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

Class 12 Physics is a crucial stage in the journey towards understanding complex scientific concepts, and Chapter 1 of the NCERT Class 12 Physics textbook, titled "Electric Charges and Fields," lays the foundation for more advanced topics. This chapter delves into the fundamental concepts of electric charges, Coulomb’s law, electric field, and Gauss's law. Understanding these concepts thoroughly is vital for students to grasp subsequent chapters effectively.

This article provides a comprehensive guide to the important questions from Chapter 1 "Electric Charges and Fields," offering detailed solutions and explanations to enhance understanding and aid in exam preparation.

1. What is Electric Charge?

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. There are two types of electric charges: positive and negative. Like charges repel each other, while opposite charges attract. The unit of electric charge is the coulomb (C).

Important Questions:

1. Define electric charge and state its SI unit.

   Solution: Electric charge is a physical property of matter that causes it to experience a force when placed in an electric field. The SI unit of electric charge is the coulomb (C).

2. Explain the concept of quantization of charge.

   Solution: The quantization of charge states that the charge on any object is always an integer multiple of the fundamental charge, e. The fundamental charge (e) is the charge of an electron, approximately equal to $1.6 \times 10^{-19}$ coulombs. Mathematically, it can be expressed as $Q = n \times e$, where $n$ is an integer.

2. Coulomb’s Law

Coulomb’s law describes the force between two point charges. According to this law, the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Important Questions:

1. State Coulomb’s law and derive the formula for the electrostatic force between two point charges.

   Solution: Coulomb’s law states that the magnitude of the electrostatic force (F) between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by:

   $$F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}$$

   where $k$ is Coulomb’s constant ($k \approx 8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2}$).

2. Two charges of +3 μC and -2 μC are placed 5 meters apart. Calculate the force between them.

   Solution: Using Coulomb’s law formula:

   $$F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}$$

   where $q_1 = 3 \times 10^{-6} \text{C}$, $q_2 = -2 \times 10^{-6} \text{C}$, $r = 5 \text{m}$, and $k = 8.99 \times 10^9 \text{N m}^2 \text{C}^{-2}$:

   $$F = \frac{8.99 \times 10^9 \times |3 \times 10^{-6} \times -2 \times 10^{-6}|}{5^2} = \frac{8.99 \times 10^9 \times 6 \times 10^{-12}}{25} \approx 2.16 \text{N}$$

3. Electric Field

The electric field ($E$) at a point in space is defined as the force ($F$) experienced by a unit positive charge placed at that point. The electric field is a vector quantity and is represented by:

$$E = \frac{F}{q}$$

Important Questions:

1. Define the electric field and derive its expression for a point charge.

   Solution: The electric field at a point due to a point charge $Q$ is given by:

   $$E = \frac{k \cdot |Q|}{r^2}$$

   where $k$ is Coulomb’s constant and $r$ is the distance from the charge to the point where the field is being calculated.

2. Calculate the electric field at a distance of 3 meters from a charge of $+5 \, \mu C$.

   Solution: Using the formula:

   $$E = \frac{k \cdot |Q|}{r^2}$$

   where $Q = 5 \times 10^{-6} \text{C}$, $r = 3 \text{m}$, and $k = 8.99 \times 10^9 \text{N m}^2 \text{C}^{-2}$:

   $$E = \frac{8.99 \times 10^9 \times 5 \times 10^{-6}}{3^2} = \frac{44.95 \times 10^3}{9} \approx 5.0 \times 10^3 \text{N/C}$$

4. Electric Field Lines

Electric field lines provide a visual representation of the electric field. They show the direction and strength of the electric field. The density of the lines represents the field's strength.

Important Questions:

1. What do electric field lines represent and what are the basic properties of these lines?

   Solution: Electric field lines represent the direction and strength of the electric field. The basic properties are:

   • Lines start from positive charges and end at negative charges.
   • The density of the lines represents the strength of the field.
   • Field lines never cross each other.
   • Lines are perpendicular to the surface of conductors.

