1.1

Electric Charge & Its Properties

📌 What is Electric Charge?

Electric charge is a fundamental property of matter. There are two types — positive (on glass rod rubbed with silk) and negative (on plastic rod rubbed with cat's fur). Like charges repel; unlike charges attract. SI unit of charge = Coulomb (C).

📌 Three Basic Properties of Charge

(i) Additivity: Total charge = algebraic sum of all charges. Charges are scalars.
(ii) Conservation: Total charge of an isolated system is always conserved. No new charge is created or destroyed — only transferred.
(iii) Quantisation: q = ne, where n is integer and e = 1.6 × 10⁻¹⁹ C (charge on electron/proton).

🧮 Key Values
Charge on electron: e = −1.602 × 10⁻¹⁹ C
Charge on proton: e = +1.602 × 10⁻¹⁹ C
Quantisation: q = ne  (n = 0, ±1, ±2, ...)
1 μC = 10⁻⁶ C  |  1 mC = 10⁻³ C
Gold-leaf electroscope diagram showing metal knob, rod, and diverging gold leaves when charged

🎯 Practice MCQs

Q1. A system has charges +3 μC, −5 μC, +2 μC, and −4 μC. What is the total charge of the system?
Answer: A) −4 μC
Total = (+3) + (−5) + (+2) + (−4) = −4 μC. Charges add algebraically.
Q2. How many electrons make up a charge of 1.6 × 10⁻¹³ C?
Answer: A) 10⁶
n = q/e = (1.6 × 10⁻¹³) / (1.6 × 10⁻¹⁹) = 10⁶ electrons.
Q3. When a glass rod is rubbed with silk, the glass rod becomes positively charged. This is because:
Answer: C) Electrons are transferred from glass to silk
No new charges are created. Electrons (mobile charges) transfer from glass to silk, making glass positive and silk negative.
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1.2

Coulomb's Law & Superposition Principle

📌 Coulomb's Law

The electrostatic force between two point charges q₁ and q₂ separated by distance r in vacuum acts along the line joining them and is: directly proportional to the product of charges, and inversely proportional to the square of the distance.

🧮 Coulomb's Law Formula
F = k |q₁q₂| / r²
or  F = (1/4πε₀) × |q₁q₂| / r²

k = 9 × 10⁹ N m² C⁻²
ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²

Vector form: F₂₁ = (1/4πε₀) × (q₁q₂/r²₂₁) r̂₂₁

📌 Superposition Principle

The force on a charge due to multiple other charges = vector sum of individual Coulomb forces. Each pair's force is unaffected by the presence of other charges.
F₁ = F₁₂ + F₁₃ + F₁₄ + ...

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Insert Image Here: Coulomb's Law geometry — two charges q₁ and q₂, position vectors r₁, r₂, force vectors F₁₂ and F₂₁ (Fig 1.3 NCERT)

🎯 Practice MCQs

Q4. The force between two point charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air is approximately:
Answer: A) 6 × 10⁻³ N
F = 9×10⁹ × (2×10⁻⁷ × 3×10⁻⁷) / (0.3)² = 9×10⁹ × 6×10⁻¹⁴ / 0.09 = 6 × 10⁻³ N.
Q5. The ratio of electric force to gravitational force between an electron and a proton is approximately:
Answer: A) 2.4 × 10³⁹
Fe/Fg = e² / (4πε₀ G mₑ mₚ) ≈ 2.4 × 10³⁹. Electric forces are enormously stronger than gravity.
Q6. If the distance between two charges is halved while the charges are kept the same, the force between them becomes:
Answer: C) 4 times
F ∝ 1/r². If r → r/2, F → F/(r/2)² × r² = 4F.
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1.3

Electric Field & Electric Field Lines

📌 Electric Field

The electric field E at a point is the force experienced by a unit positive test charge placed at that point (without disturbing the source charge). It is a vector quantity. Unit: N/C or V/m.

🧮 Electric Field Formulas
Due to point charge Q:  E = (1/4πε₀) × Q/r² r̂
Relation to force:  F = qE
Defined as:  E = lim(q→0) F/q
Due to system of n charges:  E(r) = Σ (1/4πε₀) × qᵢ/rᵢₚ² r̂ᵢₚ

📌 Properties of Electric Field Lines

(i) Field lines start at positive charges and end at negative charges.
(ii) In a charge-free region, field lines are continuous curves — no sudden breaks.
(iii) Two field lines can never cross each other (unique direction at every point).
(iv) Electrostatic field lines do not form closed loops.
(v) Density of field lines represents the strength of E.

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Insert Image Here: Electric field line diagrams (Fig 1.14 NCERT) — for positive charge, negative charge, two equal positive charges, and electric dipole (±q)

🎯 Practice MCQs

Q7. Two point charges q₁ = +3 μC and q₂ = −3 μC are placed 20 cm apart. What is the electric field at the midpoint between them?
Answer: A) 2.7 × 10⁶ N/C
At midpoint, both fields point in the same direction (toward −q). Each E = 9×10⁹ × 3×10⁻⁶ / (0.1)² = 2.7×10⁶ N/C. Total = 2 × 2.7×10⁶ = 5.4×10⁶ N/C.
Note: Correct answer is C) 5.4 × 10⁶ N/C — both fields add.
Q8. The electric field due to a point charge Q at a distance r has magnitude E. If the distance is tripled, the new field becomes:
Answer: C) E/9
E ∝ 1/r². Distance tripled → E → E/(3)² = E/9.
Q9. Which of the following is NOT a property of electric field lines?
Answer: B) Two lines can cross at 90°
Two field lines can NEVER cross each other at any angle. If they crossed, the field at that point would have two directions — which is physically impossible.
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1.4

Electric Dipole & Electric Flux

📌 Electric Dipole

An electric dipole is a pair of equal and opposite charges +q and −q separated by distance 2a. The dipole moment p = q × 2a, directed from −q to +q. Unit: C·m. Water (H₂O) is a natural polar molecule.

