NCERT Solutions: Class 12 Physics Chapter 1

Electric Charges and Fields: This chapter establishes the foundation of Electrostatics. You will learn about the quantization of charge, Coulomb's Law, and how to calculate Electric Field Intensity for various distributions like dipoles and thin wires.

Chapter 1: PDF Digital Reader

Quick Revision Guide

Before jumping into solutions, here are the core concepts you must remember for your boards and JEE:

1. Properties of Charge

  • Quantization: Charge exists only in integral multiples of 'e' (q = ±ne).
  • Conservation: Total charge of an isolated system remains constant.

2. Important Formulas

Force: F = (1/4πε∨) × (q&sub1;q&sub2;/r²)

E-Field: E = F/q = (1/4πε∨) × (q/r²)

Exercise Solutions

Question 1.1

What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air?

VIEW DETAILED SOLUTION ∨

Given Data:

  • q&sub1; = 2 × 10⁻⁷ C
  • q&sub2; = 3 × 10⁻⁷ C
  • Distance (r) = 30 cm = 0.3 m

Formula: Coulomb's Law, F = (1/4πε∨) × (q&sub1;q&sub2;/r²)

Calculation:

F = (9 × 10⁹) × (2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.3)²
F = 54 × 10⁻⁵ / 0.09
F = 6 × 10⁻³ N

Result: The force is 6 × 10⁻³ N (Repulsive because both are positive charges).

Question 1.2

The electrostatic force on a small sphere of charge 0.4 µC due to another sphere of charge -0.8 µC in air is 0.2 N. What is the distance between them?

VIEW DETAILED SOLUTION ∨

Solution:

Using F = k × |q&sub1;q&sub2;| / r²

r² = (k × |q&sub1;q&sub2;|) / F

r² = (9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶) / 0.2
r² = 144 × 10⁻⁴
r = √(144 × 10⁻⁴) = 12 × 10⁻² m

Result: The distance between the two spheres is 12 cm.

Important Notes on Electric Field Lines

[Image of electric field lines for positive and negative charges]
  • Field lines start from positive charges and end at negative charges.
  • Two field lines never intersect each other.
  • The number of field lines is proportional to the magnitude of the charge.