A dipole with dipole moment 3 x 10^-9 C m is placed in a uniform electric field of magnitude 4 x 10^4 N/C at an angle of 45° with the field. What is the torque acting on the dipole?

Correct Option: B. Solution: Torque τ = pE sin θ = 3 x 10^-9 * 4 x 10^4 * sin(45°) = 6.0 x 10^-5 N m.

Electric Charges and Fields

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Summary

Chapter One: Electric Charges and Fields

Summary

Introduction to Electrostatics: Study of forces, fields, and potentials from static charges.
Electric Charge: Discovered by Thales; involves attraction of light objects by rubbed amber.
Static Electricity: Accumulation of electric charges due to friction.
Basic Properties of Electric Charge:
- Quantisation: Total charge is an integral multiple of a basic charge (e.g., electron).
- Additivity: Total charge is the algebraic sum of individual charges.
- Conservation: Total charge in an isolated system remains constant.
Coulomb's Law: Force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
Electric Field: Force per unit charge experienced by a small positive test charge.
Electric Field Lines: Visual representation of electric fields; they cannot cross and are continuous.
Electric Dipole: Pair of equal and opposite charges separated by a distance, characterized by a dipole moment.

Key Formulas and Definitions

Physical Quantity	Symbol	Dimensions	Unit	Remarks
Electric field	E	[MLT⁻³A⁻¹]	V m⁻¹
Electric flux	Φ	[ML³ T⁻³A⁻¹]	Vm	Φ = E.AS
Dipole moment	p	[LTA]	Cm	Vector directed from negative to positive charge
Surface charge density	σ	[L⁻² TA]	C m⁻²	Charge per unit area
Linear charge density	a	[L⁻¹ TA]	C m⁻¹	Charge per unit length
Volume charge density	ρ	[L⁻³ TA]	C m⁻³	Charge per unit volume

Learning Objectives

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.

Detailed Notes

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

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.
Conservation of Charge:
- Charge cannot be created or destroyed; it can only be transferred.
- In an isolated system, total charge remains constant.
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.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Electric Charge Conservation: Students often forget that electric charge cannot be created or destroyed. When charging by rubbing, one body gains electrons while the other loses them, maintaining the total charge.
Ignoring Charge Quantization: Many overlook that electric charge is quantized and should be treated as integral multiples of the elementary charge (e.g., electrons). This is often ignored in macroscopic scenarios but is crucial in theoretical discussions.
Incorrect Application of Coulomb's Law: Students may misapply Coulomb's Law by not considering the correct signs of charges, leading to errors in calculating forces.
Field Line Misinterpretation: It's common to confuse the direction and behavior of electric field lines. Remember that field lines start from positive charges and end at negative charges, and they cannot cross each other.

Exam Tips

Always Check Units: When solving problems, ensure that all units are consistent, especially when dealing with electric fields and forces.
Draw Diagrams: Visual aids can help clarify problems involving multiple charges or electric fields. Sketching can assist in understanding the relationships between charges and their effects.
Review Key Concepts: Make sure to understand the principles of superposition and how to apply them to multiple charge systems.
Practice with Different Scenarios: Work through various problems involving conductors and insulators, as well as different configurations of charges to solidify understanding.

Practice & Assessment

Multiple Choice Questions

The glass rod becomes positively charged and the silk cloth becomes negatively charged.

The glass rod becomes negatively charged and the silk cloth becomes positively charged.

Both the glass rod and the silk cloth become positively charged.

Both the glass rod and the silk cloth become negatively charged.

Correct Answer: A

Solution:

When a glass rod is rubbed with a silk cloth, electrons are transferred from the glass rod to the silk, resulting in the glass rod becoming positively charged and the silk cloth becoming negatively charged.

0.2 N

0.02 N

0.002 N

2 N

Correct Answer: A

Solution:

The force

F

experienced by a charge

q

in an electric field

E

is given by

F = qE

. Substituting the given values,

F = (1 \times 10^{-6} \text{ C})(2 \times 10^5 \text{ N/C}) = 0.2 \text{ N}

8.5 x 10^-5 N m

6.0 x 10^-5 N m

1.2 x 10^-5 N m

1.5 x 10^-5 N m

Correct Answer: B

Solution:

Torque τ = pE sin θ = 3 x 10^-9 * 4 x 10^4 * sin(45°) = 6.0 x 10^-5 N m.

