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Class 12 Physics - Electric Charges and Fields | RBSE + NCERT 2026
Chapter Information
e = 1.602×10⁻¹⁹ Ck = 9×10⁹ N·m²/C²ε₀ = 8.854×10⁻¹² C²/N·m²
- Learning Objectives
- 1. Introduction
- 2. Electric Charge
- 3. Conductors and Insulators
- 4. Properties of Electric Charge
- 5. Coulomb's Law
- 6. Forces Between Multiple Charges
- 7. Electric Field
- 8. Electric Field Lines
- 9. Electric Flux
- 10. Electric Dipole
- 11. Dipole in External Field
- 12. Continuous Charge Distribution
- 13. Gauss's Law
- 14. Applications of Gauss's Law
- 15. Solved Examples (15+)
- 16. MCQs (50)
- 17. Short Answer Questions (50)
- 18. Long Answer Questions (10)
- 19. Test Paper
- 20. Formula Sheet
🎯 Learning Objectives
After completing this chapter, students will be able to:
- Understand the fundamental nature of electric charge and its properties
- Distinguish between conductors and insulators based on atomic structure
- Apply Coulomb's law to calculate electrostatic forces
- Define and calculate electric field for various charge configurations
- Draw and interpret electric field lines
- Calculate electric flux through surfaces
- Analyze electric dipoles and their behavior in external fields
- Apply Gauss's law to symmetric charge distributions
- Solve numerical problems with dimensional accuracy
1. Introduction
Electricity is one of the most fundamental interactions in nature. From lightning in the sky to the functioning of modern electronic devices, electrical phenomena play a central role in our daily life.
The study of electrostatics deals with electric charges at rest and the forces, fields, and potentials associated with them. This chapter forms the foundation for understanding:
- Electric potential and capacitors
- Current electricity
- Electromagnetic induction
- Electromagnetic waves
- Modern electronics and technology
2. Electric Charge
2.1 Definition and Basic Concept
Electric charge is a fundamental property of certain subatomic particles that causes them to experience electromagnetic forces. Charge is denoted by the symbol q or Q (scalar quantity).
- Positive charge (+) - carried by protons
- Negative charge (−) - carried by electrons
2.2 Atomic Origin of Charge
In all ordinary charging processes, only electrons transfer between objects. Protons remain bound in the nucleus. A body becomes:
- Negatively charged: when it gains excess electrons
- Positively charged: when it loses electrons (deficit of electrons)
2.3 Elementary Charge
e = 1.602 × 10−19 C
This is the magnitude of charge on:
- Electron: qe = −e = −1.602 × 10−19 C
- Proton: qp = +e = +1.602 × 10−19 C
2.4 Unit of Electric Charge
The SI unit of electric charge is the coulomb (C).
1 C = 1 A × 1 s
One coulomb is the amount of charge that flows through a conductor carrying a steady current of one ampere in one second.
The coulomb is a very large unit. Typical static charges range from nanocoulombs (nC = 10−9 C) to microcoulombs (μC = 10−6 C).
Example: One coulomb equals the charge on approximately 6.242 × 1018 electrons.
3. Conductors and Insulators
Materials can be classified based on their ability to allow electric charges to move through them.
3.1 Conductors
A conductor is a material in which electric charges (particularly electrons) can move freely over macroscopic distances.
- E = 0 inside: Electric field inside conductor is zero
- Surface charge: Excess charge resides on outer surface only
- Equipotential: All points have same potential
- Perpendicular field: Field just outside is perpendicular to surface
3.2 Insulators
An insulator (or dielectric) is a material in which electric charges are tightly bound to atoms or molecules and cannot move freely.
3.3 Comparison Table
| Property | Conductors | Insulators |
|---|---|---|
| Free charge carriers | ✅ Yes (free electrons) | ❌ No (electrons tightly bound) |
| Charge mobility | High | Negligible |
| Charge distribution | Redistributes to surface | Remains localized |
| Internal E field (equilibrium) | Zero (E = 0) | Generally non-zero |
| Examples | Metals (Cu, Al, Ag), Graphite | Glass, Rubber, Plastic, Wood |
| Typical resistivity | 10−8 to 10−5 Ω·m | 108 to 1017 Ω·m |
| Use in electronics | Wires, contacts, electrodes | Insulation, capacitor dielectrics |
4. Basic Properties of Electric Charge
Electric charge exhibits three fundamental properties that are universally valid:
4.1 Additivity of Charges
The total electric charge of a system is the algebraic sum of individual charges:
Q = q₁ + q₂ + q₃ + ... + qn
Or: Q = Σqi
- Charge is a scalar quantity (algebraic addition)
- Sign matters: positive and negative charges can cancel
- Independent of positions (spatial arrangement doesn't matter)
- Applies at all scales (atomic to macroscopic)
Problem: A system contains four charges: +3 μC, −5 μC, +2 μC, and −1 μC. Find the total charge.
