Unit-2. Electrostatics - Short Solutions

Part A: Definitions (1-2 marks)

Electric Field (E)

Force per unit positive charge at a point.

E = F/q₀ = kQ/r²
Unit: N/C or V/m

Electric Potential (V)

Work done to bring unit positive charge from infinity to a point.

V = W/q = kQ/r
Unit: Volt (V) or J/C

Electric Potential Difference (ΔV)

Work done to move unit charge from one point to another.

ΔV = W/q = V₂ - V₁
Unit: Volt (V)

Electric Flux (Φ)

Number of electric field lines passing through a surface.

Φ = E·A = EA cos θ
Unit: N·m²/C or V·m

Capacitor

Device that stores electric charge and energy. Two conducting plates separated by dielectric.

Capacitance (C)

Ability to store charge. Ratio of charge to potential difference.

C = Q/V
C (parallel plate) = ε₀εᵣA/d
Unit: Farad (F)

Part B: Detailed Answers (2-3 marks)

(1) Coulomb's Law

Electric force between two stationary point charges is directly proportional to product of charges and inversely proportional to square of distance.

F = kq₁q₂/r²
k = 9×10⁹ N·m²/C²

Nature:

Like charges: Repulsive
Unlike charges: Attractive

(2) Characteristics of Electric Field Lines

Start/End: Start from +ve, end at -ve charges
Direction: Tangent shows field direction
No intersection: Lines never cross
Density: Close lines = strong field, far lines = weak field
Uniform field: Parallel and equidistant
Imaginary: Lines are imaginary, field is real
Perpendicular: Always perpendicular to conductor surface
Open curves: Never form closed loops

(3) Parallel Plate Capacitor

Construction: Two parallel conducting plates of area A separated by distance d with dielectric between.

Capacitance:

C = ε₀εᵣA/d = ε₀KA/d

Factors affecting C:

C ∝ A (area)
C ∝ 1/d (distance)
C ∝ K (dielectric constant)

Applications: Energy storage, filtering, timing circuits, coupling/decoupling

(4) Series Connection of Capacitors

Characteristics:

Same charge: Q₁ = Q₂ = Q₃ = Q
Different voltages: V = V₁ + V₂ + V₃

Formula:

1/Cₛ = 1/C₁ + 1/C₂ + 1/C₃

For 2 capacitors: Cₛ = C₁C₂/(C₁+C₂)
For n equal: Cₛ = C/n

Result: Cₛ < smallest capacitance

(5) Parallel Connection of Capacitors

Characteristics:

Same voltage: V₁ = V₂ = V₃ = V
Different charges: Q = Q₁ + Q₂ + Q₃

Formula:

Cₚ = C₁ + C₂ + C₃

For n equal: Cₚ = nC

Result: Cₚ > largest capacitance

(6) Effect of Dielectric on Capacitance

When dielectric (K) inserted between plates:

Capacitance increases:

C = KC₀

Why it increases:

Dielectric molecules polarize
Reduces electric field: E = E₀/K
Reduces voltage: V = V₀/K
Same charge, lower voltage → higher capacitance

Other effects:

Increases breakdown voltage
Provides mechanical support
Makes capacitor compact

Part C: Numericals (3 marks)

(1) Coulomb force

Given: q₁ = 20 μC, q₂ = 10 μC, r = 0.02 m, k = 9×10⁹

F = kq₁q₂/r²
F = (9×10⁹)(20×10⁻⁶)(10×10⁻⁶)/(0.02)²
F = 1800×10⁻³/4×10⁻⁴ = 4500 N

Answer: 4500 N (repulsive)

(2) Potential difference

Given: W = 1600 J, q = 25 C

V = W/q = 1600/25 = 64 V

Answer: 64 V

(3) Capacitance

Given: Q = 60 μC, V = 12 V

C = Q/V = 60×10⁻⁶/12 = 5×10⁻⁶ F = 5 μF

Answer: 5 μF

(4) Series and Parallel (3 × 10μF)

Series:

Cₛ = C/n = 10/3 = 3.33 μF

Parallel:

Cₚ = nC = 3×10 = 30 μF

Answer: Series = 3.33 μF, Parallel = 30 μF

(5) Capacitance of small capacitor

Given: A = 10 mm² = 10⁻⁵ m², d = 1 mm = 10⁻³ m

C = ε₀A/d = (8.85×10⁻¹²)(10⁻⁵)/(10⁻³)
C = 8.85×10⁻¹⁴ F = 0.0885 pF

Answer: 8.85×10⁻¹⁴ F

(6) Area for 1F capacitor

Given: C = 1 F, d = 1 mm = 10⁻³ m

A = Cd/ε₀ = (1)(10⁻³)/(8.85×10⁻¹²)
A = 1.13×10⁸ m² = 113 km²

Answer: 1.13×10⁸ m² (113 km²) - Shows 1F is huge!

(7) Mixed Circuit

Example: (C₁ ∥ C₂) in series with C₃ Given: C₁ = 10 μF, C₂ = 20 μF, C₃ = 30 μF

Step 1: Parallel

C₁₂ = C₁ + C₂ = 10 + 20 = 30 μF

Step 2: Series with C₃

1/Cₜ = 1/30 + 1/30 = 2/30
Cₜ = 15 μF

Answer: 15 μF

Quick Reference

Formulas

Coulomb's Law: F = kq₁q₂/r², k = 9×10⁹
Electric Field: E = F/q = kQ/r²
Potential: V = W/q = kQ/r
Flux: Φ = EA cos θ
Capacitance: C = Q/V = ε₀εᵣA/d
Series: 1/Cₛ = 1/C₁ + 1/C₂ + 1/C₃
Parallel: Cₚ = C₁ + C₂ + C₃
With dielectric: C = KC₀
Energy: U = ½QV = ½CV² = Q²/(2C)

Constants

k = 9×10⁹ N·m²/C²
ε₀ = 8.85×10⁻¹² F/m
e = 1.6×10⁻¹⁹ C
1 μF = 10⁻⁶ F
1 nF = 10⁻⁹ F
1 pF = 10⁻¹² F

Short Solutions - Unit 2