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Electric Charges and Fields – CBSE NCERT Study Resources

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12th

12th - Physics

Electric Charges and Fields

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Electric Charges and Fields - Chapter Overview

This chapter introduces the fundamental concepts of electric charges, their properties, and the electric fields they produce. It covers Coulomb's Law, the principle of superposition, electric field lines, and Gauss's Law, providing a foundation for understanding electrostatics.

Electric Charge: A fundamental property of matter that causes it to experience a force in the presence of other charges. Charges can be positive or negative.

Key Topics

  • Properties of Electric Charges
  • Coulomb's Law and Forces Between Charges
  • Electric Field and Field Lines
  • Electric Dipole and Dipole Moment
  • Gauss's Law and Its Applications

Properties of Electric Charges

Electric charges are quantized, conserved, and exhibit additive properties. Like charges repel, while unlike charges attract.

Coulomb's Law: The force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

Electric Field and Field Lines

An electric field is a region around a charged particle where its influence can be felt. Electric field lines represent the direction and strength of the field.

Electric Dipole: A pair of equal and opposite charges separated by a small distance. The dipole moment is a vector quantity pointing from the negative to the positive charge.

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed within it. It is useful for calculating electric fields in symmetric charge distributions.

Electric Flux: The measure of the number of electric field lines passing through a given area.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define electric field.
Answer:
Definition: Force per unit positive charge at a point.
Question 2:
State Coulomb's law in vector form.
Answer:
Formula: F = k(q₁q₂/r²)r̂.
Question 3:
What is the SI unit of electric flux?
Answer:
Unit: Nm²/C or Vm.
Question 4:
Name the principle behind electrostatic shielding.
Answer:
Principle: Faraday's cage effect.
Question 5:
What is the net flux through a closed surface enclosing a dipole?
Answer:
Result: Zero (Gauss's law).
Question 6:
Define dielectric constant of a medium.
Answer:
Definition: Ratio of permittivity of medium to vacuum.
Question 7:
What happens to force between charges if distance is halved?
Answer:
Effect: Force becomes 4 times (F ∝ 1/r²).
Question 8:
Give one example of a non-polar dielectric.
Answer:
Example: O₂ or CH₄.
Question 9:
What is the direction of electric field due to a negative point charge?
Answer:
Direction: Radially inward.
Question 10:
State the number of electrons in 1 coulomb of charge.
Answer:
Value: 6.25 × 10¹⁸ electrons.
Question 11:
Name the device used to detect charge on a body.
Answer:
Device: Gold-leaf electroscope.
Question 12:
What is the work done in moving a charge on an equipotential surface?
Answer:
Work: Zero (ΔV = 0).
Question 13:
Define electric dipole moment.
Answer:
Definition: Product of charge and separation distance (p = q×2a).
Question 14:
What is the angle between electric field and equipotential surface?
Answer:
Angle: 90° (perpendicular).
Question 15:
What is the principle of conservation of charge?
Answer:

The principle of conservation of charge states that the total electric charge in an isolated system remains constant. Charge can neither be created nor destroyed, only transferred.

Question 16:
Define electric charge.
Answer:

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. It can be positive or negative.

Question 17:
State the SI unit of electric charge.
Answer:

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

Question 18:
What is the value of the permittivity of free space (ε₀)?
Answer:

The permittivity of free space (ε₀) is approximately 8.854 × 10⁻¹² C²/N·m².

Question 19:
Define electric field intensity.
Answer:

Electric field intensity at a point is the force experienced by a unit positive charge placed at that point. Its SI unit is N/C or V/m.

Question 20:
What is the direction of the electric field due to a positive point charge?
Answer:

The electric field due to a positive point charge is directed radially outward from the charge.

Question 21:
State Coulomb's law in electrostatics.
Answer:

Coulomb's law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically: F = k(q₁q₂)/r², where k is Coulomb's constant.

Question 22:
What is an electric dipole?
Answer:

An electric dipole is a pair of equal and opposite charges separated by a small distance. It is characterized by its dipole moment (p = q × 2a), where 2a is the separation between charges.

Question 23:
What is the net force on a dipole placed in a uniform electric field?
Answer:

The net force on a dipole placed in a uniform electric field is zero because the forces on the two equal and opposite charges cancel each other out.

Question 24:
Define electric flux.
Answer:

Electric flux is the measure of the number of electric field lines passing through a given area. It is a scalar quantity and its SI unit is N·m²/C.

Question 25:
State Gauss's law in electrostatics.
Answer:

Gauss's law states that the total electric flux through a closed surface is equal to 1/ε₀ times the total charge enclosed by the surface.
Mathematically: Φ = Q/ε₀.

Question 26:
What is the electric field inside a charged spherical shell?
Answer:

The electric field inside a charged spherical shell is zero because the charges distribute themselves on the outer surface, creating no field inside.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Define electric charge and state its SI unit.
Answer:

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

Question 2:
State Coulomb's law in electrostatics.
Answer:

Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
Mathematically: F = k(q₁q₂)/r², where k is Coulomb's constant.

Question 3:
What is an electric dipole? Give an example.
Answer:

An electric dipole is a pair of equal and opposite charges separated by a small distance.
Example: A water molecule (H₂O) behaves as an electric dipole due to its bent shape and unequal charge distribution.

Question 4:
Define electric field intensity and write its SI unit.
Answer:

Electric field intensity is the force experienced by a unit positive charge placed at a point in an electric field.
Its SI unit is newton per coulomb (N/C) or volt per meter (V/m).

Question 5:
State the superposition principle for electric fields.
Answer:

The superposition principle states that the net electric field at a point due to multiple charges is the vector sum of the individual fields produced by each charge at that point.

Question 6:
What is an electric flux? Write its SI unit.
Answer:

Electric flux is the measure of the number of electric field lines passing through a given area.
Its SI unit is volt-meter (V·m) or newton-meter² per coulomb (N·m²/C).

Question 7:
State Gauss's law in electrostatics.
Answer:

Gauss's law states that the total electric flux through a closed surface is equal to 1/ε₀ times the total charge enclosed by the surface.
Mathematically: Φ = Q/ε₀, where ε₀ is the permittivity of free space.

Question 8:
What is the significance of Gauss's law?
Answer:

Gauss's law simplifies the calculation of electric fields for highly symmetric charge distributions like spheres, cylinders, or planes.
It is a powerful tool in electrostatics for determining electric fields without complex integrations.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define electric flux and write its SI unit. How is it related to the number of electric field lines passing through a surface?
Answer:

Electric flux is defined as the total number of electric field lines passing through a given area. It is a measure of the distribution of the electric field over a surface.


Its SI unit is volt-meter (V·m) or newton-meter squared per coulomb (N·m²/C).


The electric flux (Φ) is directly proportional to the number of electric field lines passing through a surface. Mathematically, it is given by Φ = E · A · cosθ, where E is the electric field, A is the area, and θ is the angle between the field lines and the normal to the surface.

Question 2:
State Coulomb's law in vector form and explain the significance of the proportionality constant.
Answer:

Coulomb's law in vector form states that the electrostatic force (F) between two point charges q₁ and q₂ separated by a distance r is given by:


F = (1/4πε₀) · (q₁q₂/r²) · r̂


Here, r̂ is the unit vector along the line joining the two charges, and ε₀ is the permittivity of free space.