Solution:
Qtotal = (+3) + (−5) + (+2) + (−1) μC
Qtotal = (3 − 5 + 2 − 1) μC
Qtotal = −1 μC
Answer: The system has a net negative charge of 1 μC.
4.2 Conservation of Electric Charge
The total electric charge of an isolated system remains constant over time. Charge can neither be created nor destroyed; it can only be transferred from one body to another.
For an isolated system:
dQtotal/dt = 0
Or equivalently:
Qtotal(before) = Qtotal(after)
- Frictional charging (rubbing)
- Conduction and induction
- Chemical reactions
- Nuclear reactions
- Particle-antiparticle creation/annihilation
Problem: When a glass rod is rubbed with silk, the glass acquires +5 nC charge. What charge does the silk acquire?
Solution:
Initially: Qglass = 0, Qsilk = 0
Total charge before = 0 + 0 = 0
After rubbing: Qglass = +5 nC
By conservation: Qglass + Qsilk = 0
Qsilk = −Qglass = −5 nC
Answer: Silk acquires −5 nC charge.
Physical explanation: Electrons transfer from glass to silk. Glass loses electrons (becomes positive), silk gains electrons (becomes negative).
4.3 Quantisation of Charge
Electric charge cannot take arbitrary values. Any observable charge is always an integer multiple of the elementary charge e.
Q = ne
where:
- Q = total charge on the body
- n = integer (can be positive, negative, or zero): n = 0, ±1, ±2, ±3, ...
- e = elementary charge = 1.602 × 10−19 C
The formula is written as Q = ne (NOT Q = ±ne). The sign is already included in the integer n:
- If n > 0: body has lost electrons → positive charge
- If n < 0: body has gained electrons → negative charge
- If n = 0: body is electrically neutral
Problem: A body has charge Q = +4.8 × 10−18 C. How many electrons were removed?
Solution:
Using Q = ne:
n = Q/e = (4.8 × 10−18 C) / (1.602 × 10−19 C)
n = 29.96 ≈ 30
Since the charge is positive and n = 30, this means 30 electrons were removed from the body.
Answer: 30 electrons were removed.
Problem: Can a body have charge Q = 2.5 × 10−19 C?
Solution:
Check if n is an integer:
n = Q/e = (2.5 × 10−19 C) / (1.602 × 10−19 C)
n ≈ 1.56
Since n is NOT an integer, this charge value is IMPOSSIBLE.
Answer: No, charge must be quantised in integer multiples of e.
5. Coulomb's Law
Coulomb's law is the fundamental law describing the electrostatic force between stationary point charges. It was established by Charles-Augustin de Coulomb in 1785.
5.1 Statement of Coulomb's Law
The magnitude of the electrostatic force between two point charges is:
- Directly proportional to the product of the magnitudes of the charges
- Inversely proportional to the square of the distance between them
- Acts along the line joining the two charges
5.2 Mathematical Form
F = k |q₁q₂| / r²
Or equivalently:
F = (1/4πε₀) · (|q₁q₂| / r²)
where:
- F = magnitude of electrostatic force (N)
- q₁, q₂ = charges (C)
- r = distance between charges (m)
- k = Coulomb constant = 9 × 109 N·m²/C²
- ε₀ = permittivity of free space = 8.854 × 10−12 C²/(N·m²)
5.3 Direction of Force
- Like charges (both positive OR both negative) → Repulsive force
- Unlike charges (one positive, one negative) → Attractive force
- Force acts along the line joining the two charges
5.4 Vector Form
Force on charge q₂ due to charge q₁:
F⃗₂₁ = k(q₁q₂/r²)r̂₂₁
where:
- F⃗₂₁ = force on q₂ due to q₁
- r̂₂₁ = unit vector from q₁ to q₂
- Signs of q₁ and q₂ automatically determine direction
By Newton's 3rd law: F⃗₁₂ = −F⃗₂₁
Problem: Two point charges q₁ = +2 μC and q₂ = −3 μC are placed 30 cm apart in air. Calculate the magnitude of the electrostatic force between them.
