📋 Index
- Electric Charge — Basics
- Properties of Electric Charge
- Conductors, Insulators & Semiconductors
- Charging by Induction
- Coulomb’s Law
- Principle of Superposition
- Electric Field
- Electric Field Lines
- Electric Dipole
- Dipole in Uniform Electric Field
- Continuous Charge Distribution
- Gauss’s Theorem & Applications
- Important Formulas (Quick Revision)
- NCERT Questions & Solutions
1. Electric Charge — Basics
Electric Charge (q or Q): Physical property of matter that causes it to experience force in an electromagnetic field. SI unit = Coulomb (C).
Types of Charges
- Positive (+): Appears on glass rod rubbed with silk
- Negative (−): Appears on ebonite rod rubbed with fur
Fundamental Law: Like charges repel, unlike charges attract.
Sure Test of Electrification: Repulsion is the only sure test. Attraction can occur between charged & uncharged body (induction), but repulsion means both bodies are charged.
Methods of Charging
- Friction: Rubbing two bodies transfers electrons
- Conduction: Direct contact transfers charge
- Induction: Charging without contact
ADD_YOUR_DIAGRAM(eg: Glass rod + silk, Ebonite rod + fur, Attraction & Repulsion)_LINK_HERE
2. Properties of Electric Charge 5 MARKS
(A) Additivity
- Charges add algebraically
- Qtotal = q1 + q2 + q3 + …
- Example: (+3C) + (−5C) + (+2C) = 0 C → Neutral system
(B) Conservation of Charge
- Charge can neither be created nor destroyed
- Only transferred from one body to another
- Total charge of isolated system remains constant
- Example: Glass (+) + Silk (−) = 0 (before & after rubbing)
(C) Quantization of Charge
q = ±ne where n = 1, 2, 3, …
e = elementary charge = 1.6 × 10−19 C
- Charge exists only in integer multiples of e
- Macroscopically, n is very large → charge appears continuous
- Example: Charge −4.8 × 10−19 C means n = 3 excess electrons
Exam Tip: Quantization, Conservation & Additivity are frequently asked as 3-mark or 5-mark questions. Write definition + formula + example for full marks.
3. Conductors, Insulators & Semiconductors
| Property | Conductors | Insulators | Semiconductors |
|---|---|---|---|
| Free electrons | Very large | Negligible | Moderate |
| Conductivity | Very high | Very low | Intermediate |
| Resistivity | 10−8 Ωm | 108–1016 Ωm | 10−5–106 Ωm |
| Examples | Cu, Ag, Al, Fe, graphite | Glass, rubber, plastic, mica | Si, Ge, GaAs |
| Temp effect | Conductivity ↓ | No change | Conductivity ↑ |
Point to Remember: Insulators can be charged — charge just cannot flow through them. Charge stays at the place where it is given.
4. Charging by Induction 3 MARKS
Induction: Process of charging a neutral body by bringing a charged body near it (without contact), then grounding it, and removing the ground.
Steps to Charge Positively by Induction (using −vely charged rod)
- Bring −vely charged rod near neutral conductor → electrons repel to far side
- Ground the conductor (touch with finger) while rod is near → excess electrons escape to earth
- Remove the ground first → conductor loses electrons
- Remove the rod → conductor has net positive charge
ADD_YOUR_DIAGRAM(eg: Charging by Induction – 4 steps showing rod, sphere, grounding, net charge)_LINK_HERE
Key Points
- No contact between charging body and body being charged
- Induced charge is opposite in nature to inducing charge
- Magnitude of induced charge = magnitude of inducing charge
- Original charge on inducing body does not change
- Works best with conductors
Application: Lightning conductors on buildings work on induction. Charged cloud induces opposite charge on pointed conductor → corona discharge neutralizes cloud safely.
5. Coulomb’s Law MOST IMP
Statement
- Force between two point charges is directly proportional to product of their magnitudes
- Inversely proportional to square of distance between them
- Acts along line joining the two charges
F = k |q1q2| / r2
k = 1/4πε0 = 9 × 109 N m2 C−2
ε0 = 8.854 × 10−12 C2 N−1 m−2
Vector Form
F⃗12 = (1/4πε0) · (q1q2/r2) · r̂12
F⃗12 = Force on q1 due to q2
r̂12 = unit vector from q2 to q1
q1q2 > 0 → Repulsion | q1q2 < 0 → Attraction
Newton’s Third Law: F⃗12 = −F⃗21 (equal & opposite)
Coulomb’s Law in a Medium
Fmedium = Fvacuum / εr
εr = relative permittivity (dielectric constant)
ε = ε0εr = absolute permittivity of medium
Coulomb vs Gravitation
| Feature | Coulomb Force | Gravitational Force |
|---|---|---|
| Nature | Attractive or Repulsive | Only Attractive |
| Constant | k = 9 × 109 (very large) | G = 6.67 × 10−11 (very small) |
| Depends on | Charge (can be + or −) | Mass (always +) |
| Medium | Depends on εr | Independent of medium |
| Relative strength | Electrostatic force ≈ 1039 times stronger than gravity | |
ADD_YOUR_DIAGRAM(eg: Two point charges q1 and q2 with distance r, force vectors F12 and F21)_LINK_HERE
6. Principle of Superposition IMPORTANT
Statement: Total force on any charge due to a group of charges = vector sum of individual forces exerted by each charge, taken one at a time.
