Transformation of Electric Field | Santa Fe College

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

2024
Views:

49

Pages:

5
Author:

Jabin F.

Diagram: Two parallel plates, S and S', separated by distance d. S is at rest, S' is moving with velocity v in the y-direction. Both plates have surface charge densities σ and σ'. Electric field lines are shown perpendicular to the plates. Equations: In the rest frame (S): E_rest = σ / ε₀ In the moving frame (S'): E_moving = σ' / ε₀ σ = q / (x₀y₀) σ' = -q' / (x₀'y₀') Notes: The prime symbol (') denotes quantities measured in the moving frame. The y₀ and y₀' terms represent the distances between the plates in the rest and moving frames, respectively. The negative sign in the expression for σ' indicates that the charge density on S' is opposite in sign to that on S. Case II: Configuration: Plates are parallel to the x-y plane, moving in the x-direction. Electric field in rest frame: E_rest = (σ / ε₀) * ẑ Electric field in moving frame: E_moving = (σ / ε₀) * ẑ Conclusion: The electric field remains unchanged in both frames. Case III: Configuration: Plates are parallel to the y-z plane, moving in the x-direction. Electric field in rest frame: E_rest = (σ / ε₀) * ẑ Electric field in moving frame: E_moving = γ * (σ / ε₀) * ẑ Conclusion: The electric field is transformed by a factor of γ in the moving frame, where γ = 1 / √(1 - v²/c²) is the Lorentz factor.In the rest frame: Ē ≠ 0, 𝐵 = 0 In the moving frame: Ē ≠ 0, 𝐵 ≠ 0 Ē, 𝐵, Ē² - c²B², q are invariant. In the rest frame Ē · 𝐵 = 0 (because 𝐵 = 0) Ē' · 𝐵' = 0