The proportionality constant (1/4πε₀) ensures the law is consistent with experimental observations and accounts for the medium's effect. In free space, ε₀ ≈ 8.854 × 10⁻¹² C²/N·m².

Question 3:
What is an electric dipole? Derive the expression for the electric field due to a dipole at a point on its axial line.
Answer:

An electric dipole is a pair of equal and opposite charges (+q and -q) separated by a small distance (2a).


The electric field (E) at a point on the axial line of a dipole is derived as follows:


1. Let the distance from the dipole's center to the point be r.
2. The field due to +q: E₁ = (1/4πε₀) · q/(r - a)² (along the dipole axis).
3. The field due to -q: E₂ = (1/4πε₀) · q/(r + a)² (opposite to the dipole axis).
4. Net field: E = E₁ - E₂ = (1/4πε₀) · [q/(r - a)² - q/(r + a)²].
5. Simplifying for r >> a: E ≈ (1/4πε₀) · (2p/r³), where p = q × 2a is the dipole moment.

Question 4:
Explain why the electric field inside a conductor is zero under electrostatic conditions.
Answer:

Under electrostatic conditions (no current flow), the electric field inside a conductor is zero due to the following reasons:


1. Free electrons in a conductor redistribute themselves in response to any external electric field.
2. This redistribution creates an internal electric field that exactly cancels the external field.
3. In equilibrium, the net force on any charge inside the conductor is zero, implying E = 0.
4. Any residual field would cause further electron movement, violating the electrostatic condition.


This property is used in electrostatic shielding, where conductors protect sensitive equipment from external fields.

Question 5:
What is Gauss's law? Use it to find the electric field due to an infinitely long straight charged wire.
Answer:

Gauss's law states that the total electric flux through a closed surface is equal to the net charge enclosed divided by ε₀. Mathematically: ∮E · dA = Qenc/ε₀.


To find the electric field due to an infinitely long charged wire:


1. Choose a cylindrical Gaussian surface of radius r and length l coaxial with the wire.
2. By symmetry, E is radial and has the same magnitude at all points on the curved surface.
3. Flux through the curved surface: Φ = E × 2πrl.
4. Charge enclosed: Qenc = λl, where λ is linear charge density.
5. Applying Gauss's law: E × 2πrl = λl/ε₀.
6. Thus, E = λ/(2πε₀r), directed radially outward (for λ > 0).

Question 6:
Differentiate between uniform and non-uniform electric fields with examples.
Answer:

The key differences between uniform and non-uniform electric fields are:


Uniform Electric Field:
1. Magnitude and direction remain constant at all points.
2. Field lines are parallel and equally spaced.
3. Example: Field between two oppositely charged parallel plates.


Non-uniform Electric Field:
1. Magnitude and/or direction varies from point to point.
2. Field lines are neither parallel nor equally spaced.
3. Example: Field due to a point charge (varies as 1/r²).


In a uniform field, a dipole experiences only torque (no net force), while in a non-uniform field, it experiences both torque and net force.

Question 7:
State Coulomb's law in vector form and explain the significance of the negative sign in the expression.
Answer:

Coulomb's law in vector form is given by:

F₁₂ = (k · q₁ · q₂ · r̂₁₂) / r²

where F₁₂ is the force on charge q₁ due to q₂, k is Coulomb's constant, r̂₁₂ is the unit vector from q₁ to q₂, and r is the distance between the charges.

The negative sign indicates that the force is attractive when the charges are of opposite signs and repulsive when the charges are of the same sign. It ensures the direction of the force aligns with the nature of the interaction between charges.

Question 8:
What is Gauss's law? Use it to find the electric field due to an infinitely long straight charged wire.
Answer:

Gauss's law states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε₀). Mathematically:

Φ = ∮ E · dA = Qenc / ε₀

For an infinitely long straight charged wire with linear charge density λ, we use a cylindrical Gaussian surface:

E · (2πrl) = λl / ε₀

Solving for E:

E = λ / (2πε₀r)

The field decreases with distance (r) from the wire, characteristic of a line charge.

Question 9:
Differentiate between conductors and insulators based on their electric charge distribution properties.
Answer:

The key differences between conductors and insulators are:

  • Charge Distribution: In conductors, charges reside on the surface due to free electron movement, while in insulators, charges remain localized.
  • Electric Field: Conductors shield internal regions from external fields (E=0 inside), whereas insulators allow partial penetration of the field.
  • Polarization: Conductors exhibit induced charge separation, while insulators show dielectric polarization.

These properties make conductors essential for shielding and insulators crucial for storing charges in capacitors.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain Coulomb's Law and derive its vector form. Discuss how it differs from Newton's Law of Gravitation.
Answer:
Theoretical Framework

Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, F = k(q₁q₂)/r², where k is Coulomb's constant.

Evidence Analysis
  • Vector form: F⃗ = k(q₁q₂)/r² r̂, where r̂ is the unit vector.
  • Unlike gravity, Coulomb's force can be attractive or repulsive and is much stronger.
Critical Evaluation

Our textbook shows that while both laws are inverse-square laws, Coulomb's Law depends on charge, whereas gravity depends on mass.

Future Implications

Understanding this law is crucial for designing electrical systems and analyzing atomic interactions.

Question 2:
Define electric flux and derive its expression for a uniform electric field. How is Gauss's Law related to it?
Answer:
Theoretical Framework

Electric flux (Φ) measures the number of electric field lines passing through a surface. For a uniform field, Φ = E⋅A⋅cosθ, where E is the field strength, A is the area, and θ is the angle between E and the normal.

Evidence Analysis
  • Gauss's Law states Φ = q/ε₀, linking flux to enclosed charge.
  • Example: Flux through a closed sphere around a point charge is q/ε₀.
Critical Evaluation

We studied that Gauss's Law simplifies calculating fields for symmetric charge distributions, like spheres or cylinders.

Future Implications

This concept is vital for understanding capacitors and electromagnetic theory.

Question 3:
Describe the principle of superposition for electric fields. Illustrate with an example of two point charges.
Answer:
Theoretical Framework

The superposition principle states that the net electric field due to multiple charges is the vector sum of individual fields. For n charges, E⃗_net = Σ E⃗_i.

Evidence Analysis
  • Example: Two charges +q and -q at distance d create a dipole field.
  • At a point on the axial line, E_net = E₁ + E₂ (vector addition).
Critical Evaluation

Our textbook shows this principle applies to all electrostatic interactions, enabling complex field calculations.

Future Implications

It underpins modern technologies like semiconductor devices and MRI machines.

Question 4:
Explain electric dipole and derive the expression for its electric field on the axial line. Compare it with the equatorial line.
Answer:
Theoretical Framework

An electric dipole consists of two equal and opposite charges (+q, -q) separated by distance 2a. Its dipole moment is p = q×2a.

Evidence Analysis
  • Axial field: E = (2kp)/r³ (directed along p).
  • Equatorial field: E = (kp)/r³ (opposite to p), half the axial magnitude.
Critical Evaluation

We studied that the field is stronger along the axial line due to constructive superposition.

Future Implications

Dipole concepts are essential in polar molecules and antenna design.

