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

Introduction to Electric Charges and Fields

This chapter introduces the fundamental concepts of electrostatics, covering electric charges, Coulombs law, electric field, and electric flux. These concepts are essential for understanding electromagnetic theory and are frequently asked in JEE and NEET exams.

Key Concepts

Electric Charge

Electric charge is a fundamental property of matter. There are two types of charges:

• Positive charge: Carried by protons
• Negative charge: Carried by electrons

The SI unit of charge is Coulomb (C). The charge of an electron is e = 1.6 x 10^-19 C.

Properties of Electric Charge

1. Additivity: Total charge = algebraic sum of all charges
2. Conservation: Charge can neither be created nor destroyed
3. Quantization: Charge exists in discrete packets (q = ne)

Coulombs Law

The force between two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them.

Formula: F = k(q1 x q2)/r^2

Where k = 9 x 10^9 Nm^2/C^2

Electric Field

Electric field is the region around a charged particle where its influence can be felt.

Electric Field Intensity: E = F/q0 (Force per unit positive charge)

Due to point charge: E = kQ/r^2

Electric Field Lines

Properties of electric field lines:

• Start from positive charges and end on negative charges
• Never intersect each other
• Are continuous curves
• Tangent at any point gives the direction of electric field

Electric Flux

Electric flux is the total number of electric field lines passing through a surface.

Formula: Phi = E.A.cos(theta)

Gauss Law

The total electric flux through a closed surface is equal to 1/epsilon0 times the total charge enclosed.

Mathematical form: Phi = q/epsilon0

NCERT Exercise Solutions

Exercise 1.1

Q: What is the force between two small charged spheres having charges of 2 x 10^-7 C and 3 x 10^-7 C placed 30 cm apart in air?

Solution:

Given: q1 = 2 x 10^-7 C, q2 = 3 x 10^-7 C, r = 30 cm = 0.3 m

F = kq1q2/r^2 = (9 x 10^9)(2 x 10^-7)(3 x 10^-7)/(0.3)^2

F = 6 x 10^-3 N = 6 mN (Repulsive)

Exercise 1.2

Q: The electrostatic force on a small sphere of charge 0.4 micro-C due to another small sphere of charge -0.8 micro-C in air is 0.2 N. What is the distance between the two spheres?

Solution:

Given: q1 = 0.4 micro-C, q2 = 0.8 micro-C, F = 0.2 N

Using F = kq1q2/r^2

r^2 = kq1q2/F = (9 x 10^9)(0.4 x 10^-6)(0.8 x 10^-6)/0.2

r = 0.12 m = 12 cm

Important Formulas

• Coulombs Law: F = kq1q2/r^2
• Electric Field: E = kQ/r^2
• Electric Flux: Phi = E.A.cos(theta)
• Gauss Law: Phi = q/epsilon0
• Field due to infinite line charge: E = lambda/(2 x pi x epsilon0 x r)
• Field due to infinite plane sheet: E = sigma/(2 x epsilon0)

Tips for Board Exams and JEE/NEET

1. Master the vector nature of electric field and force
2. Practice problems on superposition principle
3. Understand Gauss law applications thoroughly
4. Draw neat diagrams for electric field lines