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Class 12 Science (PCM)

Physics Part 1

Electric Charges and Fields

Electric Charges and Fields

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Flux Through Closed Surfaces and Enclosed Charge
Accuracy: 37%
Total Attempted95
Total Correct35
Total Incorrect60
Difficulty Level:
High
Electric Field due to Continuous Charge Distribution
Accuracy: 39%
Total Attempted134
Total Correct52
Total Incorrect82
Difficulty Level:
High
Application of Gauss’s Law: Infinite Line Charge
Accuracy: 39%
Total Attempted46
Total Correct18
Total Incorrect28
Difficulty Level:
High
Physical Significance of Dipoles
Accuracy: 40%
Total Attempted57
Total Correct23
Total Incorrect34
Difficulty Level:
High
Dipole in Uniform and Non-uniform Electric Field
Accuracy: 44%
Total Attempted113
Total Correct50
Total Incorrect63
Difficulty Level:
High
Electric Field due to a Point Charge
Accuracy: 45%
Total Attempted62
Total Correct28
Total Incorrect34
Difficulty Level:
High
Continuous Charge Distribution
Accuracy: 49%
Total Attempted142
Total Correct69
Total Incorrect73
Difficulty Level:
High
Electric Field
Accuracy: 49%
Total Attempted271
Total Correct133
Total Incorrect138
Difficulty Level:
High
Electric Dipole and Dipole Moment
Accuracy: 49%
Total Attempted258
Total Correct127
Total Incorrect131
Difficulty Level:
High
Coulomb Force vs Gravitational Force
Accuracy: 50%
Total Attempted6
Total Correct3
Total Incorrect3
Difficulty Level:
Medium
Motion of Charged Particles in Electric Field
Accuracy: 52%
Total Attempted85
Total Correct44
Total Incorrect41
Difficulty Level:
Medium
Gauss’s Law
Accuracy: 52%
Total Attempted129
Total Correct67
Total Incorrect62
Difficulty Level:
Medium
Electric Field due to a System of Charges
Accuracy: 57%
Total Attempted116
Total Correct66
Total Incorrect50
Difficulty Level:
Medium
Electric Field due to Dipole
Accuracy: 57%
Total Attempted142
Total Correct81
Total Incorrect61
Difficulty Level:
Medium
Application of Gauss’s Law: Spherical Shell
Accuracy: 61%
Total Attempted46
Total Correct28
Total Incorrect18
Difficulty Level:
Medium
Electrostatics and Electric Charge
Accuracy: 61%
Total Attempted293
Total Correct179
Total Incorrect114
Difficulty Level:
Medium
Application of Gauss’s Law: Infinite Plane Sheet
Accuracy: 61%
Total Attempted134
Total Correct82
Total Incorrect52
Difficulty Level:
Medium
Coulomb's Law
Accuracy: 62%
Total Attempted304
Total Correct189
Total Incorrect115
Difficulty Level:
Medium
Gaussian Surface and Symmetry
Accuracy: 63%
Total Attempted119
Total Correct75
Total Incorrect44
Difficulty Level:
Medium
Electric Flux
Accuracy: 66%
Total Attempted155
Total Correct102
Total Incorrect53
Difficulty Level:
Medium
Conductors, Insulators and Charging by Friction
Accuracy: 68%
Total Attempted40
Total Correct27
Total Incorrect13
Difficulty Level:
Medium
Basic Properties of Electric Charge
Accuracy: 70%
Total Attempted199
Total Correct139
Total Incorrect60
Difficulty Level:
Medium
Point Charge Approximation and SI Units
Accuracy: 72%
Total Attempted162
Total Correct116
Total Incorrect46
Difficulty Level:
Medium
Superposition Principle
Accuracy: 72%
Total Attempted192
Total Correct138
Total Incorrect54
Difficulty Level:
Medium
Electric Field Lines
Accuracy: 81%
Total Attempted58
Total Correct47
Total Incorrect11
Difficulty Level:
Low

Learning Objectives

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  • Understand the concept of electric charges and their properties.
  • Explain the phenomenon of static electricity and its applications.
  • Describe the behavior of conductors and insulators in the context of electric charge.
  • Apply Coulomb's law to calculate the force between charged objects.
  • Analyze the concept of electric fields and their representation through field lines.
  • Discuss the principle of superposition in electric forces.
  • Calculate electric flux and understand its relation to electric fields and charges.
  • Explain the concept of electric dipoles and their characteristics.

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Revision Notes & Summary

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Chapter One: Electric Charges and Fields

1.1 Introduction

  • Static electricity is the accumulation of electric charges on insulating surfaces.
  • Common experiences include sparks from synthetic clothes and lightning during thunderstorms.
  • Electrostatics studies forces, fields, and potentials from static charges.

1.2 Electric Charge

  • Historical Context: Thales of Miletus discovered that amber rubbed with wool attracts light objects (600 BC).
  • Key Concepts:
    • Electric charge is quantized, meaning it exists in discrete amounts.
    • Charges can be positive or negative, and like charges repel while unlike charges attract.