Question 5:
Explain Coulomb's Law and its significance in determining the electrostatic force between two point charges. How does the force vary with the distance between them?
Answer:
Theoretical Framework

Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, F = k(q₁q₂)/r², where k is Coulomb's constant.

Evidence Analysis
  • Our textbook shows that when distance (r) doubles, the force reduces to 1/4th of its original value.
  • Example: Two charges of +1μC and -1μC separated by 1m experience a force of 9×10³ N.
Critical Evaluation

The inverse-square nature confirms the law's alignment with Newton's Law of Gravitation, emphasizing universality.

Future Implications

This principle underpins technologies like electrostatic precipitators and Van de Graaff generators.

Question 6:
Define electric flux and derive its expression for a uniform electric field passing through a plane surface. How is it affected if the surface is tilted?
Answer:
Theoretical Framework

Electric flux (Φ) measures the number of electric field lines passing through a surface. For a uniform field (E) and flat area (A), Φ = E⋅A⋅cosθ, where θ is the angle between E and the surface normal.

Evidence Analysis
  • Our textbook demonstrates that flux is maximum (Φ = EA) when θ = 0° and zero at θ = 90°.
  • Example: A 2m² plate in a 100 N/C field at 30° has Φ = 100×2×cos30° ≈ 173 Nm²/C.
Critical Evaluation

Flux depends on orientation, proving its vector nature. Gauss's Law relies on this concept.

Future Implications

Flux calculations are vital in designing capacitors and electromagnetic shielding.

Question 7:
Describe the principle of superposition for electric fields with two examples. How does it help in calculating net field due to multiple charges?
Answer:
Theoretical Framework

The superposition principle states that the net electric field at a point due to multiple charges is the vector sum of individual fields. Enet = ΣEi.

Evidence Analysis
  • Example 1: Two +1μC charges 1m apart produce zero field at the midpoint due to equal and opposite fields.
  • Example 2: A dipole’s axial field is the sum of fields from +q and -q.
Critical Evaluation

This principle simplifies complex systems, validated by experimental measurements.

Future Implications

It’s foundational for modeling molecular forces and semiconductor behavior.

Question 8:
Compare the behavior of electric field lines for a single positive charge and a dipole. What does the density of these lines indicate?
Answer:
Theoretical Framework

Field lines represent the direction and strength of the electric field. Density ∝ field strength.

Evidence Analysis
  • Single charge: Radial lines diverge symmetrically, density decreases with distance.
  • Dipole: Lines originate from +q, terminate at -q, and are denser near charges.
Critical Evaluation

Dipole fields are non-uniform, while single charges show spherical symmetry, confirmed by simulations.

Future Implications

Understanding field patterns aids in designing antennas and sensors.

Question 9:
Define electric flux and derive its expression for a uniform electric field passing through a plane surface. How is it useful in Gauss's Law?
Answer:
Theoretical Framework

Electric flux (Φ) measures the number of electric field lines passing through a surface. For a uniform field E and area A, Φ = E⋅A⋅cosθ, where θ is the angle between E and the normal.

Evidence Analysis
  • Derivation: When E is perpendicular, Φ = EA.
  • Gauss's Law uses flux to relate enclosed charge to field lines: Φ = q/ε₀.
Critical Evaluation

We studied that flux is zero for closed surfaces with no net charge, confirming symmetry in fields.

Future Implications

This concept is vital for calculating fields of symmetric charge distributions like spheres or cylinders.

Question 10:
Describe the principle of superposition for electric fields with an example. How does it apply to dipole fields?
Answer:
Theoretical Framework

The superposition principle states that the net electric field due to multiple charges is the vector sum of individual fields. For n charges, E_net = ΣE_i.

Evidence Analysis
  • Example: Two charges +q and -q at distance d create a dipole field.
  • At axial position, E = 2kp/r³, where p = qd is dipole moment.
Critical Evaluation

Our textbook shows this principle simplifies complex charge distributions, as seen in dipole calculations.

Future Implications

It underpins modern electrostatics, aiding in designing multi-charge systems like capacitors.

Question 11:
Explain electric field lines and their properties. Compare the field patterns of a point charge and a dipole.
Answer:
Theoretical Framework

Electric field lines are imaginary curves representing field direction and strength. Density indicates magnitude, and lines never intersect.

Evidence Analysis
  • Point charge: Radial lines (outward for +q, inward for -q).
  • Dipole: Curved lines from +q to -q, with zero field at midpoint.
Critical Evaluation

We studied that dipole fields decay faster (1/r³) than point charges (1/r²).

Future Implications

Visualizing fields helps predict charge behavior, crucial for applications like particle accelerators.

Question 12:
Using Gauss's law, derive an expression for the electric field due to a uniformly charged infinite plane sheet. Also, discuss the significance of the result obtained.
Answer:

To derive the electric field due to a uniformly charged infinite plane sheet using Gauss's law, follow these steps:


Step 1: Assumptions


Consider an infinite plane sheet with uniform surface charge density σ (charge per unit area). Due to symmetry, the electric field E must be perpendicular to the plane and have the same magnitude at all points equidistant from the sheet.


Step 2: Gaussian Surface


Choose a Gaussian surface as a cylinder (pillbox) with its axis perpendicular to the plane and its flat faces at equal distances from the plane. The cylinder's height is negligible compared to its radius.


Step 3: Applying Gauss's Law


Gauss's law states: ∮E · dA = Qenc/ε0.
The total flux through the Gaussian surface is the sum of the flux through the two flat faces (since the curved surface contributes zero flux as E is parallel to it).
Flux through one face = E × A (where A is the area of the face).
Total flux = 2EA.


Step 4: Enclosed Charge


The charge enclosed by the Gaussian surface is Qenc = σA.


Step 5: Solving for E


Substituting into Gauss's law:
2EA = σA/ε0
Simplifying:
E = σ/(2ε0).


Significance:


The electric field is independent of the distance from the plane sheet, which is a unique characteristic of an infinite plane. This result is widely used in electrostatics, such as in parallel plate capacitors, where the field between the plates is uniform.

Question 13:
Explain the concept of electric dipole moment. Derive an expression for the torque acting on an electric dipole placed in a uniform electric field and discuss its physical significance.
Answer:

Electric Dipole Moment:


An electric dipole consists of two equal and opposite charges (+q and -q) separated by a small distance 2a. The electric dipole moment (p) is a vector quantity defined as:
p = q × 2a (direction from -q to +q).


Torque on an Electric Dipole in a Uniform Electric Field:


Consider a dipole placed in a uniform electric field E at an angle θ with the field.


Step 1: Forces on Charges


The force on +q is F+ = qE (along E).
The force on -q is F- = -qE (opposite to E).


Step 2: Net Force and Torque


The net force is zero (F+ + F- = 0), but the forces create a torque (τ) that tends to rotate the dipole.


Step 3: Calculating Torque


Torque is given by:
τ = Force × Perpendicular Distance.
For each charge, the perpendicular distance is a sinθ.
Thus, torque due to both charges:
τ = qE × a sinθ + qE × a sinθ = 2qEa sinθ.
Since p = q × 2a, the torque can be written as:
τ = pE sinθ.
In vector form: τ = p × E.


Physical Significance:


The torque aligns the dipole with the electric field (θ = 0°), minimizing potential energy. This principle is crucial in understanding the behavior of dipoles in fields, such as in dielectric materials and molecular polarizability.

