Rbse Class 12 Physics Chapter 1: Electric Charges and Fields Solution

Explore comprehensive solutions for RBSE Class 12 Physics Chapter 1: Electric Charges and Fields. Enhance your understanding with detailed explanations and examples,notes and important concepts

The first chapter in the RBSE Class 12 Physics syllabus, “Electric Charges and Fields,” lays the foundation for understanding electrostatics. This chapter covers crucial concepts related to electric charges, the behavior of electric fields, Coulomb’s law, electric dipoles, and Gauss’s law. Here, we provide solutions to key questions, derivations, and numerical problems in this chapter, along with explanations to support a deeper understanding.

Rbse Class 12 Physics Chapter 1: Electric Charges and Fields Solution

Chapter Outline: Electric Charges and Fields

Electric Charge: Understanding the fundamental property of matter that causes it to experience a force when placed in an electric field.
Coulomb’s Law: The law describing the force between two point charges.
Electric Field: The region around a charged particle where forces are exerted on other charges.
Electric Field Lines: Visual representation of electric fields around charges.
Electric Dipole: A system of two equal and opposite charges separated by a small distance.
Gauss’s Law: A powerful method to calculate electric fields, especially with symmetric charge distributions.

Class 12 physics book

Solutions for Important Questions from Chapter 1: Electric Charges and Fields

1. Define Electric Charge and State its Properties.

Solution: Electric charge is an intrinsic property of particles that causes them to experience a force when placed in an electric or magnetic field. Key properties of electric charge are:

Quantization: Electric charge exists in discrete amounts, which are integer multiples of the elementary charge, e=1.6×10−19e = 1.6 \times 10^{-19}e=1.6×10−19 C.
Conservation: The total charge in an isolated system remains constant.
Attraction and Repulsion: Like charges repel each other, while opposite charges attract.
Additivity: Electric charges are scalar and can be added algebraically.

2. State Coulomb’s Law and Write its Mathematical Form.

Solution: Coulomb’s Law states that the electrostatic force FFF between two point charges q1q_1q1 and q2q_2q2 separated by a distance rrr is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The law is given by:F=kq1q2r2F = k \frac{{q_1 q_2}}{{r^2}}F=kr2q1q2

where kkk is the electrostatic constant, k=8.99×109 Nm2/C2k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2k=8.99×109Nm2/C2.

3. What is an Electric Field, and How is it Calculated for a Point Charge?

Solution: The electric field EEE at a point in space is defined as the force FFF experienced by a unit positive charge placed at that point. Mathematically:E=Fq=kQr2E = \frac{F}{q} = k \frac{Q}{r^2}E=qF=kr2Q

where QQQ is the source charge creating the electric field, rrr is the distance from the charge to the point where EEE is measured, and kkk is Coulomb’s constant.

4. Explain Electric Field Lines and Their Properties.

Solution: Electric field lines provide a way to visualize the electric field in a region. Properties of electric field lines include:

They originate from positive charges and terminate on negative charges.
The density of field lines indicates the strength of the electric field.
They never cross each other.
Field lines are perpendicular to the surface of a conductor.

5. Numerical Problem: Calculate the Electric Force Between Two Charges of 5μC and -3μC Separated by a Distance of 2m.

Solution: Given:

q1=5×10−6 Cq_1 = 5 \times 10^{-6} \, \text{C}q1=5×10−6C
q2=−3×10−6 Cq_2 = -3 \times 10^{-6} \, \text{C}q2=−3×10−6C
r=2 mr = 2 \, \text{m}r=2m

Using Coulomb’s law:F=k∣q1⋅q2∣r2F = k \frac{{|q_1 \cdot q_2|}}{{r^2}}F=kr2∣q1⋅q2∣

Substituting the values:F=(8.99×109)(5×10−6)(3×10−6)22F = (8.99 \times 10^9) \frac{{(5 \times 10^{-6})(3 \times 10^{-6})}}{{2^2}}F=(8.99×109)22(5×10−6)(3×10−6) F=8.99×109×15×10−124F = 8.99 \times 10^9 \times \frac{{15 \times 10^{-12}}}{{4}}F=8.99×109×415×10−12 F=33.71 NF = 33.71 \, \text{N}F=33.71N

Therefore, the force between the charges is 33.71 N.

6. Define an Electric Dipole and Derive the Formula for the Electric Field Due to a Dipole at an Axial Point.

Solution: An electric dipole consists of two equal and opposite charges +q+q+q and −q-q−q separated by a distance 2a2a2a. The dipole moment ppp is defined as p=q×2ap = q \times 2ap=q×2a.

For an axial point at distance rrr from the center of the dipole:Eaxial=14πϵ02pr3E_{\text{axial}} = \frac{1}{4 \pi \epsilon_0} \frac{2p}{r^3}Eaxial=4πϵ01r32p

This shows the electric field at an axial point is directly proportional to the dipole moment and inversely proportional to the cube of the distance from the dipole.

7. State Gauss’s Law and Explain Its Importance in Calculating Electric Fields.

Solution: Gauss’s Law states that the net electric flux ΦE\Phi_EΦE through a closed surface is equal to the total charge enclosed QQQ divided by the permittivity of free space ϵ0\epsilon_0ϵ0:ΦE=∮E⋅dA=Qϵ0\Phi_E = \oint E \cdot dA = \frac{Q}{\epsilon_0}ΦE=∮E⋅dA=ϵ0Q

Gauss’s Law is especially useful for calculating electric fields of symmetric charge distributions, such as spherical, cylindrical, and planar distributions.

Preparation Tips for Chapter 1: Electric Charges and Fields

Master Coulomb’s Law: Practice questions related to force between point charges to understand the basic concept.
Focus on Derivations: Pay special attention to the derivations of electric fields for different configurations, including dipoles and Gaussian surfaces.
Practice Numericals: Regularly practice numerical problems based on electric forces and electric fields.
Visualize Field Lines: Understanding electric field lines will help in solving conceptual problems and visualizing electric fields.

FAQs

1. What is the SI unit of Electric Charge?

The SI unit of electric charge is the Coulomb (C).

2. What are Electric Field Lines?

Electric field lines represent the direction and relative strength of the electric field in space.

3. How does Gauss’s Law help in calculating Electric Fields?

Gauss’s Law simplifies electric field calculations for symmetric charge distributions by relating field and flux over a closed surface.

4. Is the RBSE Class 12 Physics book sufficient for understanding this chapter?

Yes, the RBSE Class 12 Physics book provides detailed explanations and examples that are adequate for understanding Electric Charges and Fields.

By thoroughly understanding the concepts and practicing the solutions provided here, students will build a strong foundation in electrostatics, an essential part of Class 12 Physics.