Uniform and non-zero

Zero

Varies with distance from the center

Infinite

Correct Answer: B

Solution:

Inside a uniformly charged conducting sphere, the electric field is zero due to the symmetry and distribution of charges.

1.875 x 10¹²

3.75 x 10¹²

6.25 x 10¹²

9.375 x 10¹²

Correct Answer: A

Solution:

The charge of one electron is approximately 1.6 x 10⁻¹⁹ C. The number of electrons transferred is

\frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.875 \times 10^{12}

The paper is repelled by the comb.

The paper remains stationary.

The paper is attracted to the comb.

The paper discharges the comb.

Correct Answer: C

Solution:

The charged comb polarises the piece of paper, inducing a net dipole moment, and the paper is attracted to the comb.

The sum of the charges.

The product of the charges.

The difference of the charges.

The square of the distance between the charges.

Correct Answer: B

Solution:

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

0 N/C

3.75 x 10^4 N/C

7.5 x 10^4 N/C

1.5 x 10^5 N/C

Correct Answer: A

Solution:

Since the point is inside the conductor, the electric field is 0 N/C.

1.44 µC

14.4 µC

144 µC

1440 µC

Correct Answer: C

Solution:

The surface area of the sphere is given by

A = 4\pi r^2

. The radius

r

is half of the diameter, so

r = 1.2

m. The charge

Q

is given by

Q = \sigma A = \sigma \times 4\pi r^2 = 80.0 \times 4\pi (1.2)^2 = 144 \mu C

The leaves diverge due to the flow of charge onto them.

The leaves converge due to the flow of charge onto them.

The leaves remain stationary.

The leaves fall off.

Correct Answer: A

Solution:

When a charged object touches the metal knob, charge flows onto the leaves and they diverge.

Coulombs per square meter (C/m²)

Coulombs per cubic meter (C/m³)

Coulombs per meter (C/m)

Coulombs (C)

Correct Answer: C

Solution:

The unit of linear charge density is Coulombs per meter (C/m).

qEa

\sqrt{2} qEa

qEa/\sqrt{2}

2qEa

Correct Answer: B

Solution:

The torque

\tau

on a dipole in an electric field is given by

\tau = pE \sin \theta

, where

p = 2qa

is the dipole moment. Thus,

\tau = 2qaE \sin 45^\circ = \sqrt{2} qEa

\frac{k \cdot q}{r^2}

\frac{k \cdot q}{r}

k \cdot q \cdot r^2

k \cdot q \cdot r

Correct Answer: A

Solution:

The electric field

E

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.

It becomes negatively charged

It becomes positively charged

It remains neutral

It gains mass

Correct Answer: A

Solution:

When the glass rod is rubbed with silk, electrons are transferred from the glass to the silk, leaving the glass rod positively charged and the silk negatively charged, consistent with the conservation of charge.

Ampere

Coulomb

Newton

Volt

Correct Answer: B

Solution:

The unit of electric charge in the SI system is the Coulomb, which is defined as the amount of charge transferred by a current of one ampere in one second.

The force is halved.

The force remains the same.

The force is doubled.

The force is quartered.

Correct Answer: D

Solution:

According to Coulomb's law, the force is inversely proportional to the square of the distance. Doubling the distance reduces the force to one-fourth.

0 N/C

1.2 x 10^5 N/C

2.4 x 10^5 N/C

4.8 x 10^5 N/C

Correct Answer: B

Solution:

For a uniformly charged rod, the electric field at a point along its perpendicular bisector is given by

E = \frac{k \lambda L}{r \sqrt{L^2 + 4r^2}}

, where

\lambda = \frac{Q}{L}

is the linear charge density. Substituting the given values,

E = \frac{9 \times 10^9 N \cdot m^2/C^2 \times 20 \times 10^{-6} C/m \times 0.5 m}{0.3 m \sqrt{0.5^2 + 4 \times 0.3^2}} = 1.2 \times 10^5 N/C

1.125 N

2.25 N

4.5 N

9.0 N

Correct Answer: A

Solution:

The electrostatic force between two point charges is given by Coulomb's law:

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

where

k = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}

q_1 = 5 \times 10^{-6}

q_2 = -5 \times 10^{-6}

C, and

r = 0.4

m. Substituting these values,

F = \frac{9 \times 10^9 \times 5 \times 10^{-6} \times 5 \times 10^{-6}}{(0.4)^2} = 1.125 \text{ N}.