Question 14:
Using Gauss's law, derive an expression for the electric field due to a uniformly charged infinite plane sheet. Discuss the significance of the result.
Answer:

To derive the electric field due to a uniformly charged infinite plane sheet using Gauss's law, follow these steps:


Step 1: Assume a Gaussian surface
Consider a cylindrical Gaussian surface (pillbox) of cross-sectional area A placed symmetrically around the plane sheet. The cylinder's flat faces are parallel to the sheet.


Step 2: Apply Gauss's law
According to Gauss's law, the total electric flux (Φ) through the closed surface is:
Φ = Qenc / ε0
Here, Qenc is the charge enclosed by the Gaussian surface.


Step 3: Calculate enclosed charge
For an infinite plane sheet with surface charge density σ, the enclosed charge is:
Qenc = σA


Step 4: Determine electric flux
The electric field E is perpendicular to the plane sheet and uniform. The flux through the curved part of the cylinder is zero (as E and dA are perpendicular). The flux through each flat face is EA. Thus, total flux is:
Φ = 2EA


Step 5: Equate and solve
Using Gauss's law:
2EA = σA / ε0
Simplifying:
E = σ / (2ε0)


Significance:
The electric field is independent of the distance from the sheet, which is a unique property of an infinite plane sheet. This result is crucial in understanding the behavior of electric fields in capacitors and other applications where large charged surfaces are involved.

Question 15:
Explain the concept of electric dipole moment. Derive the expression for the electric field due to an electric dipole at a point on its axial line.
Answer:

Electric dipole moment (p) is a vector quantity defined as the product of the magnitude of either charge (q) and the separation distance (2a) between the two equal and opposite charges. Its direction is from the negative to the positive charge.
p = q × 2a


Derivation of electric field on the axial line:


Step 1: Setup
Consider an electric dipole with charges +q and -q separated by distance 2a. Let point P be at a distance r from the dipole's center along the axial line.


Step 2: Electric field due to +q
The field at P due to +q is:
E+ = kq / (r - a)2
where k = 1/(4πε0).


Step 3: Electric field due to -q
The field at P due to -q is:
E- = -kq / (r + a)2
(negative sign indicates opposite direction)


Step 4: Net electric field
The net field Eaxial is the sum of E+ and E-:
Eaxial = kq [1/(r - a)2 - 1/(r + a)2]


Step 5: Simplify for r >> a
For points far from the dipole (r >> a), the expression approximates to:
Eaxial ≈ k × 2p / r3
where p = q × 2a is the dipole moment.


Key observation:
The axial field decreases with the cube of the distance (1/r3), unlike a single charge's field (1/r2). This shows how dipole fields diminish faster with distance.

Question 16:
Using Gauss's law, derive an expression for the electric field intensity due to a uniformly charged infinite plane sheet. Also, discuss how the field varies with distance from the sheet.
Answer:

To derive the electric field intensity due to a uniformly charged infinite plane sheet using Gauss's law, follow these steps:


Step 1: Assumptions
Consider an infinite plane sheet with uniform surface charge density σ (charge per unit area). Due to symmetry, the electric field E must be perpendicular to the plane and have the same magnitude at all points equidistant from the sheet.


Step 2: Gaussian Surface
Choose a Gaussian surface in the form of a cylinder (pillbox) with its axis perpendicular to the plane and its flat faces at equal distances from the sheet. Let the area of each flat face be A.


Step 3: Flux Calculation
The total electric flux through the Gaussian surface is the sum of the flux through the two flat faces (since the curved surface contributes zero flux as E is parallel to it):
Φ = E × A (for one face) × 2 = 2EA.


Step 4: Gauss's Law Application
According to Gauss's law, Φ = qenc / ε0.
Here, qenc = σA (charge enclosed by the Gaussian surface).
Thus, 2EA = σA / ε0.


Step 5: Solve for E
Cancel A from both sides:
2E = σ / ε0
⇒ E = σ / (2ε0).


Conclusion:
The electric field intensity E is σ / (2ε0) and is independent of the distance from the sheet. This is because the infinite plane sheet creates a uniform field, and the field lines are parallel and equidistant.


Value-Added Insight:
This result holds only for an infinite sheet. For finite sheets, the field varies near the edges. The direction of E is away from the sheet if σ is positive and toward the sheet if σ is negative.

Question 17:
Explain the concept of electric dipole moment. Derive an expression for the torque acting on an electric dipole placed in a uniform electric field and discuss its significance.
Answer:

Electric Dipole Moment:


An electric dipole consists of two equal and opposite charges (+q and -q) separated by a small distance 2a. The electric dipole moment (p) is a vector quantity defined as the product of the magnitude of one of the charges and the separation between the charges. Its direction is from the negative to the positive charge.


Mathematically, p = q × 2a.


Torque on an Electric Dipole in a Uniform Electric Field:


Consider an electric dipole placed in a uniform electric field E at an angle θ with the field. The forces on the charges are +qE (in the direction of E) and -qE (opposite to E). These forces are equal in magnitude but opposite in direction, forming a couple.


Step 1: Force Analysis


Force on +q: F1 = +qE (along E).
Force on -q: F2 = -qE (opposite to E).


Step 2: Torque Calculation


The perpendicular distance between the forces is 2a sinθ. Torque (τ) is given by the product of force and perpendicular distance.


τ = qE × 2a sinθ.
Since p = q × 2a, the torque can be written as τ = pE sinθ.


In vector form, τ = p × E.


Significance:


The torque tends to align the dipole with the electric field. At θ = 0°, the torque is zero (stable equilibrium), and at θ = 180°, it is zero but unstable. This principle is used in devices like electric motors and understanding molecular behavior in electric fields.

Question 18:
Using Gauss's law, derive an expression for the electric field intensity due to a uniformly charged infinite plane sheet. Discuss the significance of the result obtained.
Answer:

To derive the expression for the electric field intensity due to a uniformly charged infinite plane sheet using Gauss's law, follow these steps:


Step 1: Assumptions


Consider an infinite plane sheet with a uniform surface charge density σ (charge per unit area). Due to symmetry, the electric field E must be perpendicular to the plane and have the same magnitude at all points equidistant from the sheet.


Step 2: Gaussian Surface


Choose a Gaussian surface in the form of a cylinder (pillbox) with its axis perpendicular to the plane. The cylinder has two flat faces of area A each, equidistant from the sheet.


Step 3: Applying Gauss's Law


According to Gauss's law, the total electric flux through the closed surface is:
Φ = ∮E · dA = Qenc/ε0
Here, Qenc = σA (charge enclosed by the Gaussian surface).


Step 4: Calculating Flux


The flux through the curved part of the cylinder is zero because E is parallel to the surface. The flux through each flat face is EA (since E and dA are parallel).
Total flux: Φ = EA + EA = 2EA.


Step 5: Equating and Solving


From Gauss's law:
2EA = σA/ε0
Simplifying:
E = σ/(2ε0)


Significance of the Result:

  • The electric field is independent of the distance from the sheet, which is a unique property of an infinite plane sheet.
  • The field is uniform and constant in magnitude and direction everywhere.
  • This result is applicable in practical scenarios like parallel plate capacitors, where the plates are assumed to be infinite for simplification.
Question 19:
Using Gauss's law, derive an expression for the electric field intensity due to a uniformly charged infinite plane sheet. Also, discuss how the field varies with distance from the sheet.
Answer:

To derive the electric field intensity due to a uniformly charged infinite plane sheet using Gauss's law, follow these steps:


1. Assumptions: Consider an infinite plane sheet with uniform surface charge density σ (charge per unit area). The sheet is infinitely large, so edge effects are negligible.