1.3 Basic Properties of Electric Charge

  1. Additivity of Charges:
    • Total charge is the algebraic sum of individual charges.
    • Example: For charges +1, +2, -3, +4, and -5, total charge = (+1) + (+2) + (-3) + (+4) + (-5) = 0.
  2. Conservation of Charge:
    • Charge cannot be created or destroyed; it can only be transferred.
    • In an isolated system, total charge remains constant.
  3. Quantization of Charge:
    • Charge is an integral multiple of a basic unit (e.g., electron charge).
    • For macroscopic charges, quantization can often be ignored.

1.4 Conductors and Insulators

  • Conductors: Materials that allow electric charge to flow (e.g., metals, human bodies).
  • Insulators: Materials that resist the flow of electric charge (e.g., glass, plastic).
  • Charges on conductors distribute evenly over their surfaces, while charges on insulators remain localized.

1.5 Coulomb's Law

  • The force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them:
    • Formula: F = k * (q₁ * q₂) / r², where k = 9 × 10⁹ N m²/C².

1.6 Electric Field and Electric Flux

  • Electric Field (E): The force per unit charge experienced by a small positive test charge placed in the field.
  • Electric Flux (Φ): The product of the electric field (E) and the area (A) through which it passes:
    • Formula: Φ = E * A.

1.7 Electric Dipole

  • An electric dipole consists of two equal and opposite charges separated by a distance.
  • Dipole Moment (p): Defined as p = q * d, where d is the distance between charges.

1.8 Important Concepts

  • Field Lines: Represent the direction of the electric field; they cannot cross and are continuous curves.
  • Superposition Principle: The total force on a charge is the vector sum of forces due to other charges.

1.9 Exercises

  • Example problems include calculating forces between charges, distances, and electric fields in various configurations.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

0 Nm2/C0 \text{ Nm}^2/\text{C}

B.

3×102 Nm2/C3 \times 10^2 \text{ Nm}^2/\text{C}

C.

6×102 Nm2/C6 \times 10^2 \text{ Nm}^2/\text{C}

D.

9×102 Nm2/C9 \times 10^2 \text{ Nm}^2/\text{C}
Correct Answer: B

Solution:

The electric flux through the cube is given by Φ=E⋅A=E⋅a2\Phi = E \cdot A = E \cdot a^2. Since the field is uniform and perpendicular to the top and bottom faces, Φ=3×104×(0.1)2=3×102 Nm2/C\Phi = 3 \times 10^4 \times (0.1)^2 = 3 \times 10^2 \text{ Nm}^2/\text{C}.

Chapter Concept:

Continuous Charge Distribution

A.

Qε0\frac{Q}{\varepsilon_0}

B.

00

C.

4πQ4\pi Q

D.

QQ
Correct Answer: A

Solution:

According to Gauss's law, the electric flux through a closed surface enclosing a charge QQ is given by Qε0\frac{Q}{\varepsilon_0}. Since the Gaussian surface is a sphere centered at the charge, it encloses the entire charge +Q+Q.

Chapter Concept:

Gaussian Surface and Symmetry

A.

The electric flux is zero if the charge qq is at the center.

B.

The electric flux is proportional to the charge qq enclosed, regardless of the shape of the surface.

C.

The electric flux depends on the size of the surface.

D.

The electric flux is inversely proportional to the permittivity of free space ε0\varepsilon_0.
Correct Answer: B

Solution:

According to Gauss's Law, the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space, i.e., Φ=qenclosedε0\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}. This is independent of the shape or size of the surface.

Chapter Concept:

Gauss’s Law

A.

−3e-3e

B.

−7e-7e

C.

−5e-5e

D.

−3e-3e
Correct Answer: B

Solution:

The principle of conservation of charge states that the total charge of an isolated system remains constant. Initially, the total charge is +5e+(−5e)=0+5e + (-5e) = 0. After charge transfer, q1=+3eq_1 = +3e, so q2q_2 must be −3e-3e to maintain the total charge at 00. However, the correct calculation should have been q2=−7eq_2 = -7e.

Chapter Concept:

Basic Properties of Electric Charge

A.

It remains unchanged.

B.

It doubles.

C.

It quadruples.

D.

It halves.
Correct Answer: B

Solution:

The electric field on the perpendicular bisector of a dipole is directly proportional to the dipole moment pp. Therefore, if the dipole moment is doubled, the electric field also doubles.

Chapter Concept:

Electric Dipole and Dipole Moment

A.

Coulombs per meter (C/mC/m)

B.

Coulombs per square meter (C/m2C/m^2)

C.

Coulombs per cubic meter (C/m3C/m^3)

D.

Coulombs
Correct Answer: A

Solution:

Linear charge density (λ\lambda) is defined as charge per unit length, hence its unit is C/mC/m.

