Obj-2 Gauss Law HC Verma Solutions Vol-2 Class-12 Ch-30

Obj-2 Gauss Law HC Verma Solutions Vol-2 Class-12 Ch-30. Step by Step Solutions of Objective -2 (MCQ-2) Questions of Chapter-30 Gauss Law (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Obj-2 Gauss Law HC Verma Solutions Vol-2 Class-12 Ch-30

Board ISC and other board
Publications Bharti Bhawan Publishers
Chapter-30 Gauss Law
Class 12
Vol  2nd
writer HC Verma
Book Name Concept of Physics
Topics Solution of Objective-2 (MCQ-2) Questions
Page-Number  140, 141

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Question for Short Answer

Objective-I

Objective-II

Exercise (update soon)


Gauss Law Objective-2 (MCQ-2) Questions

HC Verma Solutions of Ch-30  Vol-2 Concept of Physics for Class-12
(Page-140)

Question-1 :-

Mark the correct options:

(a) Gauss’s Law is valid only for symmetrical charge distributions.

(b) Gauss’s Law is valid only for charges placed in vacuum.

(c) The electric field calculated by Gauss’s Law is the field due to the charge inside the Gaussian surface.

(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.

Answer-1 :-

The option (d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.  is correct
Explanation:
The contribution of flux on the closed surface due to the charges lying outside the surface is zero because number of field line entering the closed surface is equal to the number of field lines coming out of the surface so the net contribution of the charge lying outside the closed surface to the flux is zero. Therefore, the net flux through the surface due to the charge lying outside the the closed surface is zero. The contribution that counts is only due to the charges lying within the closed surface.
Thus, the flux of the electric field through a closed surface due to all the charges (inside and outside the surface) is equal to the flux due to the charges enclosed by the surface

 

Question-2 :-

A positive point charge Q is brought near an isolated metal cube.

(a) The cube becomes negatively charged.

(b) The cube becomes positively charged.

(c) The interior becomes positively charged and the surface becomes negatively charged.

(d) The interior remains charge free and the surface gets non-uniform charge distribution.

Answer-1 :-

The option (d) The interior remains charge free and the surface gets non-uniform charge distribution. is correct
Explanation:
Since the cube is metallic, the charge gets distributed on the surface and the interior remains charge-free. However, when a positive point charge Q is brought near the metallic cube, a negative charge gets induced on the face near the positive charge Q and an equal positive charge gets induced on the face, which is away from the charge Q. Thus, the metallic surface gets non-uniform charge distribution.

Question-3 :-

A large non-conducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet as shown in the following  figure.

(a) M attracts A.
(b) M attracts B.
(c) A attracts B.
(d) B attracts A.

A large non-conducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet as shown in the following figure.

(figure 30-Q4)

Answer-3 :-

The options (a), (b), (c) and (d) are correct

Explanation:

A large non-conducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet as shown in the following figure.

Since the non-conducting sheet M is given a uniform charge, it induces a charge in the metal rod A, which further induces a charge in the metal rod B, as shown in the figure. Hence, all the options are correct.

An electric dipole is placed at the centre of a sphere. Mark the correct options.
(a) The flux of the electric field through the sphere is zero.
(b) The electric field is zero at every point of the sphere.
(c) The electric field is not zero anywhere on the sphere.
(d) The electric field is zero on a circle on the sphere.

An electric dipole is placed at the centre of a sphere. Mark the correct options.

where θ is the angle made by the point p with the centre of the dipole.
Hence, we can see that the field is not zero anywhere on the sphere.

Question-6 :-

In the figure (30-Q5) shows a charge q placed at the centre of a hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged?
(a) A
(b) B
(c) C
(d) D

In the following figure shows a charge q placed at the centre of a hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged?

(figure  30-Q5)

Answer-6 :-

The options (a) and (c) are correct

Explanation:

These are the only points in the straight line with the charge q and at the brim of the hemisphere. So, field lines emerging from these charges do not affect the flux through the hemisphere due to charge q. On the other hand, the two remaining charges are beside the surface of the hemisphere and not in line with the charge q and not at the brim. So, they will affect the flux. Hence, the correct answers are (a) and (c).

The figure 30-Q-7 shows a closed surface that intersects a conducting sphere. If a positive charge is placed at point P, the flux of the electric field through the closed surface

 

Answer-8 :-

The option (b) will become positive is correct
Explanation:

A positive charge at point P will induce a negative charge on the near face of the conducting sphere, whereas the positive charge on the farther end of the sphere. As this father end is enclosed or intersected by the closed surface, so the flux through it will become positive due to the induced positive charge.

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