2. Draw and explain the electric field lines around a positive and a negative charge.

   Solution:

   • Positive Charge: Field lines radiate outward, perpendicular to the surface of the charge.
   • Negative Charge: Field lines converge inward, pointing towards the charge.

5. Gauss’s Law

Gauss’s Law relates the electric flux through a closed surface to the charge enclosed by the surface. It is given by:

$$\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$$

where $\Phi_E$ is the electric flux, $Q_{\text{enc}}$ is the enclosed charge, and $\epsilon_0$ is the permittivity of free space ($\epsilon_0 \approx 8.85 \times 10^{-12} \text{C}^2 \text{N}^{-1} \text{m}^{-2}$).

Important Questions:

1. State Gauss’s Law and derive its application for a spherical charge distribution.

   Solution: Gauss’s Law states that the electric flux through a closed surface is proportional to the charge enclosed by the surface:

   $$\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$$

   For a spherical charge distribution, the electric field $E$ at a distance $r$ from the center of a sphere of radius $R$ and charge $Q$ is:

   • For $r > R$:

     $$E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$$

   • For $r < R$, if charge is uniformly distributed:

     $$E = \frac{1}{4 \pi \epsilon_0} \frac{Q r}{R^3}$$

2. A point charge of $+8 \text{μC}$ is placed at the center of a spherical Gaussian surface of radius 2 meters. Calculate the electric flux through the surface.

   Solution: Using Gauss’s Law:

   $$\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$$

   where $Q_{\text{enc}} = 8 \times 10^{-6} \text{C}$ and $\epsilon_0 = 8.85 \times 10^{-12} \text{C}^2 \text{N}^{-1} \text{m}^{-2}$:

   $$\Phi_E = \frac{8 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 9.04 \times 10^5 \text{N m}^2 \text{C}^{-1}$$

6. Potential Due to a Point Charge

Electric potential ($V$) at a point due to a point charge is the work done to bring a unit positive charge from infinity to that point. It is given by:

$$V = \frac{k \cdot Q}{r}$$

Important Questions:

1. Derive the expression for the electric potential due to a point charge.

   Solution: The electric potential $V$ at a distance $r$ from a point charge $Q$ is:

   $$V = \frac{k \cdot Q}{r}$$

2. Calculate the electric potential at a distance of 4 meters from a charge of $+2 \text{μC}$.

   Solution: Using the formula:

   $$V = \frac{k \cdot Q}{r}$$

   where $Q = 2 \times 10^{-6} \text{C}$, $r = 4 \text{m}$, and $k = 8.99 \times 10^9 \text{N m}^2 \text{C}^{-2}$:

   $$V = \frac{8.99 \times 10^9 \times 2 \times 10^{-6}}{4} = \frac{17.98 \times 10^3}{4} \approx 4.5 \times 10^3 \text{V}$$

7. Potential Difference and Electric Field

The potential difference ($V$) between two points in an electric field is the work done to move a unit positive charge from one point to the other. It is related to the electric field ($E$) by:

$$V = E \cdot d$$

where $d$ is the distance between the points.

Important Questions:

1. Derive the relationship between electric field and potential difference.

   Solution: The electric field $E$ is related to the potential difference $V$ by:

   $$E = - \frac{dV}{dr}$$

   This implies that:

   $$V = - \int E \, dr$$

2. If the electric field between two parallel plates is $1000 \text{V/m}$ and the distance between the plates is $0.5 \text{m}$, calculate the potential difference between them.

   Solution: Using the formula:

   $$V = E \cdot d$$

   where $E = 1000 \text{V/m}$ and $d = 0.5 \text{m}$:

   $$V = 1000 \times 0.5 = 500 \text{V}$$

Conclusion

Chapter 1 of Class 12 Physics, "Electric Charges and Fields," introduces fundamental concepts essential for understanding electrostatics. By mastering the solutions to these important questions, students can build a solid foundation in physics, preparing them for more complex topics and examinations. This article has provided a detailed explanation of key questions, offering clarity and practice opportunities for effective learning and exam preparation.