Lect07BoundaryVal - ECE 3030 Electromagnetic Fields and...

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1 ECE 3030 Electromagnetic Fields and Waves Fall 2009 Lecture 7 2009/9/11 Boundary Valued Problems Spherical Solutions 1 Swartz and Rana 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 7: Instructor: Dr. Wesley E. Swartz Capacitors with Dielectric Insulators Conductors Immersed in Electric Fields θ z a x + + + + + ++ --- - -- E o V +- ε + + + + + + + + - - - - - - - - Solutions of Laplace’s Equation in Spherical Coordinates The following are solutions to . Spherically Symmetric Solution A Constant Uniform Electric Field Solution () B r A r + = φ z A r A r = = cos z 0 2 = r + q 3 Swartz and Rana 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 7: A Dipole Oriented Along z-Axis Solution A Constant Potential Solution (a trivial solution): ( ) ( ) z A r r E ˆ = −∇ = 2 cos r A r = z + q - q A r = z x Solutions of Laplace’s Equation in Cylindrical Coordinates (1) The following are solutions to . Cylindrically Symmetric Line Charge Solution: r A r ln = 0 2 = r + λ x y 4 Swartz and Rana 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 7: A Constant Uniform Electric Field Solution x E r E r o o = = cos y x E r r E o ˆ = −∇ = x Solutions of Laplace’s Equation in Cylindrical Coordinates (2) More solutions to . A Line Dipole Oriented Along x-Axis Solution: 0 2 = r r A r cos = y - + x r ln 2 r π = d 5 Swartz and Rana 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 7: A r = r cos 2 d cos 2 d r cos 2 d r ln 2 r o o o + + A constant Potential Solution (a trivial solution): z A Non-Uniformly Charged Spherical Shell (1) Consider a charged spherical shell where the surface charge density is fixed and is given by: Looking from outside the potential looks like that of a dipole. So try a dipole-like solution for r > a : σ cos o = ( ) C cos A r + z + + + + + a 0 6 Swartz and Rana 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 7: For r < a , try something that does not go to infinity at r = 0 : How can you tell if this is the right solution?? ( ) r 2 out = D cos r B r in + = 0 Why? Why? A Non-Uniformly Charged Spherical Shell (2) Our two guessed solutions are: Boundary conditions: (1) Potential inside and outside must be continuous at the surface of the shell: cos o = 2 cos r A r out = cos r B r in = ( ) cos z + + + + + a 7 Swartz and Rana 06/5/25 Electromagnetic Fields and Waves – Fall 2009 Lecture 7: ( ) ( ) ( ) cos 2 a B a A r r a r out a r in = = = = o o o a r out o a r out o a r r in a r r out o B a A r r r r E E cos cos cos 2 cos 3 , , = + = + = = = = = (2) Discontinuity in the radial electric field at the surface must be
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Lect07BoundaryVal - ECE 3030 Electromagnetic Fields and...

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