2. Gaussian Surface: Choose a cylindrical Gaussian surface (pillbox) perpendicular to the plane sheet, with its flat faces of area A at equal distances from the sheet.


3. Flux Calculation: The electric field E is perpendicular to the sheet and has the same magnitude at all points equidistant from the sheet. The flux through the curved part of the cylinder is zero because E is parallel to this surface. The flux through each flat face is EA.


4. Total Flux: The total electric flux through the Gaussian surface is Φ = 2EA (since there are two flat faces).


5. Charge Enclosed: The charge enclosed by the Gaussian surface is q = σA.


6. Gauss's Law Application: According to Gauss's law, Φ = q/ε₀. Substituting the values, we get 2EA = σA/ε₀.


7. Electric Field: Simplifying, the electric field intensity is E = σ/(2ε₀). This shows that E is independent of the distance from the sheet, meaning the field is uniform.


Variation with Distance: The electric field due to an infinite plane sheet does not depend on the distance from the sheet. It remains constant (E = σ/(2ε₀)) at all points in space, which is a unique characteristic of an infinite plane sheet.

Question 20:
Explain the concept of electric dipole and derive an expression for the electric field intensity at a point on the axial line of an electric dipole. Compare it with the field on the equatorial line.
Answer:

Electric Dipole: An electric dipole is a pair of equal and opposite charges (+q and -q) separated by a small distance 2a. The dipole moment (p) is a vector quantity defined as p = q × 2a, directed from the negative to the positive charge.


Derivation of Electric Field on Axial Line:


1. Setup: Consider a dipole with charges +q and -q separated by distance 2a. Let the axial point be at distance r from the midpoint of the dipole.


2. Field Due to +q: The electric field at the axial point due to +q is E₁ = kq/(r - a)², directed away from +q.


3. Field Due to -q: The field due to -q is E₂ = kq/(r + a)², directed toward -q.


4. Net Field: The net field on the axial line is E_axial = E₁ - E₂ (since they are in the same direction). Substituting and simplifying, we get:


E_axial = kq[1/(r - a)² - 1/(r + a)²].


5. Approximation for r >> a: For r >> a, the expression simplifies to E_axial ≈ 2kp/r³, where p = q × 2a is the dipole moment.


Comparison with Equatorial Line:


1. Equatorial Field: The electric field on the equatorial line (perpendicular bisector) is E_equatorial = kp/r³, directed opposite to the dipole moment.


2. Key Differences:

  • The axial field is twice the equatorial field for the same distance.
  • The axial field is parallel to the dipole moment, while the equatorial field is antiparallel.
  • Both fields follow an inverse-cube law (1/r³) dependence.
Question 21:
Using Gauss's law, derive an expression for the electric field intensity due to a uniformly charged infinite plane sheet. Explain the significance of the result obtained.
Answer:

To derive the electric field intensity due to a uniformly charged infinite plane sheet using Gauss's law, follow these steps:


Step 1: Assumptions


Consider an infinite plane sheet with uniform surface charge density σ (charge per unit area). Due to symmetry, the electric field E must be perpendicular to the plane and have the same magnitude at all points equidistant from the sheet.


Step 2: Gaussian Surface


Choose a Gaussian surface in the form of a cylinder (pillbox) with its axis perpendicular to the plane and faces of area A at equal distances from the sheet.


Step 3: Flux Calculation


The total electric flux through the Gaussian surface is the sum of the flux through the two flat faces (since the curved surface contributes zero flux as E is parallel to it).


Flux through one face = E × A (since E and A are parallel).


Total flux = 2EA (as both faces contribute equally).


Step 4: Applying Gauss's Law


According to Gauss's law, the total flux is equal to the enclosed charge divided by ε₀.


Enclosed charge = σA.


Thus, 2EA = σA / ε₀.


Simplifying, E = σ / (2ε₀).


Significance:


The electric field is independent of the distance from the sheet, meaning it remains constant at all points near an infinite plane sheet. This result is crucial in understanding the behavior of electric fields near large, uniformly charged surfaces, such as capacitor plates.

Question 22:
Explain the concept of electric dipole moment. Derive an expression for the torque acting on an electric dipole placed in a uniform electric field and discuss its equilibrium positions.
Answer:

Electric Dipole Moment:


An electric dipole consists of two equal and opposite charges (+q and -q) separated by a small distance 2a. The electric dipole moment (p) is a vector quantity defined as:


p = q × 2a (direction from -q to +q).


Its SI unit is Coulomb-meter (C·m).


Torque on a Dipole in Uniform Electric Field:


When an electric dipole is placed in a uniform electric field E, the forces on the charges are:


Force on +q: F₁ = +qE (along E).


Force on -q: F₂ = -qE (opposite to E).


These forces are equal in magnitude but opposite in direction, forming a couple.


The torque (τ) is given by:


τ = Force × Perpendicular distance between forces.


Here, the perpendicular distance = 2a sinθ, where θ is the angle between p and E.


Thus, τ = qE × 2a sinθ = pE sinθ.


In vector form: τ = p × E.


Equilibrium Positions:


  • Stable Equilibrium: When θ = 0°, p is parallel to E. The torque is zero, and the dipole remains aligned with the field.
  • Unstable Equilibrium: When θ = 180°, p is antiparallel to E. The torque is zero, but any slight disturbance causes the dipole to flip.


This derivation highlights how dipoles align with electric fields, which is fundamental in understanding dielectric behavior and polarization.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A student observes that when a charged rod is brought near a neutral conducting sphere, the sphere gets attracted. However, after contact, it gets repelled. Explain the phenomenon and derive the force between them if the sphere acquires a charge of 2μC and the rod has 5μC, separated by 10 cm.
Answer:
Case Deconstruction

Initially, the neutral sphere experiences induction, causing charge separation and attraction. After contact, charge transfer occurs, leading to repulsion.

Theoretical Application

Using Coulomb's Law: F = k(q₁q₂)/r². Given q₁ = 5μC, q₂ = 2μC, r = 0.1m, and k = 9×10⁹ Nm²/C², we calculate F = (9×10⁹ × 5×10⁻⁶ × 2×10⁻⁶)/(0.1)² = 9 N (repulsive).

Critical Evaluation
  • Assumes point charges, but real objects have finite size.
  • Neglects polarization effects post-contact.
Question 2:
Two identical point charges (+Q) are fixed at a distance 'd'. A third charge (-q) is placed midway. Analyze equilibrium stability and sketch the electric field lines.
Answer:
Case Deconstruction

The system is in unstable equilibrium as any displacement of -q causes net force away from the center.

Theoretical Application

Field lines originate from +Q and terminate at -q. [Diagram: Radial lines from +Q, converging at -q with symmetrical pattern]. Force calculation shows zero net force at midpoint but non-restoring nature.