0.06 N

0.18 N

0.36 N

0.72 N

Correct Answer: A

Solution:

The force between two point charges is given by Coulomb's law:

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

, where

k = 9 \times 10^9 \text{ N m}^2/\text{C}^2

and

r = 0.3 \text{ m}

. Substituting the given values,

F = \frac{9 \times 10^9 \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^2} = 0.06 \text{ N}

The total charge of an isolated system remains unchanged over time.

Electric charge can be created or destroyed in an isolated system.

The total charge of a system can change if it is isolated.

Electric charge is not conserved in any system.

Correct Answer: A

Solution:

Conservation of electric charge means that the total charge of an isolated system remains unchanged with time.

It defines the unit of charge.

It is the gravitational constant.

It is a proportionality constant related to electric force.

It is used to measure magnetic fields.

Correct Answer: C

Solution:

In Coulomb's law,

k

is a proportionality constant that relates to the electric force between two charges.

C/m

C/m²

C/m³

Correct Answer: B

Solution:

The unit of surface charge density is C/m², as it is defined as charge per unit area.

0 N/C

1.92 x 10⁻¹⁰ N/C

3.84 x 10⁻¹⁰ N/C

4.80 x 10⁻¹⁰ N/C

Correct Answer: B

Solution:

The electric field between two parallel plates with surface charge densities

\sigma

is given by

E = \frac{\sigma}{\varepsilon_0}

. Here,

\sigma = 17.0 \times 10^{-22}

C/m² and

\varepsilon_0 = 8.854 \times 10^{-12}

C²/Nm². Thus,

E = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} = 1.92 \times 10^{-10}

N/C.

Zero everywhere inside the sphere

Uniform and non-zero inside the sphere

Varies linearly with distance from the center

Depends on the charge of the sphere

Correct Answer: A

Solution:

According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero. Since the sphere is metallic and hollow, the electric field inside it remains zero irrespective of the charge placed at the center.

8.99 N, attractive

8.99 N, repulsive

0 N

4.5 N, attractive

Correct Answer: A

Solution:

Using Coulomb's law,

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

, where

k = 9 \times 10^9 \text{ N m}^2/\text{C}^2

q_1 = 5 \times 10^{-6} \text{ C}

q_2 = -5 \times 10^{-6} \text{ C}

, and

r = 0.1 \text{ m}

. The force is attractive due to opposite charges.

Electric charge is a vector quantity.

Electric charge is quantised.

Electric charge can be fractional.

Electric charge is not conserved.

Correct Answer: B

Solution:

Electric charge is quantised, meaning it exists in discrete amounts as integer multiples of the elementary charge.

1.92 x 10⁶ Nm²/C

2.88 x 10⁶ Nm²/C

3.84 x 10⁶ Nm²/C

4.80 x 10⁶ Nm²/C

Correct Answer: A

Solution:

The total charge on the sphere is given by

Q = \sigma \cdot 4 \pi r^2

, where

\sigma = 80.0 \times 10^{-6}

C/m² and

r = 1.2

m. Thus,

Q = 80.0 \times 10^{-6} \times 4 \pi (1.2)^2 = 1.92 \times 10^{-4}

C. The electric flux is

\Phi = \frac{Q}{\varepsilon_0}

, where

\varepsilon_0 = 8.854 \times 10^{-12}

C²/Nm². Therefore,

\Phi = \frac{1.92 \times 10^{-4}}{8.854 \times 10^{-12}} = 1.92 \times 10^6

Nm²/C.

1.27 \times 10^{-5} \text{ C/m}^2

3.18 \times 10^{-5} \text{ C/m}^2

6.37 \times 10^{-5} \text{ C/m}^2

1.59 \times 10^{-5} \text{ C/m}^2

Correct Answer: A

Solution:

Surface charge density

\sigma

is given by

\sigma = \frac{Q}{4\pi r^2}

, where

Q

is the total charge and

r

is the radius. Thus,

\sigma = \frac{10^{-5}}{4\pi (0.5)^2} = 1.27 \times 10^{-5} \text{ C/m}^2

The flux remains unchanged.

The flux doubles.

The flux halves.

The flux quadruples.