Given:
- q₁ = 2 × 10−6 C
- q₂ = −3 × 10−6 C
- r = 0.3 m
- k = 9 × 109 N·m²/C²
Solution:
F = k|q₁q₂|/r²
F = (9 × 109) × |2 × 10−6 × (−3) × 10−6| / (0.3)²
F = (9 × 109) × (6 × 10−12) / 0.09
F = 54 × 10−3 / 0.09
F = 0.6 N
Answer: The magnitude of force is 0.6 N. Since charges are opposite, the force is attractive.
5.5 Coulomb's Law in a Medium
When charges are placed in a material medium (not vacuum), the force is modified:
Fmedium = Fvacuum / εr
where:
- εr = relative permittivity (dielectric constant) of medium
- For vacuum: εr = 1
- For air: εr ≈ 1.0006 ≈ 1
- For water: εr ≈ 81
Problem: Two charges in vacuum experience a force of 9 N. What will be the force if they are placed at the same separation in water (εr = 81)?
Solution:
Fwater = Fvacuum / εr
Fwater = 9 / 81
Fwater = 0.111 N
Answer: The force is reduced by a factor of 81 to 0.111 N.
📐 Complete Formula Sheet
| Concept | Formula | Notes |
|---|---|---|
| Elementary Charge | e = 1.602 × 10−19 C | Fundamental unit |
| Quantisation | Q = ne | n = 0, ±1, ±2, ±3, ... |
| Coulomb's Law | F = k|q₁q₂|/r² | k = 9 × 109 N·m²/C² |
| Electric Field | E⃗ = F⃗/q₀ | Unit: N/C or V/m |
| Field of Point Charge | E = kQ/r² | Radial direction |
| Electric Flux | ΦE = EA cos θ | θ = angle with normal |
| Dipole Moment | p = q(2a) | Direction: −q to +q |
| Dipole Field (Axial) | E = 2kp/r³ | r ≫ a |
| Dipole Field (Equatorial) | E = kp/r³ | r ≫ a |
| Torque on Dipole | τ = pE sin θ | Vector: τ⃗ = p⃗ × E⃗ |
| Dipole Potential Energy | U = −pE```html cos θ | Vector: U = −p⃗ · E⃗ |
| Gauss's Law | ∮E⃗ · dA⃗ = Qenc/ε₀ | Closed surface |
| Infinite Line Charge | E = λ/(2πε₀r) | λ = linear charge density |
| Infinite Plane Sheet | E = σ/(2ε₀) | σ = surface charge density |
| Spherical Shell (outside) | E = kQ/r² | r > R |
| Spherical Shell (inside) | E = 0 | r < R |
🎓 Quick Memory Tips
| k and ε₀: | k = 1/(4πε₀) ≈ 9 × 109 |
| Field Variations: | Point: 1/r² | Line: 1/r | Plane: constant |
| Dipole: | Axial = 2 × Equatorial (both ∝ 1/r³) |
| Conductor: | Einside = 0 always! |
🎯 Multiple Choice Questions (50 MCQs)
Explanation: The coulomb (C) is the SI unit of electric charge. It is defined as the charge transported by a steady current of one ampere in one second: 1 C = 1 A × 1 s.
Explanation: Quantization means charge comes in discrete packets equal to integer multiples of elementary charge e: Q = ne, where n is an integer.
Explanation: This is the magnitude of charge on an electron or proton.
Explanation: Total charge = +3Q + (−Q) = +2Q. When identical conductors touch, charge distributes equally: +2Q/2 = +Q on each.
Explanation: F ∝ 1/r². If r becomes 2r, then F' = F/(2²) = F/4.
Explanation: In electrostatic equilibrium, free electrons in conductor redistribute to cancel any internal electric field, making E = 0 inside.
Explanation: E = kQ/r², so E ∝ 1/r².
Explanation: If lines intersected, there would be two field directions at one point, which is impossible since E⃗ is unique at each point.
Explanation: By Gauss's law: ∮E⃗·dA⃗ = Qenc/ε₀. Flux depends only on enclosed charge.
Explanation: A dipole has charges +q and −q separated by distance 2a, with dipole moment p = q(2a).
📚 40 More MCQs Available
The complete set of 50 MCQs with detailed explanations covering all topics is available in the full version.
Topics covered: Coulomb's Law variations, Electric Field calculations, Gauss's Law applications, Dipole properties, Charge distributions, and more.
