Electrostatics - NCERT Solutions

Chapter Overview

Electrostatics is the study of electric charges at rest and the forces between them. This chapter introduces fundamental concepts of electric charge, electric field, electric potential, and Gauss’s law.

Key Concepts:

• Electric Charge: Fundamental property of matter
• Coulomb’s Law: Force between charged particles
• Electric Field: Region of influence around a charge
• Electric Potential: Work done per unit charge
• Gauss’s Law: Relationship between charge and electric flux

NCERT Exercise Solutions

Exercise 1.1

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

Solution: Given:

• Charge 1 (q₁) = +2 × 10⁻⁷ C
• Charge 2 (q₂) = +3 × 10⁻⁷ C
• Distance (r) = 30 cm = 0.3 m
• Permittivity of free space (ε₀) = 8.854 × 10⁻¹² C²/Nm²

Using Coulomb’s Law: $F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}$

$F = 9 \times 10^9 \cdot \frac{(2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2}$

$F = 9 \times 10^9 \cdot \frac{6 \times 10^{-14}}{0.09}$

$F = 9 \times 10^9 \cdot 6.67 \times 10^{-13}$

$F = 6 \times 10^{-3}$ N = 0.006 N

Answer: The force between the spheres is 0.006 N (repulsive).

Exercise 1.2

Question: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC 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?

Solution: Given:

• Charge 1 (q₁) = 0.4 μC = 0.4 × 10⁻⁶ C
• Charge 2 (q₂) = -0.8 μC = -0.8 × 10⁻⁶ C
• Force (F) = 0.2 N

(a) Finding the distance: Using Coulomb’s Law: $F = \frac{1}{4\pi\epsilon_0} \cdot \frac{|q_1 q_2|}{r^2}$

$0.2 = 9 \times 10^9 \cdot \frac{(0.4 \times 10^{-6}) \times (0.8 \times 10^{-6})}{r^2}$

$0.2 = 9 \times 10^9 \cdot \frac{3.2 \times 10^{-13}}{r^2}$

$r^2 = 9 \times 10^9 \cdot \frac{3.2 \times 10^{-13}}{0.2}$

$r^2 = 9 \times 10^9 \cdot 1.6 \times 10^{-12}$

$r^2 = 14.4 \times 10^{-3} = 0.0144$

$r = \sqrt{0.0144} = 0.12$ m = 12 cm

(b) Force on the second sphere: According to Newton’s third law, the force on the second sphere due to the first is equal in magnitude and opposite in direction to the force on the first sphere due to the second.

Answer:

• Distance between spheres = 12 cm
• Force on second sphere = 0.2 N (attractive)

Exercise 1.3

Question: Check that the ratio ke²/Gme mp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution: The ratio to check is: $\frac{ke^2}{Gm_em_p}$

Where:

• k = 1/(4πε₀) = 9 × 10⁹ Nm²/C²
• e = elementary charge = 1.6 × 10⁻¹⁹ C
• G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
• me = electron mass = 9.1 × 10⁻³¹ kg
• mp = proton mass = 1.67 × 10⁻²⁷ kg

Checking dimensions:

• ke² has dimensions: [Nm²/C²] × [C²] = [Nm²]
• Gmemp has dimensions: [Nm²/kg²] × [kg] × [kg] = [Nm²]

Since both numerator and denominator have the same dimensions [Nm²], the ratio is dimensionless.

Calculating the value: $\frac{ke^2}{Gm_em_p} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}}$

$= \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{6.67 \times 10^{-11} \times 1.52 \times 10^{-57}}$

$= \frac{2.304 \times 10^{-28}}{1.01 \times 10^{-67}}$

$≈ 2.3 \times 10^{39}$

Significance: This ratio represents the relative strength of electrostatic force to gravitational force between an electron and a proton. The value of approximately 10³⁹ shows that the electrostatic force is about 10³⁹ times stronger than the gravitational force.

NCERT Highlights for JEE/NEET

🎯 Important Concepts for JEE:

1. Electric Field due to Point Charge: $\vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r^2} \hat{r}$
2. Electric Field due to Dipole: $\vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3}$ (on axial line)
3. Gauss’s Law: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$
4. Electric Potential: $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r}$

🧬 Key Points for NEET:

1. Electric Field Lines: Start from positive charges, end at negative charges
2. Electric Flux: $\Phi = \oint \vec{E} \cdot d\vec{A}$
3. Electric Dipole Moment: $\vec{p} = q \times 2a$
4. Superposition Principle: Net field is vector sum of individual fields

JEE/NEET Practice Questions

Question 1 (JEE Level)

Statement: Two identical charged spheres are suspended by strings of equal length. The strings make an angle of 30° with each other. When suspended in a medium of dielectric constant K, the angle remains the same. Find the value of K.

Solution: For equilibrium in air: $T\sin\theta = F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q^2}{r^2}$

In medium with dielectric constant K: $T\sin\theta = F’ = \frac{1}{4\pi\epsilon_0 K} \cdot \frac{q^2}{r^2}$

Since the angle remains the same, the forces must be equal: $F = F'$

$\frac{1}{4\pi\epsilon_0} \cdot \frac{q^2}{r^2} = \frac{1}{4\pi\epsilon_0 K} \cdot \frac{q^2}{r^2}$

This gives: $1 = \frac{1}{K}$, therefore $K = 1$

Answer: K = 1 (vacuum)

Question 2 (NEET Level)

Statement: A charge +Q is placed at the center of a hollow conducting sphere. The electric field at a point inside the conductor is:

Solution: According to electrostatic shielding:

• Inside a hollow conducting sphere, the electric field due to external charges is zero
• However, since the charge +Q is inside the cavity, it will induce -Q on the inner surface
• The electric field inside the conducting material (not the cavity) is zero

Answer: Zero

Important Derivations

Electric Field due to Dipole (Axial Line)

Consider an electric dipole with charges +q and -q separated by distance 2a.

For a point P at distance r from the center (r » a):

• Distance from +q: r + a
• Distance from -q: r - a

Electric field due to +q: $E_+ = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{(r+a)^2}$

Electric field due to -q: $E_- = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{(r-a)^2}$

Net field: $E = E_+ - E_- = \frac{1}{4\pi\epsilon_0} \cdot q \left[\frac{1}{(r+a)^2} - \frac{1}{(r-a)^2}\right]$

For r » a, using binomial approximation: $E \approx \frac{1}{4\pi\epsilon_0} \cdot \frac{2qa}{r^3} = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3}$

Where p = 2qa is the dipole moment.

Important Formulas

Success Tips

1. Vector Nature: Remember electric field and force are vector quantities
2. Sign Convention: Pay attention to the sign of charges
3. Symmetry Arguments: Use symmetry to simplify complex problems
4. Gauss’s Law: Apply it for symmetric charge distributions

Master Electrostatics

Electrostatics is fundamental to understanding electric phenomena. Focus on concepts and practice regularly!

Remember: Electrostatics builds the foundation for entire electricity and magnetism. Master these concepts for success in competitive exams!

For additional practice, explore our Physics Practice Problems section.