# NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 1: Electric Charges & Fields (Part III)

CBSE Class 12 Physics NCERT Exemplar Solutions for Chapter 1 (Electric Charges and Fields) are available here. Here, you will find solutions for very short answer type questions i.e., from question number 1.14 to question number 1.19. Solutions of other questions of the chapter are available in other parts.

Created On: Jun 13, 2017 16:35 IST
Modified On: Jun 23, 2017 15:13 IST

NCERT Exemplar Solutions for Class 12 Physics, Chapter 1: Electric Charges and Fields are available here. In this article, you will get solutions for very short answer type questions (i.e., from question number 1.14 to 1.19). The detailed solutions of other questions of this chapter are available in other parts. These questions are extremely important for CBSE Class 12 Physics board exam and other competitive exams like NEET, JEE Main etc.

NCERT Exemplar Solutions for Class 12 Physics, Chapter 1: Electric Charges and Fields (from question no. 1.14 to 1.19 are given below:

Question 1.14: An arbitrary surface encloses a dipole. What is the electric flux through this surface?

Solution 1.14:

From Gauss' law, the electric flux through an enclosed surface is given by

Where, q is the net charge enclose by the surface

Now total charge enclose by the dipole = (‒ q + q) = 0

Therefore, electric flux through a surface enclosing a dipole = 0.

NCERT Solutions for Class 12 Physics: All Chapters

Question 1.15: A metallic spherical shell has an inner radius R1 and outer radius R2. A charge Q is placed at the centre of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface?

Solution 1.15:

The situation given in question is shown in figure given below

When positive charge is placed at the centre of the spherical cavity then an equal amount of negative charge (‒Q) appears on inner surface of the sphere due to induction. This charge is distributed uniformly on the inner surface. Also an equal amount of positive charge (+Q) also appears on outer surface of the sphere.

Now surface charge density on inner surface = (Total charge on inner surface/Area of the inner surface) = ‒Q/(4πR12)

Surface charge density on outer surface = (Total charge on outer surface/Area of the outer surface) = +Q/(4πR22)

Question 1.16: The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?

Solution 1.16:

In an atom, number of electrons and protons are equal and the electric fields bind the atoms to neutral entity. Fields are caused by excess charges. There can be no excess charge on the inter surface of an isolated conductor.

Question 1.17: If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.

Solution 1.17:

According to the Gauss’s law,

Here, the term q on the right side of the equation includes the sum of all charges enclosed by the surface (irrespective of the position of the charges inside the surface).

If the surface is so chosen that there are some charges inside and some outside, the electric field on the left side of equation will be due to all charges (both inside and outside)

Therefore, despite being total charge enclosed by a surface zero, it doesn't imply that the electric field everywhere on the surface is zero, the electric field may be normal to the surface.

However, if total electric field everywhere on a surface is zero then it implies that net charge inside is zero.

Question 1.18: Sketch the electric field lines for a uniformly charged hollow cylinder shown in Fig 1.8.

Solution 1.18:

Question 1.19: What will be the total flux through the faces of the cube (Fig. 1.9) with side of length a if a charge q is placed at

(a) A: a corner of the cube.

(b) B: mid-point of an edge of the cube.

(c) C: centre of a face of the cube.

(d) D: mid-point of B and C.

Solution 1.18:

(a) When charge is placed at corner A of the cube then only 1/8th portion of the charge lies incised the Gaussian surface. So, total flux through the faces of the given cube = q/8εo.

(b) When charge is placed at point B of the cube then only 1/4th portion of the charge lies incised the Gaussian surface. So, total flux through the faces of the given cube = q/8εo.

(c) When charge is placed at point C of the cube then only 1/2th portion of the charge lies incised the Gaussian surface. So, total flux through the faces of the given cube = q/2εo.

(d) When charge is placed at point D of the cube then only 1/2th portion of the charge lies incised the Gaussian surface. So, total flux through the faces of the given cube = q/2εo.

Comment ()

3 + 6 =
Post