F⃗1 = F⃗12 + F⃗13 + F⃗14 + … + F⃗1n
F⃗1 = Σ(i=2 to n) F⃗1i
- Force between any two charges is not affected by presence of other charges
- Each pair interacts as if others don’t exist
- Valid because Coulomb’s force is linear
Problem Solving Tip: For 3 charges at corners of equilateral triangle → calculate two forces, find angle between them (60°), use vector addition formula: Fnet = √(F12 + F22 + 2F1F2cosθ)
ADD_YOUR_DIAGRAM(eg: Three charges at equilateral triangle corners, force vectors on one charge, resultant)_LINK_HERE
7. Electric Field MOST IMP
Electric Field Intensity (E): Electrostatic force per unit positive test charge placed at that point.
E⃗ = F⃗/q0 (q0 → 0)
SI Unit: N/C or V/m
Dimensions: [MLT−3A−1]
Why q0 → 0?
- Large test charge would disturb original charge distribution
- Infinitesimally small q0 ensures it doesn’t alter the field being measured
Electric Field due to Point Charge
E = (1/4πε0) · |q|/r2
Direction: Radially outward (+q), Radially inward (−q)
Vector Form
E⃗ = (1/4πε0) · (q/r2) · r̂
r̂ = unit vector from source charge to field point
Superposition for Electric Field
E⃗ = E⃗1 + E⃗2 + E⃗3 + … + E⃗n
E⃗i = (1/4πε0) · (qi/ri2) · r̂i
Physical Significance: Electric field exists independently of test charge. It carries energy & momentum. It tells the force per unit charge any charge would experience at that point.
ADD_YOUR_DIAGRAM(eg: Point charge +q with radial field lines outward, -q with field lines inward)_LINK_HERE
8. Electric Field Lines 3 MARKS
Definition: Imaginary curve drawn through space such that tangent at any point gives direction of electric field vector at that point.
Properties (Very Important for Exams)
- Originate from +ve charges, terminate on −ve charges
- Tangent at any point = direction of E at that point
- Never intersect each other (would imply two directions of E at one point → impossible)
- Density of lines ∝ strength of E (closer lines = stronger field)
- Meet conductor surface at 90° (perpendicular)
- Never form closed loops in electrostatics (unlike magnetic field lines)
Field Line Patterns
| Configuration | Pattern | Special Feature |
|---|---|---|
| Single +ve charge | Radial outward | Spread out with distance |
| Single −ve charge | Radial inward | Converge toward charge |
| Two +ve charges | Repel each other | Neutral point at midpoint (E = 0) |
| Dipole (+q & −q) | Curved from + to − | Symmetric pattern |
| Parallel plates | Parallel, equally spaced | Uniform E field |
Remember: Number of field lines leaving a charge is proportional to magnitude of charge. +2q has twice as many lines as +q.
ADD_YOUR_DIAGRAM(eg: Field line patterns – point charge, two like charges, dipole, parallel plates)_LINK_HERE
9. Electric Dipole 5 MARKS
Electric Dipole: Two equal & opposite charges (+q and −q) separated by small distance 2a.
Dipole Moment
p⃗ = q × 2a⃗
Magnitude: p = q × 2a
Direction: From −q to +q
SI Unit: C·m (or Debye; 1 D = 3.336 × 10−30 C·m)
Electric Field on Axial Line (End-on Position)
Eaxial = (1/4πε0) · 2pr / (r2 − a2)2
For r >> a: Eaxial ≈ (1/4πε0) · 2p/r3
Direction: Parallel to p⃗
Electric Field on Equatorial Line (Broadside-on)
Eeq = (1/4πε0) · p / (r2 + a2)3/2
For r >> a: Eeq ≈ (1/4πε0) · p/r3
Direction: Antiparallel to p⃗
Axial vs Equatorial Comparison
| Property | Axial | Equatorial |
|---|---|---|
| Magnitude (r >> a) | E = 2kp/r3 | E = kp/r3 |
| Ratio | Eaxial = 2 × Eequatorial | |
| Direction | Parallel to p⃗ | Antiparallel to p⃗ |
General Expression (Short Dipole, r >> a)
Er = (1/4πε0) · 2p cosθ/r3
Eθ = (1/4πε0) · p sinθ/r3
Enet = (1/4πε0) · (p/r3) √(3cos2θ + 1)
For short dipole: E ∝ 1/r3 (not 1/r2 like point charge). Faster fall-off because opposite charges partially cancel each other’s field.
