Chapter One – ELECTRIC CHARGES AND FIELDS- Complete Guide to Electric Charge & Coulomb’s Law from NCERT

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Complete Guide to Electric Charge & Coulomb’s Law

1. Introduction to Electric Charge

Electric charge is a fundamental property of matter responsible for electrostatic interactions. It is quantized and conserved in nature.

2. Gold Leaf Electroscope

The gold leaf electroscope is used to detect charge on a body. It consists of a metal rod with thin gold leaves attached at the bottom. When a charged object touches the metal knob, the charge spreads and causes the leaves to diverge.

3. Coulomb’s Law

Coulomb’s law states that the force between two point charges is:

\[ F = k \frac{q_1 q_2}{r^2} \]

where \( k = 9 \times 10^9 \) Nm²/C².

4. Vector Form of Coulomb’s Law

The vector representation is:

\[ \mathbf{F_{21}} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{r}_{21} \]

5. Superposition Principle

The total force on a charge due to multiple charges is given by:

\[ \mathbf{F} = \sum_{i=1}^{n} \mathbf{F_i} \]

6. Example Calculations

Q: Two charges \( q_1 = 3 \times 10^{-6} C \) and \( q_2 = -2 \times 10^{-6} C \) are placed 0.5 m apart. Find the force between them.

Solution:

\[ F = \frac{9 \times 10^9 \times (3 \times 10^{-6}) \times (2 \times 10^{-6})}{(0.5)^2} \]

\[ F = 0.216 N \text{ (attractive)} \]

7. Assertion-Reason Questions

Assertion:

Coulomb’s force acts along the line joining two charges.

Reason:

Electrostatic force is a central force and follows Newton’s third law.

8. Multiple Choice Questions (MCQs)

Q: Which of the following statements about electric charge is correct?

(a) Charge is a vector quantity
(b) Charge is quantized
(c) Charge cannot be created or destroyed
(d) Both (b) and (c)

Answer: (d) Both (b) and (c).

9. Conclusion

This post comprehensively covers electric charge, Coulomb’s law, and practical calculations.

Complete Guide to Electric Charge & Coulomb’s Law

Understanding Electric Charge

Electric charge is a fundamental property of matter responsible for electrostatic interactions. It is quantized and conserved in nature.

Q: What does it mean when a body is charged?

(a) Excess of charge
(b) Deficit of charge
(c) Both (a) and (b)

Answer: (c) Both (a) and (b). A body is charged when it has either an excess or a deficit of electrons.

Coulomb’s Law

Coulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

\[ F = k \frac{q_1 q_2}{r^2} \]

where:

\( F \) is the electrostatic force.
\( k \) is Coulomb’s constant (\( 8.99 \times 10^9 \) Nm²/C²).
\( q_1, q_2 \) are the magnitudes of the two charges.
\( r \) is the distance between the charges.

Vector Notation in Coulomb’s Law

Since force is a vector, it is better to write Coulomb’s law in vector notation:

\[ \mathbf{F_{21}} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{r}_{21} \]

Explanation:

Position vectors: \( \mathbf{r_1} \) and \( \mathbf{r_2} \) describe the positions of charges \( q_1 \) and \( q_2 \).
Vector direction: The unit vector \( \hat{r}_{21} \) points from \( q_1 \) to \( q_2 \).
Magnitude: The force depends on \( \frac{1}{r^2} \) and the product of the charges.

Quantization of Charge

At a microscopic level, charge appears in discrete units of \( e = 1.6 \times 10^{-19} C \), but at macroscopic scales, it behaves as if continuous.

Example Calculations

Q: If \(10^9\) electrons move per second, how much time is needed to transfer 1C?

Solution: Since charge of one electron is \( e = 1.6 \times 10^{-19} C \), total number of electrons needed:

\[ n = \frac{1C}{1.6 \times 10^{-19} C} = 6.25 \times 10^{18} \]

Time required:

\[ t = \frac{6.25 \times 10^{18}}{10^9} = 6.25 \times 10^9 \text{ seconds} \]

Charge in a Cup of Water

Q: How much positive and negative charge is there in a cup of water?

Solution: Let’s assume the mass of one cup of water is 250g. The molecular mass of water is 18g. The number of molecules in one cup is:

\[ N = \left( \frac{250}{18} \right) \times 6.02 \times 10^{23} \]

Each molecule has 10 electrons and 10 protons, contributing to total charge:

\[ Q = N \times 10 \times 1.6 \times 10^{-19} C \]

Final result: \( Q = 1.34 \times 10^7 C \).

Assertion-Reason Questions

Assertion:

At macroscopic scales, the grainy nature of charge is lost and appears continuous.

Reason:

A charge of 1mC contains about \( 10^{13} \) electrons, making the quantization of charge negligible.

Correct Explanation: Since large numbers of electrons are involved, the discrete nature of charge is imperceptible.

Conclusion

This post covered fundamental concepts of electric charge, Coulomb’s law in vector form, quantization of charge, and important numerical examples.

Electrostatic Forces in an Equilateral Triangle

Problem Statement

Consider the charges \( q, q, \) and \( -q \) placed at the vertices of an equilateral triangle of side \( l \), as shown in the figure. What is the force on each charge?

Solution

Step 1: Coulomb’s Law

The force between two point charges is given by:

\[ F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{l^2} \]

Since all charges have magnitude \( q \), the force between any two charges separated by \( l \) is:

\[ F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{l^2} \]

Step 2: Forces on \( q_1 \) (Charge at \( A \))

Charge \( A \) experiences forces due to charges at \( B \) and \( C \).

Force \( F_{12} \) due to charge at \( B \): Repulsion along \( AB \), magnitude \( F \).
Force \( F_{13} \) due to charge at \( C \): Attraction along \( AC \), magnitude \( F \).

Step 3: Resolving Forces

Resolving \( F_{13} \):

\[ F_{13x} = F \cos 60^\circ = \frac{F}{2} \] \[ F_{13y} = F \sin 60^\circ = \frac{\sqrt{3}F}{2} \]

Net Force:

\[ F_{\text{net},x} = F + \frac{F}{2} = \frac{3F}{2} \] \[ F_{\text{net},y} = \frac{\sqrt{3}F}{2} \] \[ F_1 = \sqrt{ \left( \frac{3F}{2} \right)^2 + \left( \frac{\sqrt{3}F}{2} \right)^2 } = \sqrt{3}F \]

Angle:

\[ \theta = \tan^{-1} \left( \frac{\sqrt{3}}{3} \right) = 30^\circ \]

Step 4: Forces on \( q_2 \) and \( q_3 \)

Force on \( q_2 \) is identical in magnitude \( \sqrt{3}F \) but at \( 30^\circ \) below the horizontal.
Force on \( q_3 \) is \( \sqrt{3}F \) directed downward.

Assertion and Reason

Assertion: In an equilateral triangle configuration of charges \( q, q, -q \), each charge experiences a net force of magnitude \( \sqrt{3}F \), and the system is in equilibrium.

Reason: The forces between charges follow Coulomb’s law and vector addition. Due to symmetry, the forces on each charge have equal magnitude but are directed in such a way that they maintain equilibrium.

Conclusion: The assertion is correct, and the reason correctly explains it.