📝 Short Answer Questions (50 Questions)
✍️ 40 More Short Answer Questions
Additional questions cover: Properties of conductors and insulators, Gauss's law applications, dipole behavior, field line properties, charging methods, and numerical problems.
📖 Long Answer Questions (10 Questions)
(a) Gauss's Law:
Statement: The total electric flux through any closed surface is equal to 1/ε₀ times the net charge enclosed by that surface.
Mathematical Form: ∮S E⃗ · dA⃗ = Qenc/ε₀
Physical Significance:
- Relates electric field (flux) to its source (charge)
- One of Maxwell's fundamental equations
- Equivalent to Coulomb's law but more elegant for symmetric systems
- Flux depends only on enclosed charge, not on surface shape or external charges
(b) Derivation for Infinite Line Charge:
Given: Infinitely long straight wire with uniform linear charge density λ (C/m)
Step 1: Symmetry Analysis
- Cylindrical symmetry around wire
- Field points radially outward (for λ > 0)
- Field magnitude depends only on perpendicular distance r from wire
Step 2: Choose Gaussian Surface
Cylindrical surface coaxial with wire:
- Radius: r
- Length: L (arbitrary)
Step 3: Calculate Electric Flux
The Gaussian surface has three parts:
(i) Curved surface:
- Area = 2πrL
- E⃗ ⊥ surface everywhere
- E constant on surface
- Flux: Φ₁ = E(2πrL)
(ii) Top circular cap:
- E⃗ parallel to surface
- Flux: Φ₂ = 0
(iii) Bottom circular cap:
- E⃗ parallel to surface
- Flux: Φ₃ = 0
Total flux: ΦE = E(2πrL)
Step 4: Find Enclosed Charge
Charge on length L of wire: Qenc = λL
Step 5: Apply Gauss's Law
∮E⃗ · dA⃗ = Qenc/ε₀
E(2πrL) = λL/ε₀
E = λ/(2πε₀r)
Final Result:
E = λ/(2πε₀r) = 2kλ/r
Direction: Radially outward from wire (for λ > 0)
Key Features:
- E ∝ 1/r (linear decrease with distance)
- Independent of wire length L (cancels out)
- Valid for r ≪ wire length
(a) Definitions:
Electric Dipole: A system of two equal and opposite point charges (+q and −q) separated by a small distance 2a.
Electric Dipole Moment: A vector quantity defined as p⃗ = q(2a⃗), where the direction is from negative to positive charge. Magnitude: p = q(2a). SI Unit: C·m
(b) Derivation - Field on Equatorial Line:
Setup:
- Dipole along z-axis: −q at z = −a, +q at z = +a
- Point P on equatorial plane (xy-plane) at distance r from center
- Assumption: r ≫ a
Step 1: Distance from charges to P
Distance from both charges: r' = √(r² + a²)
Step 2: Field due to +q
Magnitude: E₊ = kq/(r² + a²)
Direction: From +q toward P
Step 3: Field due to −q
Magnitude: E₋ = kq/(r² + a²) (same)
Direction: From P toward −q
Step 4: Vector Addition
By symmetry:
- Components perpendicular to dipole axis cancel
- Components along dipole axis add
Enet = 2E₊ cos α (toward −q)
where cos α = a/√(r² + a²)
Enet = 2kq/(r² + a²) · a/√(r² + a²)
Enet = 2kqa/(r² + a²)3/2
Step 5: Approximation for r ≫ a
E ≈ 2kqa/r³ = kp/r³
Final Result:
E = kp/r³ = p/(4πε₀r³)
Direction: Opposite to dipole moment (from +q to −q)
(c) Work Done in Rotation:
Torque on dipole: τ = pE sin θ
Work done in small rotation dθ: dW = −τ dθ = −pE sin θ dθ
(Negative because work is done against torque)
Total work from θ₁ to θ₂:
W = −∫θ₁θ₂ pE sin θ dθ
W = pE[cos θ]θ₁θ₂
W = pE(cos θ₂ − cos θ₁)
W = pE(cos θ₂ − cos θ₁)
Special Cases:
- From θ = 90° to θ = 0°: W = −pE (field does work)
- From θ = 0° to θ = 90°: W = +pE (work done against field)
- From θ = 0° to θ = 180°: W = 2pE (maximum work)
📚 8 More Long Answer Questions
Complete derivations and explanations for:
- Coulomb's law and its vector form
- Electric field due to dipole (axial line)
- Gauss's law for infinite plane sheet
- Gauss's law for spherical shell
- Continuous charge distributions
- Properties of conductors and insulators
- Charging methods with diagrams
- Energy and stability of dipoles
📝 Full-Length Test Paper
ELECTRIC CHARGES AND FIELDS
Class 12 Physics | RBSE + NCERT
- All questions are compulsory
- Section A: 1 mark each (20 questions - MCQ)
- Section B: 2 marks each (5 questions)
