NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT Solutions for Class 12 Electric Charges and Fields: With the implementation of the National Education Policy 2020, educational boards are revising their curricula. A good textbook follows the revised curriculum, and thus, NCERT has also revised its textbooks in accordance with the revised curriculum. 

Read: NCERT Class 12 Physics Solutions

Read: CBSE Class 12 Physics Syllabus 2023-24

Read: RBSE Class 12 Physics Syllabus 2023-24

Read: Punjab Board Class 12 Physics Syllabus 2023-24

This article will provide you with the complete solution guide for NCERT Class 12 Physics Chapter 1, Electric Charges and Fields. The solutions follow a step-by-step approach to explain the concepts properly. The complete Electric Charges and Fields NCERT solutions are provided below, along with a compiled PDF. The solution PDF is free to download. Read and download the PDF below.

The questions and their solutions discussed in this article don’t follow the revised NCERT textbook, as there are few educational boards that still follow the old NCERT textbooks. Thus, those who follow the revised textbooks must check the list of deleted topics provided below and accordingly solve the questions.

Electric Charges and Fields Solutions

1.1 What is the force between two small charged spheres having charges of 2x10^-7 C and 3x10^-7 C placed 30 cm apart in air?

Sol.

Given:

Q1 = 2x10^-7 C

Q2 = 3x10^-7 C

r = 30x10^-2 m

We know,

F=(Q_1 Q_2)/(4πε_o r² )

Substituting the given values, we get

F = 6x10^-3 N

2.1 The electrostatic force on a small sphere of charge 0.4 micro coulomb due to another small sphere of charge -0.8 micro coulomb in air is 0.2 N.

a) What is the distance between the two spheres?

b) What is the force on the second sphere due to the first?

Sol.

a) From the formula

F=(Q_1 Q_2)/(4πε_o r² )

Substituting the values, we get

r = 0.12 m = 12 cm

b) Force on second sphere due to first

F21 = (Q_1 Q_2)/(4πε_o r² )

This comes to be -0.2 N, negative sign implying the attractive nature of force. This conforms to Newton’s third law.

3.1 Check that the ratio ke2/Gmemp is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Sol.

We know very well

k = 9x10⁹ Nm²/C²

e = 1.6x10^-19 C

G = 6.6x10^-11 Nm²/kg²

me = 9.11x10^-31 kg

mp = 1.6x10^-27 kg

Ratio = (Nm²)/C² .C².kg²/Nm² .1/kg² =dimensionless

And value = 2.4x10³⁹

The ratio signifies that electrical forces are immensely stronger than the gravitational forces.

4.1 a) Explain the meaning of the statement 'electric charge of a body is quantized'.

b) Why can one ignore quantization of electric charge when dealing with macroscopic, i.e., large-scale charges?

Sol.

a) Quantization is one of the three basic properties of electric charge. It means that every charge is an integral multiple of e, i.e., ne...-2, -1, 0, 1, 2... The addition of charges, subtraction of charges, being an integer always gives integer result. Thus a charge can always be incremented or decremented in terms of e.

b) Macroscopic charges have very large number of electrons.

The quantization here can be taken as a continuous phenomenon, analogous to closely spaced dots resembling to line from a distance.

5.1 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how the observation is consistent with the law of conservation of charge.

Sol.

When two bodies are rubbed with each other transfer of charge takes place. One body receives charge and other loses, becoming negatively and positively charged respectively. In the whole process no new charge is created or destroyed. This implies that in an isolated system the total charge is always conserved.

6.1 Four point charges qA=2 µC, qB=-5 µC, qC= 2 µC, and qD=-5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Sol.

The charges of equal magnitude and same sign are at the corners of same diagonal. So they will exhibit equal and opposite forces at the charge situated at center, cancelling out each other. So the force is zero Newton.

7.1 a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

b) Explain why two field lines never cross each other at any point?

Sol.

a) The direction of electric field is given by tangent at each point on the curve. At sudden breaks, the field will have more than one direction which is not possible. That's why electrostatic field line is a continuous curve

b) At the crossing point there will be two directions of electric field at that point given by the two tangents. This cannot happen, and so two field lines never cross each other at any point.

8.1 Two point charges qA=3 μC and qB=-3 µC are located 20 cm apart in vacuum.

a) Two point charges qA = 3 uC and qB=-3 µC are located 20 cm apart in vacuum.

b) If a negative test charge of magnitude 1.5 x 10^-9C is placed at this point, what is the force experienced by the test charge?

Sol.

a) Given:

qA = 3 microC

qB = -3 microC

r = 10 cm (point is in middle)

We know, electric field is given by formula

E=Q/(4π€_o r²)

Electric field due to charge A (by use of above formula)

E1 = 2.7x10⁶ N/C

Electric field due to charge B

E2=2.7x10⁶ N/C

Since both the fields have the same direction, the electric field at O will be summation of above field

So E=5.4x10⁶ N/C

b) Given:

qA = 3 microC

qB= -3 microC

q0 = -1.5 nanoC

r = 10 cm (point is in middle)

By the formula

F=(Q_1Q_2)/(4лε_о г²)

F1=(3x10^-6).(-1.5x10^-9).(9x10⁹)/(10x10^-2)2 = -4.05x10^-3 N (attractive)

F2 = +4.05x10^-3 N (repulsive)

So the net force experienced by test charge 

F=F1-F2=8.1x10^-3 N towards charge A.

9.1 A system has two charges qA = 2.5 x 10^-7 C and qB = -2.5 x 10^-7 C located at points A: (0, 0, -15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

Sol.

Total charge of electric dipole = zero coulomb

Now, given

2a = 30 cm

q=2.5x10^-7 C

Dipole moment, p = q.2a

= (2.5x10^-7).(30)

= 7.5x10^-8 Cm

10.1 An electric dipole with dipole moment 4 x 10^-9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 x 10⁴ NC^-1. Calculate the magnitude of the torque acting on the dipole.

Given

Dipole moment, p = 4x10^-9 Cm

θ=30

E=5x10⁴ N/C

We know

Torque = p E sinθ

= (4x10^-9)(5x10⁴)(sin30)

= 10^-4 Nm

NCERT Rationalised Content Chapter 1: Electric Charges and Fields

NCERT has now revised its textbooks in accordance with the changes made in the CBSE Class 12 Physics syllabus. Students following the new NCERT textbooks should be aware of the topics and chapters that are no longer part of the revised NCERT textbooks and CBSE curriculum. A few educational boards still have not made any changes to their syllabus, thus following the old NCERT textbooks. Below is the list of topics which are now removed from NCERT Class 12 Physics textbook Electric Charges and Fields. Check and do the solutions accordingly.

CBSE Class 12 Physics Updated Question Paper Design 2023-24

Also Check: 

• CBSE Class 12 Physics Sample Paper 2023-24 with Solution

• NCERT Class 12 Textbook PDFs

• CBSE Class 12 Physics Previous Year Question Papers with Solutions PDF