Critical Evaluation
  • Assumes vacuum conditions; medium effects alter forces.
  • Practical systems require damping for stability.
Question 3:
A Gaussian surface encloses charges +3q, -2q, and +q. Calculate flux and explain why the result is independent of surface shape.
Answer:
Case Deconstruction

Net enclosed charge = (+3q) + (-2q) + (+q) = +2q. Gauss's Law states Φ = Qenc/ε₀ = 2q/ε₀.

Theoretical Application

Flux depends only on total enclosed charge, not distribution or surface geometry. Our textbook shows this through symmetry arguments.

Critical Evaluation
  • Assumes closed surface; open surfaces yield different results.
  • Practical measurements require ideal conditions.
Question 4:
Compare electric field patterns for: (i) infinite plane sheet (σ = 5 C/m²) and (ii) dipole (p = 10⁻⁷ Cm). Derive expressions and contrast their spatial dependence.
Answer:
Case Deconstruction

(i) Plane sheet: E = σ/2ε₀ (constant, 2.82×10¹¹ N/C). (ii) Dipole: E ∝ p/r³ (varies with distance).

Theoretical Application

Sheet field is uniform, while dipole field decays rapidly. [Diagram: Parallel lines for sheet, curved asymmetric lines for dipole].

Critical Evaluation
  • Infinite sheet is idealized; real sheets have edge effects.
  • Dipole approximation valid only for r >> charge separation.
Question 5:
A student observes that a charged glass rod attracts tiny paper pieces but repels a similarly charged ebonite rod. Explain the phenomenon and derive the mathematical expression for Coulomb's force between two point charges.
Answer:
Case Deconstruction

The glass rod induces opposite charges on paper pieces, causing attraction. Like charges (glass-ebonite) repel.

Theoretical Application

Coulomb's law states: F = k(q₁q₂)/r², where k is Coulomb's constant, q₁,q₂ are charges, and r is separation.

Critical Evaluation
  • Example 1: Balloon sticking to a wall shows induction.
  • Example 2: Repulsion between two electrons confirms Coulomb's law.
Question 6:
An electric dipole of moment p is placed in a uniform electric field E. Analyze the torque acting on it and compare its potential energy in stable vs. unstable equilibrium.
Answer:
Case Deconstruction

Torque (τ) rotates the dipole: τ = p×E.

Theoretical Application

Potential energy: U = -p·E (stable when θ=0°, unstable at θ=180°).

Critical Evaluation
  • Example 1: Compass needle aligns with Earth's field (stable).
  • Example 2: Dipole antiparallel to field has maximum energy.
Question 7:
Gauss's law states ∮E·dA = q/ε₀. Apply this to derive the electric field due to an infinite plane sheet of charge and contrast it with a spherical shell.
Answer:
Case Deconstruction

For a plane sheet, symmetry gives E = σ/2ε₀ (σ: surface charge density).

Theoretical Application

Spherical shell: E = q/4πε₀r² outside, zero inside (shell theorem).

Critical Evaluation
  • Example 1: Capacitor plates approximate infinite sheets.
  • Example 2: Van de Graaff generator demonstrates shell behavior.
Question 8:
A conductor has a cavity with a point charge +q inside. Prove that the net charge on the cavity wall is -q and explain why the external field remains unaffected if the conductor is grounded.
Answer:
Case Deconstruction

Induced -q on cavity wall ensures zero field inside conductor (electrostatic shielding).

Theoretical Application

Grounding neutralizes the conductor’s exterior charge, leaving cavity charges isolated.

Critical Evaluation
  • Example 1: Faraday cages block external fields.
  • Example 2: Sensitive equipment uses shielding for protection.
Question 9:
A student observes that when a glass rod is rubbed with silk, it acquires a positive charge, while an ebonite rod rubbed with wool gains a negative charge. Explain the phenomenon using electron transfer theory and discuss how this aligns with the law of conservation of charge.
Answer:
Case Deconstruction

When a glass rod is rubbed with silk, electrons transfer from the rod to the silk, leaving the rod positively charged. Conversely, ebonite gains electrons from wool, becoming negatively charged.

Theoretical Application

Our textbook shows that charge is neither created nor destroyed, only transferred. The total charge remains zero, validating the law of conservation of charge.

Critical Evaluation
  • Example 1: Combing hair with a plastic comb transfers electrons, making the comb negative.
  • Example 2: Balloon sticking to a wall after rubbing demonstrates charge transfer.
Question 10:
Two identical point charges +Q are placed 2 meters apart. Calculate the electric field at the midpoint and analyze why the field is zero despite the presence of charges.
Answer:
Case Deconstruction

The electric field due to each +Q charge at the midpoint is equal in magnitude but opposite in direction.

Theoretical Application

We studied that electric fields are vectors. The fields cancel out at the midpoint due to symmetry, resulting in a net zero field.

Critical Evaluation
  • Example 1: Two equal and opposite charges also produce zero field at their midpoint.
  • Example 2: A dipole's field is non-zero except along the perpendicular bisector.
Question 11:
A Gaussian surface encloses a dipole. Using Gauss's law, explain why the net flux through the surface is zero, even though the dipole creates an electric field.
Answer:
Case Deconstruction

A dipole consists of equal and opposite charges, so the net enclosed charge is zero.

Theoretical Application

Our textbook shows Gauss's law states flux depends only on enclosed charge. Since net charge is zero, flux is zero, despite the field.

Critical Evaluation
  • Example 1: A single charge enclosed yields non-zero flux.
  • Example 2: A uniformly charged sphere behaves like a point charge for flux calculations.
Question 12:
An electric dipole is placed in a uniform electric field. Derive the expression for torque acting on it and discuss how the dipole aligns with the field at equilibrium.
Answer:
Case Deconstruction

Torque (τ) on a dipole is given by τ = pE sinθ, where p is dipole moment and E is field strength.

Theoretical Application

We studied that at equilibrium (θ = 0°), torque becomes zero, aligning the dipole with the field.

Critical Evaluation
  • Example 1: A compass needle aligns with Earth's magnetic field similarly.
  • Example 2: Water molecules align in an electric field due to their dipole nature.
Question 13:
A student observes that a charged glass rod attracts tiny paper pieces but loses its charge over time. Explain the polarization of paper pieces and why the rod loses charge. Include a comparison with a metallic conductor.
Answer:
Case Deconstruction

The glass rod induces polarization in paper pieces, aligning their charges oppositely, causing attraction. Charge leaks due to air's weak conductivity.

Theoretical Application
  • Unlike paper, a metallic conductor would redistribute charges instantly, neutralizing the rod's effect faster.
  • Example: A charged balloon sticks to a wall (polarization) but falls when discharged.
Critical Evaluation

Our textbook shows polarization depends on material. Air humidity accelerates charge loss, as seen in rainy vs. dry conditions.

Question 14:
Two identical point charges placed 3 cm apart exert a 4 N force. Calculate the force if one charge is doubled and distance is halved. Discuss how Coulomb's law validates this change.
Answer:
Case Deconstruction

Initial force F = 4 N. New force F' = (2q₁q₂)/(r/2)² = 8F = 32 N.

Theoretical Application
  • Coulomb's law: F ∝ q₁q₂/r². Doubling charge and halving distance multiplies force by 8.
  • Example: Increasing phone charger voltage reduces charging time non-linearly.
Critical Evaluation

We studied that force changes are predictable only in vacuum; medium's permittivity alters results practically.

