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Question 28

(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.

 

                                                                

Answer

(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 \(\epsilon_{0}\) is the permittivity of free space.



 

According to Gauss’s law,

 

Flux, \(\phi = \overrightarrow{E}.\overrightarrow{ds} = \frac{q}{\epsilon_{0}}\)

 

Here, E = 0

 

\(\frac{q}{\epsilon_{0}} = 0\)

 

\(\because\epsilon_{0} \neq 0\)

 

\(\therefore q=0\)

 

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.

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