Electrostatics – Stage 3

Electric Charges & Fields – Page 7

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1. Electric Field due to a Uniformly Charged Ring

Consider a ring of radius R carrying total charge Q. Find electric field at a point on the axis at distance x from center.

• Radial components cancel by symmetry
• Only axial components add

Electric Field:
E = (1 / 4π ε₀) · (Q x) / (x² + R²)^3/2

Special cases:

• x = 0 → E = 0 (center)
• x ≫ R → behaves like point charge

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2. Electric Field due to a Uniformly Charged Disc

Disc of radius R, surface charge density σ.

Method: Divide disc into concentric rings and integrate.

Electric Field on Axis:
E = (σ / 2ε₀) · [1 − x / √(x² + R²)]

Limiting Cases:

• x = 0 → E = σ / (2ε₀)
• R → ∞ → behaves like infinite sheet

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3. Electric Field due to Infinite Sheet (Revision)

E = σ / (2ε₀)

• Independent of distance
• Direction normal to sheet

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4. Electric Field due to Thin Spherical Shell

Total charge Q, radius R.

(a) Outside the Shell (r ≥ R)

E = 1 / (4π ε₀) · Q / r²

(b) Inside the Shell (r < R)

E = 0

Key Result: Electric field inside hollow charged shell is zero.

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5. Comparison: Ring vs Disc vs Shell

Distribution	Field at Center
Ring	0
Disc	σ / (2ε₀)
Spherical Shell	0

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6. JEE Advanced Thinking Points

• Ring & disc → integration based problems
• Shell → Gauss law shortcut
• Disc → infinite sheet as limiting case

JEE Trap:
Do not apply Gauss law to ring or disc directly.

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Stage 3 – Page 7 Completed ✅

Next → Stage 3 – Page 8 (Electric Dipole, Field & Torque)