Electric Charges and Fields
1. Introduction
Electrostatics deals with stationary electric charges and the forces between them. This chapter is fundamental to understanding all electromagnetic phenomena.
2. Electric Charge
Charge is an intrinsic property of matter. Like charges repel, unlike charges attract. Charge is quantized (Q = ne) and conserved.
3. Coulomb's Law
F = (1/4πε₀) × (q₁q₂/r²), where 1/4πε₀ = 9 × 10⁹ Nm²/C².
'Coulomb's law applies strictly to point charges. Forces are vectors — use vector addition for multiple charges.'
4. Electric Field
E = F/q₀ (force per unit positive test charge).
4.1 Field Due to a Point Charge
E = (1/4πε₀) × q/r² (radially outward for positive, inward for negative).
4.2 Electric Field Lines
- Start at positive charges, end at negative charges.
- Tangent at any point gives field direction.
- Number of lines is proportional to charge magnitude.
- Lines never cross.
5. Electric Dipole
Two equal and opposite charges separated by a small distance.
5.1 Dipole Moment
p = q × 2a, directed from negative to positive.
5.2 Field on Axial Line
E = (1/4πε₀) × 2p/r³ (along p).
5.3 Field on Equatorial Line
E = (1/4πε₀) × (-p/r³) (opposite to p).
5.4 Torque on Dipole in Uniform Field
τ = p × E, magnitude τ = pE sin θ.
6. Gauss's Theorem
The total electric flux through a closed surface equals 1/ε₀ times the net charge enclosed: Φ = q_enclosed/ε₀.
6.1 Applications
Field due to infinite line charge: E = λ/(2πε₀r).
Field due to infinite plane sheet: E = σ/(2ε₀).
Field due to spherical shell: Inside: E = 0. Outside: E = (1/4πε₀) × q/r².
Field due to solid sphere: Inside: E = (1/4πε₀) × qr/R³. Outside: E = (1/4πε₀) × q/r².
6.2 Detailed Derivations
Gauss's Law for a Spherical Shell
Consider a spherical shell of radius R with total charge Q.
Outside point (r > R): Choose spherical Gaussian surface of radius r. By symmetry, E is radial and constant on the surface. Φ = E × 4πr² = Q/ε₀ ⇒ E = Q/(4πε₀r²). The field behaves as if all charge is concentrated at the centre.
Inside point (r < R): Choose spherical Gaussian surface inside the shell. Charge enclosed = 0 ⇒ E = 0. This is the basis of electrostatic shielding.
Gauss's Law for a Solid Sphere
For a uniformly charged sphere of radius R and total charge Q:
Outside (r > R): Same as point charge: E = Q/(4πε₀r²).
Inside (r < R): Charge enclosed = Q × (r³/R³). E × 4πr² = (Qr³/R³)/ε₀ ⇒ E = Qr/(4πε₀R³).
Field varies linearly inside and follows inverse square outside.
Gauss's Law for an Infinite Plane Sheet
Consider a thin infinite sheet of surface charge density σ. Choose a cylindrical Gaussian surface (pillbox) crossing the sheet.
Flux: 2EA = σA/ε₀ ⇒ E = σ/(2ε₀). The field is independent of distance — uniform on both sides.
7. Worked Problems
Problem 1: Two charges 2 μC and -4 μC are placed 10 cm apart. Find where the electric field is zero. Solution: Field zero between charges (since opposite signs). If x from 2 μC, then (1/4πε₀)×2/x² = (1/4πε₀)×4/(10-x)². So 2/x² = 4/(10-x)² ⇒ √2/x = 2/(10-x) ⇒ x = 10√2/(2+√2) = 5.86 cm from 2 μC.
Problem 2: An electric dipole of moment 4 × 10^{-9} Cm is placed at 30° to a uniform field of 5 × 10⁴ N/C. Find torque. Solution: τ = pE sin θ = 4×10^{-9} × 5×10⁴ × sin 30° = 4×10^{-9} × 5×10⁴ × 0.5 = 10^{-4} Nm.
Problem 3: Using Gauss's theorem, find E at a distance r from a uniformly charged infinite wire carrying linear charge density λ. Solution: Consider a cylindrical Gaussian surface of radius r and length L. Flux through curved surface = E × 2πrL. Charge enclosed = λL. By Gauss: E × 2πrL = λL/ε₀ ⇒ E = λ/(2πε₀r).
8. Common Mistakes
'Students often forget the direction of the dipole moment — it points from negative to positive, which is counterintuitive.'
'While applying Gauss's law, choose a Gaussian surface that exploits symmetry — sphere for point charges, cylinder for line charges, pillbox for sheets.'
9. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| Coulomb's law | 3 | 2 |
| Electric field | 3 | 2 |
| Dipole | 4 | 2 |
| Gauss's theorem | 5 | 3 |
10. Self-Test Questions
- Two charges q and -4q are placed 20 cm apart. Where should a third charge be placed so it experiences no net force?
- Calculate E at the centre of a square of side 2 m with charges +1, -1, +2, -2 μC at corners.
- Derive an expression for the torque on an electric dipole in a uniform electric field.
- Using Gauss's theorem, find E due to an infinite plane sheet of charge.
- Find the electric flux through a sphere of radius 1 m containing a charge of 10 μC at its centre.