Question 15:
An electric dipole of moment 5×10⁻⁸ C·m aligns at 30° to a 10³ N/C field. Compute torque and potential energy. Contrast this with a single charge in the same field.
Answer:
Case Deconstruction

Torque τ = pEsinθ = 2.5×10⁻⁵ N·m. PE = -pEcosθ = -4.33×10⁻⁵ J.

Theoretical Application
  • A single charge experiences force (F = qE) but no torque or orientation-based PE.
  • Example: Compass needle (dipole) aligns in Earth's field, unlike a lone magnet pole.
Critical Evaluation

Our textbook shows dipoles stabilize at θ = 0°, whereas single charges accelerate indefinitely.

Question 16:
A Gaussian surface encloses charges +2q, -q, and +3q. Determine flux and justify using Gauss's law. How does flux change if surface size doubles?
Answer:
Case Deconstruction

Net charge = (+2q - q + 3q) = +4q. Flux Φ = 4q/ε₀. Size change doesn't alter flux.

Theoretical Application
  • Gauss's law: Flux depends only on enclosed charge, not surface dimensions.
  • Example: A lamp inside a box gives constant brightness (flux) regardless of box size.
Critical Evaluation

We studied that Gauss's law simplifies calculations but requires high symmetry, unlike Coulomb's law.

Question 17:

A student performs an experiment to study the electric field due to a uniformly charged thin spherical shell. The student measures the electric field at various distances from the center of the shell, both inside and outside. Based on the observations:

  • Inside the shell, the electric field is zero.
  • Outside the shell, the electric field varies inversely with the square of the distance from the center.

Using Gauss's law, explain these observations and derive the expression for the electric field outside the shell.

Answer:

Explanation using Gauss's Law:

For a uniformly charged thin spherical shell, the charge is distributed symmetrically on the surface. According to Gauss's law, the electric flux through a closed surface is proportional to the charge enclosed.

Inside the shell:
Consider a Gaussian surface inside the shell (radius r < R). Since no charge is enclosed (Qenc = 0), the electric field is zero.

Outside the shell:
For a Gaussian surface outside the shell (radius r > R), the total charge enclosed is Q.
Using Gauss's law:
ΦE = ∮E · dA = Qenc/ε0
Due to symmetry, E is radial and constant over the Gaussian surface.
E × 4πr2 = Q/ε0
Thus, E = (1/4πε0) × (Q/r2)
This shows the electric field varies as 1/r2 outside the shell.

Key takeaway: The shell behaves as if all its charge is concentrated at the center for points outside it, while the field inside is zero due to symmetry.

Question 18:

Two point charges, +4μC and -2μC, are placed 10 cm apart in air. A student calculates the electric field at a point 6 cm from the +4μC charge along the line joining the two charges. However, the student is unsure whether the field is zero at any point between the charges. Clarify this doubt and determine the exact location where the electric field is zero.

Answer:

Step-by-Step Solution:

Let the charges be:
q1 = +4μC, q2 = -2μC, separated by d = 10 cm.

Condition for zero electric field:
The electric fields due to q1 and q2 must cancel each other. Since q2 is negative, the zero point must lie outside the charges, on the side of the smaller magnitude charge (q2).

Let the zero field point be at distance x from q1, beyond q2.
E1 = E2
(1/4πε0) × (q1/x2) = (1/4πε0) × (|q2|/(x - d)2)
Simplifying:
4/x2 = 2/(x - 10)2
2(x - 10)2 = x2
Solving the quadratic equation:
x = 10(2 + √2) cm ≈ 34.14 cm from q1.

Conclusion:
The electric field is zero at approximately 34.14 cm from the +4μC charge, on the side opposite to the -2μC charge. There is no point between the charges where the field is zero because the fields due to both charges add up in the same direction in that region.

Question 19:
A student performs an experiment to study the electric field due to a dipole. The student places a dipole consisting of charges +q and -q separated by a distance 2a along the x-axis. The student then calculates the electric field at a point P located at a distance r from the midpoint of the dipole along its axial line (where r >> a).

Explain the steps the student should follow to derive the expression for the electric field at point P due to the dipole. Also, state the direction of the electric field at this point.

Answer:

The student should follow these steps to derive the expression for the electric field at point P along the axial line of the dipole:


Step 1: Identify the configuration of the dipole and the point P.
The dipole consists of charges +q and -q separated by distance 2a, placed along the x-axis. Point P is at distance r from the midpoint of the dipole, where r >> a.

Step 2: Calculate the electric field due to +q at point P.
The distance of +q from P is (r - a).
Using Coulomb's law, the electric field E+ is:
E+ = kq / (r - a)2, directed away from +q (towards +x-axis).

Step 3: Calculate the electric field due to -q at point P.
The distance of -q from P is (r + a).
The electric field E- is:
E- = kq / (r + a)2, directed towards -q (also towards +x-axis).

Step 4: Find the net electric field at P.
Since both fields are in the same direction, the net field Enet is:
Enet = E+ - E- = kq [1/(r - a)2 - 1/(r + a)2].

Step 5: Simplify the expression for r >> a.
Using binomial approximation, (r ± a)-2 ≈ r-2 (1 ∓ 2a/r).
Substituting, we get:
Enet ≈ kq [ (1 + 2a/r)/r2 - (1 - 2a/r)/r2 ] = kq (4a/r3).
Since dipole moment p = q × 2a, the final expression is:
Eaxial = 2kp / r3.

Direction: The electric field at point P is along the +x-axis (from -q to +q).
Question 20:
A charged spherical conductor of radius R carries a total charge Q. A student is asked to determine the electric field and electric potential at a point P located at a distance r from the center of the sphere, where r > R.

Explain how the student can derive the expressions for the electric field and electric potential at point P. Also, compare the behavior of the electric field and potential inside the conductor (for r < R).

Answer:

The student can derive the expressions for the electric field and electric potential at point P (r > R) using Gauss's law and the properties of a spherical conductor:


Electric Field (r > R):
Step 1: Choose a Gaussian surface as a sphere of radius r concentric with the conductor.
Step 2: Apply Gauss's law: ∮E · dA = Qenc/ε0.
Since the field is radial and symmetric, E is constant over the Gaussian surface.
Thus, E × 4πr2 = Q/ε0.
Step 3: Solve for E:
E = Q / (4πε0r2), directed radially outward.

Electric Potential (r > R):
Step 1: The potential at infinity is taken as zero.
Step 2: Integrate the electric field from infinity to r:
V = -∫∞r E · dr = -∫∞r (Q / 4πε0r2) dr.
Step 3: Solve the integral:
V = [Q / 4πε0r]∞r = Q / 4πε0r.

Behavior Inside the Conductor (r < R):
  • Electric Field: Inside a conductor in electrostatic equilibrium, the electric field is zero (E = 0) because charges redistribute to cancel any internal field.
  • Electric Potential: The potential inside the conductor is constant and equal to its value at the surface (V = Q / 4πε0R), as no work is done to move a charge inside the conductor.

Comparison:
The electric field drops to zero inside the conductor, while the potential remains constant. Outside the conductor, both the field and potential decrease with distance, but the field follows an inverse-square law, whereas the potential follows an inverse relationship.
Question 21:
A student performs an experiment to study the electric field due to a dipole. The student places an electric dipole consisting of two equal and opposite charges (±q) separated by a distance 2a along the x-axis. The dipole moment p is directed from -q to +q.

