Physics Class 12 NCERT Solutions: Chapter 1 Electric Charges and Fields Part 9 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q: 28. (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 [Figure (B) ] . (C) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

(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 the permittivity of free space. According to Gauss՚s law,

Flux,

Here,

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 is kept inside conductor A and it is insulated from A. Hence, a charge of amount will be induced in the inner surface of conductor A and is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is .

(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.

Q: 29. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is where is the unit vector in the outward normal direction, and is the surface charge density near the hole.

Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.

Let E is the electric field just outside the conductor, q is the electric charge, is the charge density, and is the permittivity of free space.

Charge

According to Gauss՚s law,

Flux,

Therefore, the field due to the rest of the conductor is .

Hence, proved.

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