Electrostatics – Stage 1
Chapter 1: Electric Charges & Fields – Page 9
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1. Electric Field due to a Uniformly Charged Ring
Consider a ring of radius R, total charge Q, and point P on its axis at distance x from center.
Due to symmetry:
- Transverse components cancel
- Only axial components add
E = (1 / 4πε₀) · (Qx / (x² + R²)3/2)
- E = 0 at center (x = 0)
- E is maximum at x = R / √2
2. Electric Field due to a Uniformly Charged Disc
Disc radius = R, surface charge density = σ Point P lies on axis at distance x
Disc is treated as collection of concentric rings.
E = (σ / 2ε₀) · [1 − x / √(x² + R²)]
Special Case:
As R → ∞ (infinite plane),
E = σ / (2ε₀)
3. Electric Field due to a Thin Infinite Plane Sheet
- Independent of distance
- Perpendicular to sheet
E = σ / (2ε₀)
JEE Insight: Field does not change even if observer moves away.
4. Electric Field due to a Finite Line Charge
Line charge with linear density λ Point P makes angles θ₁ and θ₂ with ends
E = (λ / 4πε₀r) · (sinθ₁ + sinθ₂)
- Direction perpendicular to line
- Used frequently in JEE Advanced
5. Strategy for Continuous Charge Problems
- Divide charge into small dq
- Write dE using Coulomb’s law
- Resolve components
- Use symmetry to cancel terms
- Integrate over entire body
Golden Tip: Always check which components cancel before integrating.
6. Comparison Summary (Very Important)
| Charge Distribution | Electric Field |
|---|---|
| Ring (on axis) | ∝ x / (x² + R²)3/2 |
| Disc | σ/2ε₀ · (1 − x/√(x²+R²)) |
| Infinite plane | Constant |
| Finite line | ∝ (sinθ₁ + sinθ₂) |
Stage 1 – Page 9 Takeaway
- Ring & disc problems test integration logic
- Symmetry simplifies calculations
- Foundation for capacitors & advanced electrostatics
Next → Stage 1, Page 10: Electric Flux, Solid Angle & Gauss Law Master Problems
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