Symmetry of Charges and Electric Field | Lecture Note - Edubirdie

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

2024
Views:

430

Pages:

4
Author:

Jabin F.

Top Section E₁ = 1/(4πε₀) * (2q/d²) E₂ = -1/(4πε₀) * (2q/(d+2a)²) Middle Section Ent = E₁ + E₂ = 0 Ec = Esin30° + Esin60° Ecos30° = Esin60° |EA| = |EB| = |Ec| OA = √(d² + a²) = √(13)a Bottom Section |EA| = |EB| = (Ec sin30° + Es sin60°)(+y) EBC = (Ec sin30° + Es sin60°)(+y) Diagram: A hexagon with charges +q placed at vertices A, B, C, D, and F. Vertex E is empty. Electric field at the center (O): EO = 2/(4πε₀a²) * (cos60°x̂ + sin30°ŷ) EO = q/(4πε₀a²) * (1/2x̂ + √3/2ŷ) Net electrostatic potential at the center: Vnet = V1 + V2 + V3 + V4 + V5 V = q/(4πε₀a) Vnet = 5q/(4πε₀a) Conclusion The magnitude of the electric field at the center is q/(4πε₀a²). The electrostatic potential at the center is 5q/(4πε₀a). Top Section E+ = PR(r-2)/3ε₀ E- = PR(r+2)/3ε₀ E = PR/(3ε₀) * (1+2/r) Middle Section Enct = E+ + E- = PR(2)/3ε₀ E = PR/(3ε₀) V = (1/8πε₀) * ∫(PR³/r²) dr = PR²/6ε₀ Bottom Section Vout = PR²/3ε₀ V+ = PR²/6ε₀ V- = -PR²/3ε₀ Vnet = V+ + V- = PR²/3ε₀

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