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

2024
Views:

182

Pages:

4
Author:

Jabin F.

Header: "# Boundary Condition on E⊥V" External E.F.: "E_ext = E_0y + E_0z" E_above: "E_above = E_0y + (E_0 + σ/(2ε_0))z" E_below: "E_below = E_0y + (E_0 - σ/(2ε_0))z" E_above_parallel: "E_above_parallel = E_0" E_below_parallel: "E_below_parallel = E_0" E_above_perpendicular: "E_above_perpendicular = E_0 + σ/(2ε_0)" E_below_perpendicular: "E_below_perpendicular = E_0 - σ/(2ε_0)" Tangential Component of E.F. is Continuous across boundary surface charge σ: "E_above_parallel - E_below_parallel = 0" Eabove + Ebelow = Eo Eabove|z=0 = Ebelow|z=0 = Eo Parallel component of E.f is discontinuous across boundary surface. if E is given and p = ? we will use ∇.E = ρ/εo and if σ = ? than we required boundary condition. for example: Ein = 0 Eout = q/4πεor^2 Ein + = 0 Ein|rR = q/4πεor^2 Vin|rR = q/4ππεor^2 q/4ππεor^2 - 0 = σ/εo σ = q/4πR^2 Vabove - Vbelow = 0 Potential is continuous across the boundary surface. Now from Eqn. (D)/Back pages Eabove - Ebelow = σ/εo E = -∇V -∂Vabove/∂z|_{z=0} + ∂Vbelow/∂z|_{z=0} = σ/εo -∂Vabove/∂r|_{r=R} + ∂Vbelow/∂r|_{r=R} = σ/εo Q.No. 33: A conducting sphere of Radius R is placed in a uniform E.F Eo directed along +z axis. The electric potential for outside point is given as Vout = -Eo(1 - R^3/r^3)r cos θ Where r is the distance from the centre and θ is the polar angle. The charge density on the surface of the sphere is: V is given σ = ? -∂Vabove/∂r|_{r=R} + ∂Vbelow/∂r|_{r=R} = σ/εo σ = -εo * ∂Vabove/∂r|_{r=R} = -εo * d/dr[-Eo(1 - R^3/r^3)r cos θ] σ = 3εoEo cos θ Q.No. 34: θ = 30°, σ = 3√3/2 εoEo Charge on upper hemisphere: Q = ∫(∫(3εoEo cos θ)r^2 sin θ dθ dφ)|_{r=R} = 3ππεoEoR^2 Charge on lower hemisphere: Q = -3ππεoEoR^2