Correct Answer: A

Solution:

According to Gauss's law, the electric flux through a closed surface is given by the charge enclosed divided by the permittivity of free space, i.e.,

\Phi = \frac{Q}{\varepsilon_0}

. Since the charge is doubled and the radius of the sphere does not affect the flux, the flux remains unchanged.

5.76 N

2.88 N

7.2 N

3.6 N

Correct Answer: A

Solution:

Using Coulomb's law, the force

F

is given by:

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

where

k = 9 \times 10^9 \text{ N m}^2/\text{C}^2

q_1 = 4 \times 10^{-7} \text{ C}

q_2 = -4 \times 10^{-7} \text{ C}

, and

r = 0.5 \text{ m}

F = 9 \times 10^9 \times \frac{(4 \times 10^{-7})(4 \times 10^{-7})}{(0.5)^2} = 5.76 \text{ N}

They can form closed loops.

They start at negative charges and end at positive charges.

They never cross each other.

They are discontinuous curves.

Correct Answer: C

Solution:

Electric field lines never cross each other. They start at positive charges and end at negative charges, and they are continuous curves.

\sigma = \frac{Q}{V}

\sigma = \frac{Q}{S}

\sigma = \frac{Q}{l}

\sigma = \frac{Q}{m}

Correct Answer: B

Solution:

Surface charge density (

\sigma

) is defined as the charge (

Q

) per unit area (

S

) on the surface of a conductor.

1.45 µC

1.81 µC

2.30 µC

3.62 µC

Correct Answer: B

Solution:

The surface area of a sphere is given by

A = 4\pi r^2

. The radius

r

is half of the diameter, so

r = 1.2

m. Thus,

A = 4\pi (1.2)^2 = 18.1 \text{ m}^2

. The total charge

Q

is the product of the surface charge density

\sigma

and the surface area

A

Q = \sigma A = 80.0 \times 10^{-6} \times 18.1 = 1.448 \text{ µC}

. Rounding to two decimal places,

Q = 1.45 \text{ µC}

Conservation of charge

Quantisation of charge

Superposition principle

Additivity of charges

Correct Answer: A

Solution:

A gold-leaf electroscope operates on the principle of conservation of charge. When a charged object touches the metal knob, the charge is transferred to the leaves, causing them to diverge.

Zero

Non-zero and directed towards

q_A

Non-zero and directed towards

q_B

Non-zero and directed perpendicular to the line joining

q_A

and

q_B

Correct Answer: A

Solution:

At the midpoint, the electric fields due to

q_A

and

q_B

are equal in magnitude but opposite in direction, thus cancelling each other out, resulting in a net electric field of zero.

Zero

80.0 µC/m²

9 x 10⁴ N/C

Depends on the charge of the particle

Correct Answer: A

Solution:

According to Gauss's Law, the electric field inside a uniformly charged spherical shell is zero, regardless of the charge at the center.

They can cross each other.

They form closed loops.

They start at positive charges and end at negative charges.

They are discontinuous curves.

Correct Answer: C

Solution:

Electric field lines start at positive charges and end at negative charges. They cannot cross each other and are continuous curves.

7.2 x 10^4 N/C

3.6 x 10^4 N/C

1.8 x 10^4 N/C

9.0 x 10^4 N/C

Correct Answer: A

Solution:

The electric field due to an infinite line charge is E = (2kλ)/r, where r is the perpendicular distance. Substituting values, E = (2 * 9 x 10^9 * 2 x 10^-6) / 0.5 = 7.2 x 10^4 N/C.

1.87 x 10¹²

2.08 x 10¹²

3.75 x 10¹²

4.80 x 10¹²

Correct Answer: A

Solution:

The charge of one electron is approximately

1.6 \times 10^{-19}

C. The number of electrons transferred is

n = \frac{Q}{e} = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.87 \times 10^{12}

electrons.

