Electrostatics

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Gauss' Law: The total electric flux through any closed surface is equal to the total charge enclosed by the surface divided by ε₀. Mathematical Expression: ∮ E⃗ · ds⃗ = q/ε₀ Gaussian Surface: An ima

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Using Gauss' Law with spherical Gaussian surface of radius r: (i) Outside (r > R): E = q/(4πε₀r²) [same as point charge] (ii) On surface (r = R): E = q/(4πε₀R²) = σ/ε₀ (iii) Inside (r < R): E = 0 [no

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Given: R = 10 cm = 0.1 m, q = 1 μC = 1×10⁻⁶ C (a) At r = 30 cm = 0.3 m (outside): E = q/(4πε₀r²) = (9×10⁹ × 1×10⁻⁶)/(0.3)² = 10⁵ N/C (b) On surface (r = R = 0.1 m): E = q/(4πε₀R²) = (9×10⁹ × 1×10⁻⁶)

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For infinite line charge with linear charge density λ: 1. Choose cylindrical Gaussian surface of radius r and length l 2. By symmetry, E is radial and constant on curved surface 3. Flux through end c

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Electric Potential (V): Work done per unit positive charge in bringing it from infinity to that point. V = U/q = W∞/q Relation with Electric Field: E = -dV/dx (for 1D) E⃗ = -∇V (gradient of potential

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Consider bringing unit positive charge from ∞ to distance r from charge q: Work done against electric force: dW = -F⃗ · dr⃗ = -E dr (taking radial path) Since E = q/(4πε₀r²): dW = -q/(4πε₀r²) dr To

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Given: R = 10 cm = 0.1 m, q = 250 μC = 250×10⁻⁶ C For any point on the circular wire, distance to center = R Potential at center due to entire charge: V = q/(4πε₀R) V = (9×10⁹ × 250×10⁻⁶)/(0.1) V =

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Consider dipole with charges +q and -q separated by 2l, dipole moment p = q×2l: For point P at distance r (r >> l) making angle θ with axis: r₁ ≈ r - l cos θ (distance from +q) r₂ ≈ r + l cos θ (dist