🧮 Dipole Field Formulas
On axis (r >> a): E = 2p / (4πε₀ r³)  [along p̂]
On equatorial plane (r >> a): E = −p / (4πε₀ r³)  [opposite p̂]
Torque in uniform field: τ = p × E = pE sinθ
Note: Dipole field ∝ 1/r³ (vs 1/r² for point charge)

📌 Electric Flux

Electric flux through area element ΔS:  Δφ = E · ΔS = E ΔS cosθ
where θ is angle between E and the normal to surface. Unit: N·C⁻¹·m² or V·m.
Total flux: φ = Σ E · ΔS

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Insert Image Here: Electric dipole diagram (Fig 1.17 NCERT) — showing charges +q and −q, dipole axis, dipole moment vector p, and field on axis vs equatorial plane

🎯 Practice MCQs

Q10. Two charges ±10 μC are placed 5 mm apart. The dipole moment is:
Answer: A) 5 × 10⁻⁸ C·m
p = q × 2a = 10⁻⁵ C × 5 × 10⁻³ m = 5 × 10⁻⁸ C·m.
Q11. An electric dipole of moment p is placed in a uniform electric field E at angle 30°. The torque on the dipole is:
Answer: A) pE/2
τ = pE sinθ = pE sin30° = pE × ½ = pE/2.
Q12. The electric flux through a surface of area 0.02 m² placed perpendicular to a uniform electric field of 500 N/C is:
Answer: A) 10 N·m²/C
φ = E × ΔS × cosθ = 500 × 0.02 × cos0° = 10 N·m²/C.
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1.5

Gauss's Law & Applications

📌 Gauss's Law

The total electric flux through any closed surface S is equal to the total charge enclosed by S divided by ε₀.
φ = q_enc / ε₀
This holds for any closed surface (Gaussian surface) of any shape or size. The Gaussian surface must not pass through discrete charges.

🧮 Applications of Gauss's Law
Infinite line charge (linear density λ):
  E = λ / (2πε₀ r)  [radially outward]

Infinite plane sheet (surface density σ):
  E = σ / (2ε₀)  [perpendicular, both sides]

Spherical shell (charge Q, radius R):
  E = Q / (4πε₀ r²) for r ≥ R  [outside]
  E = 0 for r < R  [inside — zero field]

📌 Key Result: Field Inside a Shell = Zero

For a uniformly charged thin spherical shell, the electric field at all interior points is exactly zero. This is a direct consequence of Gauss's Law and the 1/r² nature of Coulomb's law. Experimentally verified — confirms inverse-square law.

Gaussian surface diagrams — cylindrical surface around a line charge, parallelepiped around plane sheet, and spherical surface around thin shell

🎯 Practice MCQs

Q13. A point charge of 2 μC is at the centre of a cubic Gaussian surface of side 9 cm. The net electric flux through the surface is:
Answer: A) 2.26 × 10⁵ N·m²/C
φ = q/ε₀ = (2 × 10⁻⁶) / (8.854 × 10⁻¹²) = 2.26 × 10⁵ N·m²/C. Shape of surface is irrelevant.
Q14. The electric field inside a uniformly charged thin spherical shell is:
Answer: D) Zero
By Gauss's law, for any Gaussian surface inside the shell, charge enclosed = 0 → E × 4πr² = 0 → E = 0.
Q15. An infinite line charge with linear charge density λ = 2 × 10⁻⁸ C/m. The electric field at a perpendicular distance of 0.1 m is approximately:
Answer: A) 3600 N/C
E = λ/(2πε₀r) = (2×10⁻⁸) / (2π × 8.854×10⁻¹² × 0.1)
= 2×10⁻⁸ / (5.56×10⁻¹²) ≈ 3600 N/C.
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📚 Chapter Summary — Electric Charges & Fields

Two types of charge: positive and negative; like charges repel, unlike attract
Properties of charge: Additivity, Conservation, Quantisation (q = ne)
e = 1.6 × 10⁻¹⁹ C; Coulomb (C) is the SI unit of charge
Coulomb's Law: F = kq₁q₂/r² = (1/4πε₀) q₁q₂/r² ; k = 9×10⁹ N m² C⁻²
Superposition Principle: Net force = vector sum of individual Coulomb forces
Electric Field: E = F/q = (1/4πε₀) Q/r²; direction outward for +Q, inward for −Q
Field lines: start at +q, end at −q; never cross; no closed loops; density ∝ |E|
Dipole moment: p = q × 2a ; torque τ = p × E = pE sinθ; field ∝ 1/r³
Electric flux: φ = E · ΔS = E ΔS cosθ ; unit: N m² C⁻¹
Gauss's Law: φ_total = q_enc/ε₀ ; valid for any closed surface
Field inside spherical shell = zero; outside = same as point charge at centre
Infinite plane sheet: E = σ/2ε₀ (both sides); Infinite line charge: E = λ/2πε₀r

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