(i) Derive the expression for the electric field at a point on the equatorial line of the dipole.
(ii) How does the field vary with distance r from the dipole when r >> a?

Answer:

(i) Electric field on the equatorial line of a dipole:
Consider a point P on the equatorial line at distance r from the center of the dipole.
The electric field due to +q at P is: E+ = kq / (r2 + a2) along the line joining +q and P.
The electric field due to -q at P is: E- = kq / (r2 + a2) along the line joining P and -q.
The vertical components cancel out, and the horizontal components add up.
Net electric field E = 2E+cosθ, where cosθ = a / √(r2 + a2).
Substituting, we get: E = kp / (r2 + a2)3/2, where p = 2aq is the dipole moment.

(ii) Variation of field with distance when r >> a:
For r >> a, a2 becomes negligible compared to r2.
Thus, the expression simplifies to: E ≈ kp / r3.
This shows that the electric field due to a dipole on the equatorial line decreases with the cube of the distance (E ∝ 1/r3), unlike a point charge where E ∝ 1/r2.

Question 22:
A thin conducting spherical shell of radius R carries a charge Q uniformly distributed on its surface. Using Gauss's law, answer the following:

(i) Find the electric field at a point outside the shell at distance r from the center (where r > R).
(ii) What is the electric field inside the shell (where r < R)? Justify your answer.

Answer:

(i) Electric field outside the shell (r > R):
By Gauss's law, the electric flux through a Gaussian surface (a sphere of radius r) is: Φ = Qenc / ε0.
Since the charge Q is uniformly distributed on the shell, Qenc = Q.
The electric field E is radial and uniform over the Gaussian surface, so: Φ = E × 4πr2.
Equating the two expressions: E × 4πr2 = Q / ε0.
Solving for E, we get: E = Q / (4πε0r2).
This shows that the field outside the shell behaves as if all the charge were concentrated at the center.

(ii) Electric field inside the shell (r < R):
For a Gaussian surface inside the shell (r < R), the enclosed charge Qenc = 0 because all charge resides on the surface.
Applying Gauss's law: Φ = E × 4πr2 = 0.
Thus, E = 0 inside the shell.
Physical interpretation: The electric field inside a charged conducting shell is zero due to electrostatic shielding, as the charges redistribute themselves to cancel any internal field.

Question 23:
A student performs an experiment to study the electric field due to a uniformly charged thin spherical shell. She observes that inside the shell, the electric field is zero, while outside, it behaves as if the entire charge is concentrated at the center. Based on this, answer the following:

(i) Explain why the electric field inside the shell is zero using Gauss's law.
(ii) If the shell has a radius of 10 cm and a charge of 5 μC, calculate the electric field at a point 15 cm from the center.
Answer:

(i) According to Gauss's law, the electric flux through a closed surface is proportional to the charge enclosed by that surface. For a uniformly charged spherical shell, if we consider a Gaussian surface inside the shell (radius r < shell radius R), the enclosed charge is zero because all the charge lies on the outer surface. Thus, the electric field inside the shell is zero.


(ii) Given:
Charge (Q) = 5 μC = 5 × 10-6 C
Distance from center (r) = 15 cm = 0.15 m
Using the formula for the electric field outside a spherical shell:
E = (1/4πε0) * (Q/r2)
Substituting the values:
E = (9 × 109 Nm2/C2) * (5 × 10-6 C) / (0.15 m)2
E = 2 × 106 N/C (outward direction).

Question 24:
Two point charges, +4 μC and -2 μC, are placed 12 cm apart in air. A third charge of +1 μC is placed midway between them. Analyze the situation and answer:

(i) Calculate the net force experienced by the +1 μC charge due to the other two charges.
(ii) What would be the direction of the net force? Justify your answer.
Answer:

(i) Given:
Charge 1 (q1) = +4 μC = 4 × 10-6 C
Charge 2 (q2) = -2 μC = -2 × 10-6 C
Charge 3 (q3) = +1 μC = 1 × 10-6 C
Distance between q1 and q2 = 12 cm ⇒ distance of q3 from each = 6 cm = 0.06 m.

Force on q3 due to q1 (repulsive, away from q1):
F1 = (9 × 109) * (4 × 10-6 * 1 × 10-6) / (0.06)2 = 10 N
Force on q3 due to q2 (attractive, toward q2):
F2 = (9 × 109) * (2 × 10-6 * 1 × 10-6) / (0.06)2 = 5 N
Net force = F1 + F2 = 10 N + 5 N = 15 N (since both forces act in the same direction).


(ii) The net force direction is toward the -2 μC charge because:

  • The repulsive force from +4 μC pushes q3 away (toward -2 μC).
  • The attractive force from -2 μC pulls q3 toward it.
  • Both forces add up in the same direction.

Question 25:
A student performs an experiment to study the electric field due to a dipole. The student places a dipole consisting of charges +q and -q separated by a distance 2a along the x-axis. Using a test charge, the student measures the electric field at a point P located at a distance r from the midpoint of the dipole along its axial line.

Based on the experiment, derive the expression for the electric field at point P due to the dipole. Also, explain how the direction of the electric field is determined.

Answer:

The electric field at point P along the axial line of a dipole can be derived as follows:


1. The dipole consists of charges +q and -q separated by distance 2a.


2. The electric field due to +q at point P (distance r - a from +q) is:
E+ = kq / (r - a)2, directed away from +q.


3. The electric field due to -q at point P (distance r + a from -q) is:
E- = kq / (r + a)2, directed toward -q.


4. The net electric field Eaxial is the difference between E+ and E- since they are in the same direction along the axial line:
Eaxial = E+ - E- = kq [1/(r - a)2 - 1/(r + a)2].


5. For r >> a, simplifying gives:
Eaxial ≈ 2kp / r3, where p = q × 2a is the dipole moment.


The direction of the electric field is along the dipole axis from -q to +q for points on the axial line. This is because the field due to +q dominates at point P (as it is closer).

Question 26:
A charged spherical conductor of radius R carries a total charge Q. A student wants to determine the electric field and potential at a point P located at a distance r from the center of the conductor, where r > R.

Using Gauss's law, derive the expressions for the electric field and electric potential at point P. Also, explain why the entire charge on the conductor can be assumed to be concentrated at its center for this calculation.

Answer:

The electric field and potential at point P (r > R) can be derived using Gauss's law as follows:


1. Electric Field:
By Gauss's law, for a spherical Gaussian surface of radius r enclosing the conductor:
∮E · dA = Qenc / ε0.
Since the field is radial and uniform:
E × 4πr2 = Q / ε0.
Thus, E = Q / (4πε0r2), directed radially outward.


2. Electric Potential:
The potential at P is the work done to bring a unit charge from infinity to P:
V = -∫∞r E · dr.
Substituting E:
V = -∫∞r (Q / 4πε0r2) dr = Q / (4πε0r).


3. Charge Concentration at Center:
A spherical conductor has symmetrical charge distribution on its surface. By the shell theorem, the field outside a charged sphere is identical to that of a point charge at its center. Hence, the entire charge Q can be treated as concentrated at the center for calculations outside the sphere.

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