ADD_YOUR_DIAGRAM(eg: Dipole with +q and -q, distance 2a, dipole moment vector p, axial and equatorial lines)_LINK_HERE
10. Dipole in Uniform External Electric Field 5 MARKS
Forces on Dipole
- Force on +q: F⃗+ = +qE⃗
- Force on −q: F⃗− = −qE⃗
- Net Force = 0 (equal & opposite)
- But Net Torque ≠ 0 (forces act at different points)
Torque on Dipole
τ⃗ = p⃗ × E⃗
Magnitude: τ = pE sinθ
θ = angle between p⃗ and E⃗
Torque at Different Orientations
| θ | Torque | State |
|---|---|---|
| 0° (parallel) | τ = 0 | Stable Equilibrium |
| 90° (perpendicular) | τ = pE (max) | Maximum torque |
| 180° (antiparallel) | τ = 0 | Unstable Equilibrium |
Potential Energy of Dipole
U = −pE cosθ = −p⃗ · E⃗
Work done in rotating from θ1 to θ2: W = pE(cosθ1 − cosθ2)
Potential Energy at Different Orientations
| θ | U | State |
|---|---|---|
| 0° | U = −pE (minimum) | Most stable |
| 90° | U = 0 | Reference |
| 180° | U = +pE (maximum) | Most unstable |
Application: Microwave ovens — water molecules are permanent dipoles. Oscillating E-field makes them rotate, creating heat through molecular friction.
ADD_YOUR_DIAGRAM(eg: Dipole in uniform E field at angle theta, forces F+ and F-, torque direction)_LINK_HERE
11. Continuous Charge Distribution
Types of Charge Density
| Type | Symbol | Formula | Unit |
|---|---|---|---|
| Linear | λ | λ = dq/dl | C/m |
| Surface | σ | σ = dq/dS | C/m2 |
| Volume | ρ | ρ = dq/dV | C/m3 |
Total Charge
- q = ∫ λ dl (line)
- q = ∫ σ dS (surface)
- q = ∫ ρ dV (volume)
Example: 2m wire with total charge 4 μC uniformly distributed.
λ = q/L = 4μC/2m = 2 μC/m
λ = q/L = 4μC/2m = 2 μC/m
12. Gauss’s Theorem & Applications MOST IMP
Electric Flux
ΦE = E⃗ · S⃗ = ES cosθ
S⃗ = area vector (normal to surface)
θ = angle between E⃗ and S⃗
SI Unit: N·m2/C or V·m
- Flux through closed surface: ΦE = ∮ E⃗ · dS⃗
- Positive flux = field lines leaving surface
- Negative flux = field lines entering surface
Gauss’s Theorem
Statement: Total electric flux through any closed surface = (1/ε0) × total charge enclosed by that surface.
∮ E⃗ · dS⃗ = qenclosed / ε0
Closed surface = Gaussian surface
Key Point: Only enclosed charge matters. External charges contribute zero net flux (field lines enter & exit, canceling out).
