Electrostatics – Stage 1
Chapter 1: Electric Charges & Fields – Page 8
1. Strategy for Applying Gauss’s Law
- Identify symmetry of charge distribution
- Choose correct Gaussian surface
- Find electric flux Φ = ∮𝐄·d𝐀
- Use Φ = Qenclosed/ε₀
Golden Rule: Choose Gaussian surface where E is constant or zero.
2. Spherical Charge Distribution
(a) Point Charge at Center
Gaussian Surface: Sphere of radius r
E · 4πr² = q / ε₀
E = (1 / 4πε₀) · (q / r²)
✔ Same as Coulomb’s law ✔ Field radial and symmetric
(b) Uniformly Charged Solid Sphere (Radius R)
Case 1: Outside Sphere (r ≥ R)
E = (1 / 4πε₀) · (Q / r²)
Entire charge acts as if concentrated at center.
Case 2: Inside Sphere (r < R)
Enclosed charge:
Qenc = Q · (r³ / R³)
E = (1 / 4πε₀) · (Q r / R³)
Important: Inside a solid sphere, E ∝ r
3. Infinite Line Charge (Linear Charge Density λ)
Gaussian Surface: Cylindrical surface (radius r, length L)
Flux = E · (2πrL)
Qenc = λL
E = λ / (2πε₀r)
- E ∝ 1/r
- Independent of length
4. Infinite Plane Sheet of Charge
Surface charge density = σ
Gaussian Surface: Pillbox
2EA = σA / ε₀
E = σ / (2ε₀)
Key Result: Electric field is independent of distance.
5. Conducting Sphere (Electrostatic Equilibrium)
- Electric field inside conductor = 0
- All charge resides on surface
- Field just outside: E = σ/ε₀
JEE Favorite: Inside any conductor → E = 0
6. Comparison Table (Must Remember)
| System | E Dependence |
|---|---|
| Point charge | 1 / r² |
| Infinite line | 1 / r |
| Infinite plane | Constant |
| Solid sphere (inside) | ∝ r |
Stage 1 – Page 8 Takeaway
- Gauss law simplifies symmetric systems
- Correct surface choice is crucial
- Most JEE field questions come from this page
Next → Stage 1, Page 9: Electric Field due to Ring, Disc & Non-Uniform Systems
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