5 x 10^-6 N

8 x 10^-6 N

2 x 10^-6 N

10 x 10^-6 N

Correct Answer: D

Solution:

The force on a charge in an electric field is given by

\mathbf{F} = q \mathbf{E}

. Substituting the given values,

\mathbf{F} = 1 \times 10^{-6} C \times (5 \times 1^2 \hat{i} + 3 \times 1 \hat{j}) N/C = (5 \hat{i} + 3 \hat{j}) \times 10^{-6} N

. The magnitude of the force is

\sqrt{(5 \times 10^{-6})^2 + (3 \times 10^{-6})^2} = 10 \times 10^{-6} N

Zero

Non-zero, directed towards the positive charge

Non-zero, directed towards the negative charge

Non-zero, perpendicular to the line joining the charges

Correct Answer: C

Solution:

The electric field due to a positive charge is directed away from the charge, while for a negative charge, it is directed towards the charge. At the midpoint, the fields due to both charges add up in the direction towards the negative charge, resulting in a non-zero field directed towards the negative charge.

0.216 N, attractive

0.216 N, repulsive

0.108 N, attractive

0.108 N, repulsive

Correct Answer: A

Solution:

Using Coulomb's law:

F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} = \frac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{0.5^2} = 0.216

N. The force is attractive because the charges are opposite.

They do not allow electric charge to move through them.

They have electric charges that are free to move inside the material.

They are always negatively charged.

They can only be charged by induction.

Correct Answer: B

Solution:

Conductors have electric charges (electrons) that are comparatively free to move inside the material, allowing electricity to pass through them easily.

0 N

0.9 N

1.8 N

3.6 N

Correct Answer: A

Solution:

The forces due to charges at opposite corners of the square cancel each other out because they are equal in magnitude and opposite in direction. Thus, the net force on the charge at the center is 0 N.

7.5 x 10⁻⁸ C m

1.25 x 10⁻⁸ C m

2.5 x 10⁻⁸ C m

5.0 x 10⁻⁸ C m

Correct Answer: D

Solution:

The dipole moment

p

is given by

p = q \cdot d

, where

q = 2.5 \times 10^{-7}

C and

d = 0.3

m (the distance between the charges). Thus,

p = 2.5 \times 10^{-7} \times 0.3 = 7.5 \times 10^{-8}

C m.

0.06 N

0.18 N

0.09 N

0.12 N

Correct Answer: C

Solution:

Using Coulomb's law, the force is calculated as

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

where

k = 9 \times 10^9 \text{N m}^2 \text{C}^{-2}

q_1 = 2 \times 10^{-7} \text{C}

q_2 = 3 \times 10^{-7} \text{C}

, and

r = 0.3 \text{m}

. Substituting these values gives

F = 0.09 \text{N}

0 N/C

9 x 10^5 N/C towards the negative charge

9 x 10^5 N/C away from the negative charge

4.5 x 10^5 N/C towards the negative charge

Correct Answer: B

Solution:

The electric field due to a point charge is given by

E = \frac{kq}{r^2}

. At the midpoint, the fields due to the two charges add up. The field due to the positive charge is

9 x 10^5 N/C

towards the negative charge, and the field due to the negative charge is also

9 x 10^5 N/C

towards the negative charge. Thus, the total field is

9 x 10^5 N/C

towards the negative charge.

Electric charge can be created but not destroyed.

Electric charge can be destroyed but not created.

Electric charge can neither be created nor destroyed.

Electric charge is always conserved only in closed systems.

Correct Answer: C

Solution:

The conservation of electric charge states that the total charge of an isolated system remains constant over time, meaning charge can neither be created nor destroyed.

5 \times 10^{-6} C

5 \times 10^{-5} C

5 \times 10^{-4} C

5 \times 10^{-3} C

Correct Answer: B

Solution:

The total charge

Q

on the rod is given by

Q = \lambda \cdot L

, where

L

is the length of the rod. Thus,

Q = 5 \times 10^{-6} \text{ C/m} \times 1 \text{ m} = 5 \times 10^{-5} \text{ C}

3.6 x 10^6 N/C

1.8 x 10^6 N/C

Zero

9 x 10^5 N/C

Correct Answer: A

Solution:

The electric field due to an infinite line charge is given by

E = \frac{\lambda}{2\pi\varepsilon_0 r}

, where

\lambda = 2 \times 10^{-6} \text{ C/m}

r = 0.01 \text{ m}

, and

\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2

. Substituting these values gives

E = 3.6 \times 10^6 \text{ N/C}

1 x 10⁻⁴ Nm

2 x 10⁻⁴ Nm

3 x 10⁻⁴ Nm

4 x 10⁻⁴ Nm

Correct Answer: A

Solution:

The torque

\tau

on a dipole in an electric field is given by

\tau = pE \sin \theta

. Here,

p = 4 \times 10^{-9}

C m,

E = 5 \times 10^4

N/C, and

\theta = 30^\circ

. Thus,

\tau = 4 \times 10^{-9} \times 5 \times 10^4 \times \sin 30^\circ = 1 \times 10^{-4}

Nm.

