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

2024
Views:

251

Pages:

2
Author:

Jabin F.

Problem Statement: Find the electric field at the center of a hemisphere with a given charge density on its surface. Solution: Given: σ = σ₀cosθ, R (radius of hemisphere) Steps: Calculate the electric field component dEz due to a small charge element dq. Use integration to find the total electric field Ez at the center. Substitute the given charge density and integrate over the appropriate limits. Simplify the expression to obtain the final result. Mathematical Equations: σ = σ₀cosθ dEz = dEcosθ(-z) dE = dq / (4πε₀R²) dq = σ₀cosθR²sinθdθdφ dEz = (σ₀cosθR²sinθdθdφcosθ(-z)) / (4πε₀R²) dEz = (σ₀cos²θsinθdθdφ(-z)) / (4πε₀) Ez = (σ₀ / (4πε₀)) ∫₀²π ∫₀π/₂ cos²θsinθdθdφ(-z) Let u = cosθ, -sinθdθ = du Ez = (σ₀ / (4πε₀)) ∫₀¹ -u²du Ez = (σ₀ / (4πε₀)) [-u³/3]₀¹ Ez = (σ₀ / (4πε₀)) (-1/3) Ez = -σ₀ / (12πε₀) Problem and Solution: Problem: Calculate the potential at the center of a hemisphere with a given charge density. Solution: Calculate the potential difference dV due to a small charge element dq. Use integration to find the total potential V at the center. Substitute the given charge density and integrate over the appropriate limits. Simplify the expression to obtain the final result. Mathematical Equations: dV = dq / (4πε₀R) V = ∫dV = ∫(σ₀cosθR²sinθdθdφ) / (4πε₀R) V = (σ₀R / (4πε₀)) ∫₀²π ∫₀π/₂ sinθcosθdθdφ V = (σ₀R / (2ε₀)) ∫₀π/₂ sinθcosθdθ V = (σ₀R / (4ε₀)) Total charge on hemisphere: Φ = ∫σdS Φ = ∫σ₀cosθR²sinθdθdφ Φ = σ₀πR²