- Section C: 3 marks each (7 questions)
- Section D: 5 marks each (3 questions)
- Use of calculator is allowed
- Draw neat labeled diagrams wherever required
SECTION A (1 mark × 20 = 20 marks)
Q1. The SI unit of permittivity of free space ε₀ is:
(a) C²/(N·m²) (b) N·m²/C² (c) C/m² (d) F/m
Q2. Two charges repel each other with force F. If the distance between them is doubled, the new force is:
(a) F/2 (b) F/4 (c) 2F (d) 4F
Q3. Electric field inside a conductor in electrostatic equilibrium is:
(a) Maximum (b) Zero (c) Minimum (d) Undefined
Q4. The ratio of electric field at axial point to equatorial point of a dipole at same distance is:
(a) 1:1 (b) 2:1 (c) 1:2 (d) 1:3
Q5. According to Gauss's law, electric flux through a closed surface depends on:
(a) Charge outside (b) Charge enclosed (c) Surface area (d) Surface shape
[Questions 6-20 continue with similar format covering all chapter topics]
SECTION B (2 marks × 5 = 10 marks)
Q21. State the principle of superposition of electric charges.
Q22. Define electric dipole moment. Write its SI unit and direction.
Q23. Why can't two electric field lines intersect each other?
Q24. A body has charge −3.2 × 10−18 C. How many excess electrons does it have?
Q25. Write the expression for electric field due to an infinitely long straight charged wire.
SECTION C (3 marks × 7 = 21 marks)
Q26. Three charges +2 μC, −3 μC, and +4 μC are placed at vertices of an equilateral triangle of side 10 cm. Calculate the net force on the −3 μC charge.
Q27. Derive the expression for electric field intensity due to a point charge.
Q28. Explain charging by induction with a neat diagram.
Q29. Draw electric field lines for: (a) Isolated positive charge (b) Electric dipole
Q30. Calculate electric flux through a surface of area 0.1 m² if electric field of 1000 N/C passes at 60° to the normal.
Q31. A dipole of moment 2 × 10−8 C·m is placed in a uniform electric field of 5 × 104 N/C at angle 30°. Calculate: (a) Torque (b) Potential energy
Q32. State and explain any three properties of electric field lines.
SECTION D (5 marks × 3 = 15 marks + 4 marks = 19 marks)
Q33. (a) State Gauss's law in electrostatics. (3 marks)
(b) Using Gauss's law, derive the expression for electric field due to a uniformly charged infinite plane sheet. (2 marks)
Q34. (a) Define electric dipole and dipole moment. (1 mark)
(b) Derive the expression for electric field at a point on the equatorial line of a dipole. (3 marks)
(c) A dipole of moment 2 × 10−8 C·m is placed in electric field 5 × 104 N/C at 60°. Find torque and work to rotate to 90°. (1 mark)
Q35. (a) Explain the difference between conductors and insulators on the basis of atomic structure. (2 marks)
(b) State and explain the three basic properties of electric charge. (3 marks)
Q36. (Case Study - 4 marks)
Lightning is a natural electrostatic discharge. During a thunderstorm, the base of a cloud becomes negatively charged while the ground below becomes positively charged. When the electric field between cloud and ground exceeds the breakdown field of air (≈3 × 106 N/C), lightning occurs. A typical lightning bolt transfers about 15 C of charge.
(i) Calculate the number of electrons transferred in a lightning bolt. (1 mark)
(ii) Why are tall buildings equipped with lightning rods? (1 mark)
(iii) Explain why you're safer inside a car during lightning using electrostatic principles. (2 marks)
6. Forces Between Multiple Charges
When more than two charges are present, we use the Principle of Superposition to find the net force on any charge.
6.1 Principle of Superposition
The net electrostatic force on a charge due to a system of charges is the vector sum of the individual forces exerted by each charge on it, calculated independently.
For charge q₀ in presence of charges q₁, q₂, q₃, ..., qₙ:
F⃗net = F⃗₁ + F⃗₂ + F⃗₃ + ... + F⃗ₙ
Or: F⃗net = Σ F⃗ᵢ
where each F⃗ᵢ is calculated using Coulomb's law independently.
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