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NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 1: Electric Charges and Fields (Part II)

Class 12th Physics NCERT Exemplar solutions for Chapter 1 - Electric Charges and Fields are available here. With this article, you will get solutions from question number 1.8 to question number 1.13. Basically, these questions are multiple choice questions with multiple correct answers. All the questions of NCERT Exemplar are important for CBSE 12th board exam and other competitive exams.

Jun 6, 2017 16:00 IST
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NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 1: Electric Charges and Fields (Part II) MCQs II

NCERT Exemplar solutions for Class 12 Physics, Chapter 1: Electric Charges and Fields are available here. This chapter of NCERT Exemplar has variety questions. We have provided the solutions in several parts. This is Part II, here you will find solutions for MCQs II (or multiple choice questions with multiple correct answer). Solutions for rest of the questions are available in other parts. These questions are very important for CBSE 12th Physics board exam & competitive examinations like JEE Main, NEET etc.

NCERT Exemplar Solution for 12th Physics, Chapter 1: Electric Charges and Fields (MCQs II) are given below:

Question 1.8: If

over a surface, then

(a) the electric field inside the surface and on it is zero

(b) the electric field inside the surface is necessarily uniform

(c) the number of flux lines entering the surface must be equal to the number of flux lines leaving it

(d) all charges must necessarily be outside the surface

Solution 1.8: (c, d)

Given,

It means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.
Now, from Gauss' law,

Where q is charge enclosed by the surface.

Now,

q = 0 i.e., net charge enclosed by the surface must be zero.

Hence, all other charges must necessarily be outside the surface.

Question 1.9: The electric field at a point is

(a) always continuous

(b) continuous if there is no charge at that point

(c) discontinuous only if there is a negative charge at that point

(d) discontinuous if there is a charge at that point

Solution 1.9: (b, d)

As we know, electric field lines emanate from positive charges and terminate on negative charges. Thus, electric field due to the charge Q will be continuous, if there is no charge at that point. It will be discontinuous if there is a charge at that point.

This can be understood with the help of figures shown below

Question 1.10: If there were only one type of charge in the universe, then

Solution 1.10: (c, d)

According to the Gauss' law

Where q is the charge enclosed by the surface. If the charge is outside the surface, then charge

If the charge is outside the surface, then charge enclosed by the surface is q = 0, therefore,

Electric flux doesn't depend on the nature or type of charge.

Question 1.11: Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region,

(a) the electric field is necessarily zero

(b) the electric field is due to the dipole moment of the charge distribution only

(c) the dominant electric field is ∝ 1/r3  for large r, where r is the distance from a origin in this regions

(d) the work done to move a charged particle along a closed path, away from the region, will be zero

Solution 1.11: (c, d)

Here, net charge is zero in the region. We can assume that contain one or more than one electric dipoles.

So, at points outside the region, the dominant electric field for ∝ 1/r3 for large r.

Also, electric field is conservative field, work done to move a charged particle along a closed path, away from the region will be zero.

Question 1.12: Refer to the arrangement of charges in Fig. 1.6 and a Gaussian surface of radius R with Q at the centre. Then

(a) total flux through the surface of the sphere is ‒Qo

(b) field on the surface of the sphere is ‒Q/(4πεoR2)

(c) flux through the surface of sphere due to 5Q is zero

(d) field on the surface of sphere due to –2Q is same everywhere

Solution 1.12: (a, c)

Gauss' law states that total electric flux of an enclosed surface is given by qo. Here, q is the net charge enclosed by the Gaussian surface.

From the figure,

Net charge inside the surface is = Q ‒ 2Q = ‒ Q.

Net flux through the surface of the sphere = ‒ Qo

Here, charge 5Q lies outside the Gaussian surface, so it will not make no contribution to electric flux through the given surface.

Question 1.13: A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring (Fig. 1.7). Then

(a) If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.

(b) If q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring.

(c) If q < 0, it will perform SHM for small displacement along the axis.

(d) q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0.

Solution 1.13: (a, b, c)

Here, the positive charge Q is uniformly distributed on the outer surface of the ring. Thus, the electric field inside the ring is zero. So, the effect of electric force on charge q due to the positive charge Q is at the centre zero.

When q < 0, force between Q and q is attractive. If q is displaced away from the centre in the plane of the ring, the net force on one side decrease and other side increase by the same amount. So, it will be pushed back towards the centre.

When this charge is displaced towards the axis of the ring, it will gain potential energy and will be pulled by the ring (towards its centre). Due to increase in kinetic energy, it will cross the centre and again gains potential energy. Afterwards, it will be again pulled towards the centre and this process (SHM) will continue due to interconversion of energies.

When q > 0, force between Q and q is repulsive. If q is displaced away from the centre in the plane of the ring, again, the net force on one side decrease and other side increase by the same amount. So, it will be pushed back towards the centre.

Within the plane of the ring, the equilibrium is stable as the charge q tends to move back to its position.

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