Application 1: Infinite Line Charge
E = λ / (2πε0r)
Direction: Radially outward (for +λ)
E ∝ 1/r (not 1/r2)
Application 2: Infinite Plane Sheet
E = σ / (2ε0)
Direction: Perpendicular to sheet
Independent of distance! (E = constant everywhere)
Application 3: Charged Spherical Shell
| Region | Electric Field | Key Result |
|---|---|---|
| Outside (r > R) | E = (1/4πε0) · Q/r2 | Behaves like point charge at center |
| Inside (r < R) | E = 0 | Zero field everywhere inside |
Application 4: Uniformly Charged Solid Sphere
| Region | Electric Field | Key Result |
|---|---|---|
| Outside (r > R) | E = (1/4πε0) · Q/r2 | Behaves like point charge at center |
| Inside (r < R) | E = (1/4πε0) · Qr/R3 = ρr/(3ε0) | E ∝ r (linear increase from center) |
ADD_YOUR_DIAGRAM(eg: Gaussian surfaces – cylinder for line charge, pillbox for plane, sphere for shell/solid sphere)_LINK_HERE
ADD_YOUR_DIAGRAM(eg: E vs r graph for solid sphere – linear inside, 1/r2 outside, max at r=R)_LINK_HERE
13. Important Formulas — Quick Revision LAST MINUTE
Constants
- k = 1/4πε0 = 9 × 109 N m2 C−2
- ε0 = 8.854 × 10−12 C2 N−1 m−2
- e = 1.6 × 10−19 C
All Formulas at a Glance
| Concept | Formula |
|---|---|
| Coulomb’s Force | F = k|q1q2|/r2 |
| Electric Field (point charge) | E = k|q|/r2 |
| Dipole Moment | p = q × 2a |
| Axial Field (dipole) | E = 2kp/r3 (r >> a) |
| Equatorial Field (dipole) | E = kp/r3 (r >> a) |
| Torque on Dipole | τ = pE sinθ |
| PE of Dipole | U = −pE cosθ |
| Work Done | W = pE(cosθ1 − cosθ2) |
| Electric Flux | Φ = ES cosθ |
| Gauss’s Law | ∮E⃗·dS⃗ = q/ε0 |
| Infinite Line | E = λ/(2πε0r) |
| Infinite Plane | E = σ/(2ε0) |
| Shell (outside) | E = kQ/r2 |
| Shell (inside) | E = 0 |
| Solid Sphere (inside) | E = kQr/R3 |
Field Dependence Summary:
Point charge → E ∝ 1/r2
Line charge → E ∝ 1/r
Plane sheet → E = constant
Dipole → E ∝ 1/r3
Shell inside → E = 0
Point charge → E ∝ 1/r2
Line charge → E ∝ 1/r
Plane sheet → E = constant
Dipole → E ∝ 1/r3
Shell inside → E = 0
14. NCERT Questions & Solutions EXAM PATTERN
Q1. What is the force between two small charged spheres having charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air?
Given: q1 = 2 × 10−7 C, q2 = 3 × 10−7 C, r = 30 cm = 0.30 m
Formula: F = k|q1q2|/r2
F = (9 × 109) × (2 × 10−7) × (3 × 10−7) / (0.30)2
F = (9 × 109) × (6 × 10−14) / 0.09
F = 54 × 10−5 / 0.09 = 6 × 10−3 N (Repulsive)
Formula: F = k|q1q2|/r2
F = (9 × 109) × (2 × 10−7) × (3 × 10−7) / (0.30)2
F = (9 × 109) × (6 × 10−14) / 0.09
F = 54 × 10−5 / 0.09 = 6 × 10−3 N (Repulsive)
Tip: Convert cm to m. Mention attractive/repulsive for full marks.
Q2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N. (a) Find the distance. (b) What is the force on the second sphere due to the first?
(a) q1 = 0.4 μC, q2 = −0.8 μC, F = 0.2 N
r2 = k|q1q2|/F = (9×109)(0.4×10−6)(0.8×10−6)/0.2
r2 = 144 × 10−4 → r = 0.12 m = 12 cm
(b) By Newton’s Third Law: 0.2 N, attractive (equal & opposite)
r2 = k|q1q2|/F = (9×109)(0.4×10−6)(0.8×10−6)/0.2
r2 = 144 × 10−4 → r = 0.12 m = 12 cm
(b) By Newton’s Third Law: 0.2 N, attractive (equal & opposite)
Tip: Part (b) is conceptual. Don’t recalculate — use Newton’s Third Law.
Q3. Check that ke2/Gmemp is dimensionless and find its value. What does it signify?
Dimensions:
[k] = [ML3T−4A−2], [e2] = [A2T2]
[ke2] = [ML3T−2]
[G] = [M−1L3T−2], [memp] = [M2]
[Gmemp] = [ML3T−2]
Ratio = [M0L0T0] → Dimensionless ✓
Value: ≈ 2.27 × 1039
Significance: Electrostatic force between proton & electron is ~1039 times stronger than gravitational force. Gravity is negligible at atomic scale.
[k] = [ML3T−4A−2], [e2] = [A2T2]
[ke2] = [ML3T−2]
[G] = [M−1L3T−2], [memp] = [M2]
[Gmemp] = [ML3T−2]
Ratio = [M0L0T0] → Dimensionless ✓
Value: ≈ 2.27 × 1039
Significance: Electrostatic force between proton & electron is ~1039 times stronger than gravitational force. Gravity is negligible at atomic scale.
Tip: Always explain physical meaning of the result.