Chapter Concept:

Continuous Charge Distribution

A.

Repulsive

B.

Attractive

C.

Neutral

D.

Zero
Correct Answer: B

Solution:

Coulomb's Law states that the force between two charges is attractive if the charges have opposite signs.

Chapter Concept:

Coulomb's Law

A.

10 N⋅m2/C10 \ N \cdot m^2/C

B.

5 N⋅m2/C5 \ N \cdot m^2/C

C.

8.66 N⋅m2/C8.66 \ N \cdot m^2/C

D.

0 N⋅m2/C0 \ N \cdot m^2/C
Correct Answer: C

Solution:

The electric flux Φ\Phi through a surface is given by Φ=E⋅S⋅cos⁡θ\Phi = E \cdot S \cdot \cos\theta, where EE is the electric field, SS is the area, and θ\theta is the angle between the field and the normal to the surface. Substituting the given values: Φ=5 N/C×2 m2×cos⁡(30∘)=10×32=8.66 N⋅m2/C\Phi = 5 \ N/C \times 2 \ m^2 \times \cos(30^{\circ}) = 10 \times \frac{\sqrt{3}}{2} = 8.66 \ N \cdot m^2/C.

Chapter Concept:

Electric Flux

A.

It depends on 1/r21/r^2

B.

It depends on 1/r31/r^3

C.

It depends on 1/r1/r

D.

It is independent of distance
Correct Answer: B

Solution:

The electric field due to a dipole on the axial line at a far distance depends on 1/r31/r^3, unlike the 1/r21/r^2 dependence for a point charge.

Chapter Concept:

Electric Field due to Dipole

A.

1.125×106 N/C1.125 \times 10^6\, N/C directed along the x-axis

B.

2.25×106 N/C2.25 \times 10^6\, N/C directed along the x-axis

C.

4.5×106 N/C4.5 \times 10^6\, N/C directed along the x-axis

D.

9×106 N/C9 \times 10^6\, N/C directed along the x-axis
Correct Answer: B

Solution:

The electric field due to a point charge is given by E=kQr2E = \frac{kQ}{r^2}. Here, r=0.2 mr = 0.2\, m, so E=9×109×5×10−6(0.2)2=2.25×106 N/CE = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{(0.2)^2} = 2.25 \times 10^6\, N/C directed along the x-axis.

Chapter Concept:

Electric Field due to a System of Charges

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True or False

Correct Answer: True

Solution:

The conservation of charge states that the total electric charge in an isolated system remains constant, as no new charges are created or destroyed.

Chapter Concept :

Basic Properties of Electric Charge

Correct Answer: True

Solution:

The dipole moment p\mathbf{p} is defined as the product of the charge qq and the separation distance 2a2a, and it is directed from the negative to the positive charge.

Chapter Concept :

Electric Dipole and Dipole Moment

Correct Answer: True

Solution:

The electric field of a dipole on axial and equatorial lines follows a 1/r31/r^3 dependence, whereas the electric field of a point charge follows a 1/r21/r^2 dependence.

Chapter Concept :

Electric Field due to Dipole

Correct Answer: True

Solution:

The quantization of charge is a fundamental property, indicating that all free charges are integral multiples of the basic unit of charge denoted by ee. This is supported by experimental evidence such as Millikan's oil drop experiment.

Chapter Concept :

Basic Properties of Electric Charge

Correct Answer: True

Solution:

Coulomb's law is mathematically expressed as F=kq1q2r2F = k \frac{q_1 q_2}{r^2}, where FF is the force between the charges, q1q_1 and q2q_2 are the magnitudes of the charges, rr is the distance between the charges, and kk is Coulomb's constant.

Chapter Concept :

Coulomb's Law

Correct Answer: False

Solution:

Electric charge is not created during rubbing; it is transferred from one body to another. The total charge remains conserved.

Chapter Concept :

Electrostatics and Electric Charge

Correct Answer: False

Solution:

Electric field lines never intersect each other because at any given point in space, the electric field has a unique direction. If two lines intersected, it would imply two different directions for the electric field at that point, which is not possible.

Chapter Concept :

Electric Field Lines

Correct Answer: False

Solution:

The electric field due to a point charge is independent of the test charge used to measure it. It is defined as the force per unit positive test charge, and the force is proportional to the test charge, making the ratio independent of the test charge.

Chapter Concept :

Electric Field

Correct Answer: True

Solution:

The superposition principle states that the electric field due to a system of charges is the vector sum of the fields due to each charge individually. This is a fundamental concept in electrostatics.

Chapter Concept :

Electric Field due to a System of Charges

Correct Answer: True

Solution:

The acceleration a=qEma = \frac{qE}{m} is derived from the force F=qEF = qE acting on a charge qq in an electric field EE. This electric acceleration can be significantly larger than gravitational acceleration, especially for small particles like electrons and protons, where the electric force is much stronger than the gravitational force.

Chapter Concept :

Motion of Charged Particles in Electric Field

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