Electric Charges and Fields — Class 12 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 12 Physics chapter "Electric Charges and Fields" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Electric Charges and Fields" — 8 important questions with detailed answers for CBSE bo…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Define electric charge. State the SI unit and symbol for electric charge.
State Coulomb's law. Two charges of 2 μC and 3 μC are separated by 30 cm. Cal…
What is electric field? Write its definition and SI unit.
A point charge of +5 μC is placed at a point. What is the electric field at a…
What is an electric dipole? Define dipole moment and give its SI unit.
Derive the expression for electric field on the axial line of a dipole at dis…
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Define electric charge. State the SI unit and symbol for …	✓ Solved
State Coulomb's law. Two charges of 2 μC and 3 μC are sep…	✓ Solved
What is electric field? Write its definition and SI unit.	✓ Solved
A point charge of +5 μC is placed at a point. What is the…	✓ Solved
What is an electric dipole? Define dipole moment and give…	✓ Solved
Derive the expression for electric field on the axial lin…	✓ Solved

Showing 6 of 8 questions

Q1: Define electric charge. State the SI unit and symbol for electric charge.

Electric charge is the fundamental property of matter responsible for electromagnetic interactions. It is conserved and quantized. SI Unit: Coulomb (C) Symbol: q or Q Quantization: q = ne, where n is an integer and e = 1.6 × 10⁻¹⁹ C (elementary charge)

Q2: State Coulomb's law. Two charges of 2 μC and 3 μC are separated by 30 cm. Calculate the electrostatic force between them.

Coulomb's Law: F = k(q₁q₂)/r² where k = 9 × 10⁹ N⋅m²/C² (Coulomb's constant) Given: q₁ = 2 μC = 2 × 10⁻⁶ C q₂ = 3 μC = 3 × 10⁻⁶ C r = 30 cm = 0.3 m Step 1: F = (9 × 10⁹) × (2 × 10⁻⁶) × (3 × 10⁻⁶) / (0.3)² Step 2: F = (9 × 10⁹) × (6 × 10⁻¹²) / 0.09 Step 3: F = 54 × 10⁻³ / 0.09 Step 4: F = 0.6 N Final Answer: 0.6 N (repulsive if charges are like, attractive if opposite)

Q3: What is electric field? Write its definition and SI unit.

Electric Field (E) is the region around a charge where another charge experiences an electrostatic force. Definition: E = F/q₀ (force per unit positive test charge) SI Unit: N/C or V/m (Newton per Coulomb or Volt per meter) Electric field due to point charge q at distance r: E = kq/r² = q/(4πε₀r²) where ε₀ = 8.85 × 10⁻¹² F/m (permittivity of free space)

Q4: A point charge of +5 μC is placed at a point. What is the electric field at a distance of 10 cm from this charge?

Given: q = +5 μC = 5 × 10⁻⁶ C r = 10 cm = 0.1 m k = 9 × 10⁹ N⋅m²/C² Formula: E = kq/r² Step 1: E = (9 × 10⁹) × (5 × 10⁻⁶) / (0.1)² Step 2: E = (45 × 10³) / 0.01 Step 3: E = 4.5 × 10⁶ N/C Final Answer: 4.5 × 10⁶ N/C (directed radially outward from positive charge)

Q5: What is an electric dipole? Define dipole moment and give its SI unit.

Electric Dipole: System of two equal and opposite charges separated by a small distance. Dipole Moment (p): Product of charge magnitude and separation distance. p = q × d where q = magnitude of each charge d = separation between charges (vector from negative to positive charge) SI Unit: C⋅m (Coulomb-meter) Direction: From negative charge toward positive charge Alternatively: Can be expressed in terms of electron charge as Debye unit: 1 D = 3.33 × 10⁻³⁰ C⋅m

Q6: Derive the expression for electric field on the axial line of a dipole at distance r from center.

Consider dipole with charges +q and -q separated by distance 2a. Point P on axial line at distance r from center. Field due to +q at distance (r - a): E₊ = kq/(r - a)² Field due to -q at distance (r + a): E₋ = -kq/(r + a)² Net field (E = E₊ + E₋): E = kq/(r - a)² - kq/(r + a)² E = kq[(r + a)² - (r - a)²] / [(r - a)²(r + a)²] E = kq[4ar] / [(r - a)²(r + a)²] For r >> a (dipole approximation): E = 4kqa/r³ = 2kp/r³ = p/(2πε₀r³) where p = 2qa (dipole moment)

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