Q4. An electric dipole with dipole moment 4 × 10−9 C m is aligned at 30° with a uniform electric field of magnitude 5 × 104 N/C. Calculate the torque.
Given: p = 4 × 10−9 C m, E = 5 × 104 N/C, θ = 30°
Formula: τ = pE sinθ
τ = (4 × 10−9) × (5 × 104) × sin(30°)
τ = (4 × 10−9) × (5 × 104) × 0.5
τ = 1.0 × 10−4 N m
Formula: τ = pE sinθ
τ = (4 × 10−9) × (5 × 104) × sin(30°)
τ = (4 × 10−9) × (5 × 104) × 0.5
τ = 1.0 × 10−4 N m
Tip: Use sinθ for torque, not cosθ.
Q5. A polythene piece rubbed with wool has a negative charge of 3 × 10−7 C. (a) Estimate number of electrons transferred. (b) Is there mass transfer?
(a) q = −3 × 10−7 C, e = −1.6 × 10−19 C
n = q/e = (−3×10−7)/(−1.6×10−19) = 1.88 × 1012 electrons
(b) Yes. Mass transferred = n × me
= (1.88 × 1012) × (9.1 × 10−31)
= 1.71 × 10−18 kg
Electrons transferred from wool to polythene. Polythene gains mass; wool loses mass.
n = q/e = (−3×10−7)/(−1.6×10−19) = 1.88 × 1012 electrons
(b) Yes. Mass transferred = n × me
= (1.88 × 1012) × (9.1 × 10−31)
= 1.71 × 10−18 kg
Electrons transferred from wool to polythene. Polythene gains mass; wool loses mass.
Tip: Don’t say “no mass transfer” just because amount is tiny. Electrons have mass.
Q6. State Gauss’s theorem and use it to find E due to an infinitely long straight uniformly charged wire.
Statement: Total electric flux through any closed surface = (1/ε0) × net charge enclosed.
∮ E⃗ · dS⃗ = qenclosed/ε0
Derivation for Infinite Line Charge:
• Symmetry: E is radial, same magnitude at distance r
• Gaussian Surface: Coaxial cylinder of radius r, length L
• Flux through curved surface: E × (2πrL)
• Flux through top & bottom: 0 (E parallel to surface)
• Total flux = E(2πrL)
• Enclosed charge: q = λL
By Gauss’s theorem: E(2πrL) = λL/ε0
E = λ/(2πε0r)
Direction: Radially outward for +λ
∮ E⃗ · dS⃗ = qenclosed/ε0
Derivation for Infinite Line Charge:
• Symmetry: E is radial, same magnitude at distance r
• Gaussian Surface: Coaxial cylinder of radius r, length L
• Flux through curved surface: E × (2πrL)
• Flux through top & bottom: 0 (E parallel to surface)
• Total flux = E(2πrL)
• Enclosed charge: q = λL
By Gauss’s theorem: E(2πrL) = λL/ε0
E = λ/(2πε0r)
Direction: Radially outward for +λ
Tip: Mention symmetry, draw Gaussian surface, explain why flux through caps = 0. Structured derivations get full marks.
Q7. A hollow charged conductor has a tiny hole cut into its surface. Show that E in the hole is (σ/2ε0) n̂.
• Just outside conductor: E = σ/ε0 (standard result)
• This E is sum of: Epatch + Erest
• Small patch behaves like infinite plane: Epatch = σ/(2ε0) on both sides
• Outside: Epatch + Erest = σ/ε0
→ Erest = σ/ε0 − σ/(2ε0) = σ/(2ε0)
• Inside conductor: Erest − Epatch = 0 ✓
• In the hole (patch removed): Ehole = Erest = σ/(2ε0) n̂
• This E is sum of: Epatch + Erest
• Small patch behaves like infinite plane: Epatch = σ/(2ε0) on both sides
• Outside: Epatch + Erest = σ/ε0
→ Erest = σ/ε0 − σ/(2ε0) = σ/(2ε0)
• Inside conductor: Erest − Epatch = 0 ✓
• In the hole (patch removed): Ehole = Erest = σ/(2ε0) n̂
Tip: This is a conceptual superposition problem. Break the field into contributions from the patch and the rest.
🎯 Exam Writing Tips
- ✅ Draw diagram for every problem — carries marks
- ✅ Write formula first, then substitute values
- ✅ Always mention direction of force / field
- ✅ Use vector notation where required
- ✅ Check dimensional consistency
- ✅ For Gauss’s law: state symmetry + Gaussian surface explicitly
- ✅ Remember: Point → 1/r2 | Line → 1/r | Plane → constant | Dipole → 1/r3
All the best for your exams! 🌟