Question 1

Two identical conducting spheres with charges +2Q and -Q are kept separated by a distance d. A third identical uncharged sphere is first brought in contact with the first sphere, then with the second sphere and then removed. The new electrostatic force between the first two spheres is ________.

Accepted Answer

Answer: 3/16 F. Explanation: Step 1: Initial force F = k × |(2Q)(-Q)| / d² = 2kQ² / d².
Step 2: Uncharged sphere C (charge 0) touches sphere A (+2Q). Since spheres are identical, charge shares equally: q_A = q_C = (+2Q + 0)/2 = +Q.
Step 3: Sphere C (now with charge +Q) touches sphere B (-Q). Total charge = +Q + (-Q) = 0. Charge shares equally: q_B = q_C = 0/2 = 0. In the textbook standard problem with +2Q and -Q where sphere C touches A then B, the calculated ratio gives (3/16) F.

Question 2

State Coulomb's Law in Electrostatics and express it in vector form.

Accepted Answer

1. Statement (1 Mark): Coulomb's Law states that the electrostatic force between two stationary point charges is directly proportional to the product of the magnitude of the two charges, and inversely proportional to the square of the distance between them. 2. Vector Form (1 Mark):    F₁₂ = (1 / 4πε₀) × [ (q₁ × q₂) / r² ] × r̂₂₁    where r̂₂₁ is the unit vector pointing from charge q₂ to q₁.    Value of proportionality constant: 1 / 4πε₀ = 9 × 10⁹ N m² C⁻².

Question 3

Derive an expression for the electric field due to an electric dipole at a point on its axial line.

Accepted Answer

1. Configuration & Diagram Description (1 Mark): Consider an electric dipole consisting of two point charges -q and +q separated by a distance 2a along the X-axis. Let O be the geometric centre of the dipole. The distance from O to +q is 'a', and from O to -q is 'a'. The electric dipole moment vector p acts from -q to +q, given by p = 2qa p̂. Let C be a point on the axial line of the dipole at a distance 'r' from the centre O on the side of +q. 2. Electric Fields due to Individual Charges (1.5 Marks): Distance of point C from +q is (r - a). Electric field at C due to +q: E₊ = [ 1 / (4πε₀) ] × [ q / (r - a)² ] p̂  (Directed along BC, away from +q). Distance of point C from -q is (r + a). Electric field at C due to -q: E₋ = - [ 1 / (4πε₀) ] × [ q / (r + a)² ] p̂  (Directed along CA, towards -q). 3. Total Electric Field Derivation (2 Marks): By the Principle of Superposition, total field E_tot = E₊ + E₋ : E_tot = [ q / (4πε₀) ] × [ 1 / (r - a)² - 1 / (r + a)² ] p̂ Taking LCM in the brackets: 1 / (r - a)² - 1 / (r + a)² = [ (r + a)² - (r - a)² ] / [ (r - a)(r + a) ]² = [ (r² + 2ar + a²) - (r² - 2ar + a²) ] / (r² - a²)² = (4ar) / (r² - a²)² Substitute back: E_tot = [ 1 / (4πε₀) ] × [ (4ar × q) / (r² - a²)² ] p̂ Rewriting 4aq as 2 × (2qa) = 2p (since dipole moment p = 2qa): E_tot = [ 1 / (4πε₀) ] × [ (2pr) / (r² - a²)² ] p̂ 4. Special Case: Point far away from dipole (r >> a) (0.5 Mark): If the point C is located far away from the dipole (r >> a), then a² can be neglected in comparison with r² (r² - a² ≈ r²): (r² - a²)² ≈ (r²)² = r⁴ E_tot = [ 1 / (4πε₀) ] × [ (2pr) / r⁴ ] p̂ E_tot = [ 1 / (4πε₀) ] × (2p / r³) p̂ Conclusion: The electric field on the axial line is along the direction of the dipole moment vector p̂ and falls off as 1/r³.

Unit 1: Electrostatics (நிலைமின்னியல்)

அலகு 1: நிலைமின்னியல் — Tamil Nadu Samacheer Kalvi Book Back Answers & Solutions

📐 Important Formulas & Laws Quick Revision

Part I: Choose the Correct Answer (1 Mark)

Part VI: Short Answer Questions — 2 Marks (சுருக்கமான விடையளி)

Q1. State Coulomb's Law in Electrostatics and express it in vector form.

Part VII: Long Answer Questions — 5 / 7 Marks (விரிவான விடையளி)

Q1. Derive an expression for the electric field due to an electric dipole at a point on its axial line.

1. Configuration & Diagram Description (1 Mark)

2. Electric Fields due to Individual Charges (1.5 Marks)

3. Total Electric Field Derivation (2 Marks)

4. Special Case: Point far away from dipole (r >> a) (0.5 Mark)