1.00 µC/m

0.50 µC/m

0.10 µC/m

0.05 µC/m

Correct Answer: A

Solution:

The electric field

E

due to an infinite line charge with linear charge density

\lambda

is given by

E = \frac{\lambda}{2\pi\varepsilon_0 r}

. Solving for

\lambda

, we have

\lambda = 2\pi\varepsilon_0 r E

. Substituting

\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2

r = 0.02 \text{ m}

, and

E = 9 \times 10^4 \text{ N/C}

, we get

\lambda = 2\pi \times 8.854 \times 10^{-12} \times 0.02 \times 9 \times 10^4 = 1.00 \text{ µC/m}

0.06 N

0.12 N

0.18 N

0.24 N

Correct Answer: A

Solution:

Using Coulomb's law,

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

, where

k = 9 \times 10^9

N m² C⁻²,

q_1 = 2 \times 10^{-7}

q_2 = 3 \times 10^{-7}

C, and

r = 0.3

m. Substituting these values,

F = \frac{9 \times 10^9 \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{0.3^2} = 0.06

It is zero.

It is constant.

It is equal to the charge density.

It varies with distance.

Correct Answer: A

Solution:

For a neutral atom, the total charge enclosed by the Gaussian surface is zero, hence the electric field is zero.

0.4 x 10^-4 N m

0.8 x 10^-4 N m

1.6 x 10^-4 N m

3.2 x 10^-4 N m

Correct Answer: B

Solution:

The torque on a dipole in a uniform electric field is given by

\tau = pE \sin \theta

. Substituting the given values,

\tau = 4 \times 10^{-9} C \cdot m \times 2 \times 10^4 N/C \times \sin 60° = 0.8 \times 10^{-4} N \cdot m

Electric charge is a vector quantity.

Electric charge is quantised.

Electric charge can be created or destroyed.

Electric charge is always negative.

Correct Answer: B

Solution:

Electric charge is quantised, meaning it is always an integral multiple of a basic quantum of charge.

Electrostatic induction

Magnetic attraction

Gravitational force

Nuclear force

Correct Answer: A

Solution:

The charged comb polarizes the pieces of paper, inducing a net dipole moment in the direction of the electric field. This results in an attractive force between the comb and the paper.

0.06 N

0.18 N

0.36 N

0.72 N

Correct Answer: A

Solution:

Using Coulomb's law,

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

, where

k = 9 \times 10^9 N m^2 C^{-2}

and

r = 0.3

m. Substituting the values,

F = \frac{9 \times 10^9 \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^2} = 0.06

True or False

Correct Answer: False

Solution:

Quantisation of electric charge is a basic law of nature, but there is no analogous law for the quantisation of mass.

Correct Answer: True

Solution:

For discrete charge configurations, the electric field is undefined at the exact location of the charges because the field would theoretically be infinite.

Correct Answer: True

Solution:

The surface charge density

\sigma

is defined as the charge

Q

on an area element

S

Solution:

The force between two point charges depends on the medium between them, as the constant of proportionality

k

in Coulomb's law is affected by the permittivity of the medium.

Correct Answer: True

Solution:

The surface charge density, denoted by

\sigma

, is defined as the charge

Q

on an area element

S

, i.e.,

Solution:

The electric field due to a charge configuration with total charge zero is not zero; for distances large compared to the size of the configuration, its field falls off faster than

n

is an integer.

Correct Answer: True

Solution:

In SI units, the unit of charge (coulomb) is defined by

1 \, C = 1 \, A \, s

, which is based on its magnetic effect (Ampere's law). This makes the value of the constant

k

in Coulomb's law approximately

9 \times 10^9 \, N \, m^2 \, C^{-2}

Correct Answer: True

Solution:

For a continuous charge distribution, the electric field is defined at any point in the distribution.

