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

2024
Views:

390

Pages:

5
Author:

Jabin F.

Net Charge on r=a Φnet | r=a = (+q) - (ε₀/4πε₀) * (Q/a²) Φnet = q - q + ε₀Q/ε = ε₀Q/ε Φnet | r=a = ε₀Q/ε = Q/K Net Bound Charge Φbound | r=b = (ε - ε₀)Q / 4πε₀b² Φbound | r=b = (ε - ε₀)Q/ε = -Q' Conclusion: Net bound charge of any system is always zero. Therefore, Φc + Φb = 0, where Φc is the charge at r=a and Φb is the bound charge. This leads to the equation: Φc = -Φb = Q/K Top Section Φb = q(k-1)/k V = (1/4πε₀)[(ε-ε₀)Φ/εb + ε₀Φ/ε₀] ε/ε₀ = κ V = -q(k-1)/k Middle Section ∮s D·ds = Φfree Φfree = q D = q/(4πr²) D = εE = ε₀εrE = ε₀(1+χe)E E = q/(4πε₀(1+χe)r²) P = εχeE P = 2Me/(4π(1+χe)r²) Bottom Section r > R: Φfree = q D = q/(4πr²) Top Section E² = (1/4πε₀)²(q/r⁴) - (V/r) In free space, V = 0, so E² = (1/4πε₀)²(q/r⁴) P = 0 for r > R (from (vii)) qf + qb = (q + qe)/κ Middle Section qf = (q + qe)/κ - qb qb = -(qe/(1+χe)) qb|r=0 = -(qe/(1+χe)) qb|r=R = -(qe/(1+χe)) Bottom Section Φb|r=R = P·n̂ = (2χe/(4π(1+χe)R²)) * (r̂/R) Φb|r=R = 2χe/(4π(1+χe)R³) Top Section Pb = -∇·P Pb = -(2χe/(1+χe)) * ∇(δ³(r)) Here we can't do this integral equal to zero because r=0 is in its range. Bottom Section Φb|r=0 = ∫V Pb dτ Φb|r=0 = -(2χe/(1+χe)) * ∫V δ³(r) dτ Φb|r=0 = -(2χe/(1+χe))