge given to the conductor, Q will be distributed on its surface only.
(b) By placing conductor B with charge .Q. in cavity of conductor A, - q charge is induced on the cavity surface. And + q charge is induced on the outer surface of conductor A. Now, total charge surface on of A will become Q + q.
(c) Suppose, we have electrically neutral metallic conductor A having some cavity (empty space) and we have another electrically neutral metallic conductor C which is to be completely protected against external electric field. For this, we place < inside the cavity of A but at the same tim isolated from A (i.e. A and C are not inte connected).
Now, suppose A is subjected to somi external electric field as shown in abov figure. Here as soon as we apply thir-external electric field, free electrons on th. outer surface of A displace immediately ii the direction opposite to externally appliec electric field Ea because of this, anothe: electric field gets developed inside A called induced electric field `vecE_(i)` in a directior opposite to `vecE_(a)` . When `E_(i) = E_(a)` free electrons
do not move at all and thus become static on the outer surface of A. Here, A is said to be in "Electrostatic equilibrium condition".
Now, for all the points on the Gaussian surface 1 (surrounding to cavity and very close to cavity) we have E = 0, because `q_("enclosed")=0` surface of cavity remains chargeless as it was before the application of external electric field. It means that surface of cavity remains unaffected though conductor A is exposed to external electric field.
Moreover, inside the cavity there is no charge as conductor C is electrically neutral. Hence, for all the points on Gaussian surface 2 also, E = 0. Thus, C gets complete protection by the surface of cavity. Here, however strong applied electric field may be there is no induction of charge on the surface of C. That is why C is said to be "Electrostatically shielded" against external electrical conditions outside A.
Note : During thunderstorm with lightening (especially in rainy season) when we are inside the car with doors and windows closed we are in the cavity of a metallic body of a car and so we remain safe in the car due to "electrostatic shielding". Thus, during lightening in the atmosphere, one should not stand on the open road. Instead, it is always preferable to be inside the closed car.