Electric Charges and Fields

AI Learning Assistant

Summary

Chapter One: Electric Charges and Fields

Summary

Key Formulas and Definitions

Learning Objectives

Detailed Notes

Chapter One: Electric Charges and Fields

1.1 Introduction

1.2 Electric Charge

1.3 Basic Properties of Electric Charge

1.4 Conductors and Insulators

1.5 Coulomb's Law

1.6 Electric Field and Electric Flux

1.7 Electric Dipole

1.8 Important Concepts

1.9 Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.What happens when a glass rod is rubbed with a silk cloth?

Correct Answer: A

Q2.A charged particle with charge q=1×10−6q = 1 \times 10^{-6}q=1×10−6 C is placed in a uniform electric field of magnitude E=2×105E = 2 \times 10^5E=2×105 N/C. What is the force experienced by the particle?

Correct Answer: A

Q3.A dipole with dipole moment 3 x 10^-9 C m is placed in a uniform electric field of magnitude 4 x 10^4 N/C at an angle of 45° with the field. What is the torque acting on the dipole?

Correct Answer: B

Q4.What is the electric field inside a uniformly charged conducting sphere?

Correct Answer: B

Q5.A polythene sheet is rubbed with wool and gains a charge of -3 x 10⁻⁷ C. How many electrons are transferred to the polythene?

Correct Answer: A

Q6.What happens when a charged comb is brought near a piece of paper?

Correct Answer: C

Q7.According to Coulomb's law, the force between two charges is proportional to:

Correct Answer: B

Q8.A spherical conductor of radius 1.2 m carries a total charge of 5.0 µC. Calculate the electric field at a point 0.6 m from the center of the sphere.

Correct Answer: A

Q9.What is the charge on a sphere with a surface charge density of 80.0 µC/m² and a diameter of 2.4 m?

Correct Answer: C

Q10.What happens when a charged object touches the metal knob of a gold-leaf electroscope?

Correct Answer: A

Q11.What is the unit of linear charge density?

Correct Answer: C

Q12.An electric dipole consists of charges +q+q+q and −q-q−q separated by a distance 2a2a2a. If the dipole is placed in a uniform electric field EEE, what is the torque acting on it when it is aligned at an angle θ=45∘\theta = 45^\circθ=45∘ with the field?

Correct Answer: B

Q13.What is the electric field at a point due to a point charge qqq?

Correct Answer: A

Q14.A glass rod is rubbed with silk, resulting in the glass rod becoming positively charged. What happens to the silk?

Correct Answer: A

Q15.What is the unit of electric charge in the International System of Units (SI)?

Correct Answer: B

Q16.If two point charges are separated by a distance, what happens to the force between them if the distance is doubled?

Correct Answer: D

Q17.A uniformly charged insulating rod of length 0.5 m carries a total charge of 10 µC. Calculate the electric field at a point 0.3 m away from the midpoint of the rod along its perpendicular bisector.

Correct Answer: B

Q18.Consider two point charges, q1=5×10−6q_1 = 5 \times 10^{-6}q1​=5×10−6 C and q2=−5×10−6q_2 = -5 \times 10^{-6}q2​=−5×10−6 C, placed 0.4 m apart in a vacuum. What is the magnitude of the electrostatic force between them?

Correct Answer: A

Q19.Two point charges, q1=2×10−7q_1 = 2 \times 10^{-7}q1​=2×10−7 C and q2=3×10−7q_2 = 3 \times 10^{-7}q2​=3×10−7 C, are placed 30 cm apart in air. What is the magnitude of the force between them?

Correct Answer: A

Q20.Which of the following statements is true about the conservation of electric charge?

Correct Answer: A

Q21.What is the significance of the constant kkk in Coulomb's law?

Correct Answer: C

Q22.What is the unit of surface charge density?

Correct Answer: B

Q23.Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude 17.0 x 10⁻²² C/m². What is the electric field between the plates?

Correct Answer: B

Q24.A charged particle is placed at the center of a hollow metallic sphere. What will be the electric field inside the sphere?

Correct Answer: A

Q25.What is the force between two charges, 5 µC and -5 µC, placed 10 cm apart in air?

Correct Answer: A

Q26.Which of the following is a property of electric charge?

Correct Answer: B

Q27.A uniformly charged conducting sphere of diameter 2.4 m has a surface charge density of 80.0 µC/m². What is the total electric flux leaving the surface of the sphere?

Correct Answer: A

Q28.A spherical conductor of radius 0.5 m carries a total charge of 10−510^{-5}10−5 C. What is the surface charge density on the sphere?

Correct Answer: A