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

2024
Views:

340

Pages:

5
Author:

Jabin F.

Potential Energy of the System: U = work required to assemble the system. If given ρ or λ, then continuous charge system. Equation 1: U = (ε₀ / 2) * ∫(E² * dτ) (all space) Case (i): a < r < b: U = (ε₀ / 2) * ∫(E² * 4πr² dr) from 0 to b U = (ε₀ / 2) * ∫(k² * (r - a)² * 4πr² dr) from 0 to b U = (ε₀ * 4πk²) / 2 * ∫((r² - 2ar + a²) * r² dr) from 0 to b U = (ε₀ * 4πk²) / 2 * [(r⁴/4) - (2a/3) * r³ + (a²/2) * r²] from 0 to b U = (ε₀ * 4πk²) / 2 * [(b⁴/4) - (2a/3) * b³ + (a²/2) * b²] U = (ε₀ * 4πk * a²) / 4 Case (ii): r > b: U = (ε₀ / 2) * ∫(E² * 4πr² dr) from 0 to ∞ U = (ε₀ / 2) * ∫((k * b²)² * 4πr² dr) from 0 to ∞ U = (ε₀ * 4πk² * b⁴) / 2 * ∫(r² dr) from 0 to ∞ U = (ε₀ * 4πk² * b⁴) / 2 * (r³/3) from 0 to ∞ The integral diverges to infinity. Case (ii): r > b: U = (ε₀ / 2) * ∫(E² * 4πr² dr) from a to ∞ U = (ε₀ / 2) * ∫((k * b²)² * 4πr² dr) from a to ∞ U = (ε₀ * 4πk² * b⁴) / 2 * ∫(r² dr) from a to ∞ U = (ε₀ * 4πk² * b⁴) / 2 * (r³/3) from a to ∞ U = (ε₀ * 4πk² * b⁴) / 6 Case (iii): a < r < b: U₁ = (ε₀ / 2) * ∫(E² * 4πr² dr) from a to b E = (k * (r - a)) / (ε₀ * r²) U = (ε₀ / 2) * ∫((k * (r - a))² / (ε₀ * r²))² * 4πr² dr from a to b U = (4π * ε₀ * k²) / 2 * ∫((r - a)⁴ / r⁴) * r² dr from a to b U = (4π * ε₀ * k²) / 2 * ∫((r - a)⁴ / r²) dr from a to b U = (4π * ε₀ * k²) / 2 * ∫(r⁴ - 4r³a + 6r²a² - 4ra³ + a⁴) / r² dr from a to b U = (4π * ε₀ * k²) / 2 * ∫(r² - 4ra + 6a² - 4a³/r + a⁴/r²) dr from a to b U = (4π * ε₀ * k²) / 2 * [(r³/3) - (4a/2) * r² + (6a²/1) * r - (4a³/1) * ln(r) + (a⁴/1) * (1/r)] from a to b U = (4π * ε₀ * k²) / 2 * [(b³/3) - (2a) * b² + (6a²) * b - (4a³) * ln(b) + (a⁴/1) * (1/b) - (a³/3) + (2a) * a² - (6a²) * a + (4a³) * ln(a) - (a⁴/1) * (1/a)] Simplify further to obtain the final expression for U. Equation 20: |Eout| = (ρ₀ / ε₀) * ((R/3) - (ar/4)) Calculation: (ρ₀ / ε₀) * ((R/3) - (ar/4)) = (ρ₀ / ε₀) * ((4R - 3ar) / 12) Check whether E.F is continuous or discontinuous at r = R (surface): Ein|r=R = (ρ₀R) / (12ε₀) Eout|r=R = (ρ₀R) / (12ε₀) Conclusion: E is continuous at the boundary (r = R). Now, |Eout| = ? Qline = ∫(ρ₀ * (1 - ar/R)) * 4πr² dr from 0 to R Qline = 4πρ₀ * ∫((r³ - r⁴/R)) dr from 0 to R Qline = 4πρ₀ * [(r⁴/4) - (r⁵/(5R))] from 0 to R Qline = 4πρ₀ * [(R⁴/4) - (R⁵/(5R))] Qline = 4πρ₀ * (R⁴/4 - R⁴/5) Qline = 4πρ₀ * (R⁴/20)