University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

325

Pages:

7
Author:

Jabin F.

A charge -q' is placed at a distance x from the center of a uniformly charged ring of radius R and charge density λ. If x << R, calculate the time period of oscillation of the charge particle. E_ax = (λπR) / (2ε₀(x² + R²)^(3/2)) F = -λq'R(x) / (2ε₀(x² + R²)^(3/2)) If x << R: F = -λq'R / (2ε₀R²) For spring: F = -kx K = λq' / (2ε₀R²) (b) λ = 1 q₁, q₂ ⇒ q₁ + q₂ = Q constant F = (q₁q₂) / (4πε₀r²) F = (q₁(Q-q₁)) / (4πε₀r²) For maximum: dF/dq₁ = 0 dF/dq₁ = (1 / (4πε₀r²)) * ((Q-q₁) + q₁) = 0 q₁ = Q/2 q₂ = Q/2 If sum of two charges is constant, then for maximum force both charges should be same. (i) q₁ = λl, q₂ = λl λ = 1 q₁ = l, q₂ = l q₁y₁ = q₂y₂ P = 1 F = q₃E_net = 0 E_net = 0 Then q₃ = 0 or -q₃ both in A 100 + x² + 20x = 2x² x² + 20x - 100 = 0 x = (-20 ± √(400 + 400)) / 2 x = -20 ± 20√2 / 2 x = 10(√2 - 1) x = 10 * 0.41 x = 4.1 cm (equilibrium) E_net = (2 / (4πε₀R²)) * (1/√2 + 1/√2) * (-ŷ) E_net = (2 / (4πε₀R²)) * (ŷ) E_BC = (1 / (4πε₀R²)) * (1/√2 + 1/√2) * (-ŷ) E_BC = (1 / (4πε₀R²)) * (ŷ) E_net = E_A + E_BC = 0 Φ√2 = 2 Φ = 2/√2 (Q/2) = 2/√2 ρ = (Q/2)δ(x-x₀) - (Q/2)δ(x+x₀) E_at (x₀, 0, 0) = (1/4πε₀) * (Q/2) * (1/x₀² + 1/x₀²) * ŷ at x = x₀, ρ = ∞ at x = -x₀, ρ = ∞ E_net = (1/4πε₀) * (Q/2) * (1/x₀² + 1/x₀²) * ŷ + (1/4πε₀) * (Q/2) * (1/x₀² + 1/x₀²) * ŷ = (2Q) / (9πε₀x₀²) * ŷ (Q/2) = 2/√2 E = σ / (2ε₀) = 20x10⁻⁹ * πT / (2(4πε₀)) = 40π * 10⁻⁹ * 9x10⁹ = 360π (-ẑ) V/m (d) (New book) Ques. 45 P-8 E_net = σ₁ / (2ε₀) + σ₂ / (2ε₀) = ((6.8 + 4.3) * 10⁻⁶) / (2 * 8.8571 * 10⁻¹²) = 6.3 * 10⁵ N/C