MIT6_641s09_sol_quiz2006_2

MIT6_641s09_sol_quiz2006_2 - MIT OpenCourseWare...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 6.641 — Electromagnetic Fields, Forces, and Motion Spring 2006 Quiz 2- Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 1 R ε 2 , σ 2 ε 1 , σ 1 z θ z lim r →∞ E = E i z System is in time independent steady state E = E i z Figure 1: Lossy dielectric sphere within a lossy dielectric with an imposed uniform electric field E = E o i z . (Image by MIT OpenCourseWare.) A Question: What are the necessary boundary conditions to solve for the electrostatic scalar potential Φ( r, θ ) and electric field E ( rθ ) inside and outside the sphere in the time independent steady state? Solution: Φ( r = 0) → finite Φ( r = R − ) = Φ( r = R + ) 1 Spring 2006 Quiz 2 6.641, Spring 2005 J r ( r = R − ) = J r ( r = R + ) ⇒ σ 1 E r ( r = R − ) = σ 2 E r ( r = R + ) Φ( r → ∞ ) = − E z = − E r cos θ B Question: Find Φ( r, θ ) and E ( r, θ ) in the time independent steady state. Solution: Φ = R ( r ) F ( θ ) Solution for n = 1 case: R ( r ) = Ar + B r 1 2 , F ( θ ) = C cos θ. Ar + B 1 cos θ r < R Φ = r 2 Cr + D 1 cos θ r > R 2 r BC I Φ( r = 0)finite ⇒ B = 0 BC IV Φ( r → ∞ ) = − E r cos θ ⇒ C = − E 1 BC II Φ( r = R − ) = Φ( r = R + ) ⇒ AR = − E R + D R 2 ∂ ∂ 2 D BC III J r ( r = R − ) = J r ( r = R + ) ⇒ − σ 1 Φ( r = R − ) = − σ 2 Φ( r = R + ) ⇒ σ 1 A = + σ 2 − E − ∂r ∂r R 3 1 σ 2 2 D σ 2 2 σ 2 = = − E − D − E + D − E − R 3 σ 1 R 3 σ 1 R 3 σ 1 1 2 σ 2 1 σ 2 R 3 E 1 − σ σ 1 2 σ 1 − σ 2 D R 3 + R 3 = E − E ⇒ D = 2 σ 2 ⇒ D = R 3 E σ 1 + 2 σ 2 σ 1 σ 1 1 + σ 1 σ 1 − σ 2 σ 1 − σ 2 − σ 1 − 2 σ 2 3 σ 2 ⇒ A = − E + E = E = − E = A σ 1 + 2 σ 2 σ 1 + 2 σ...
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This note was uploaded on 12/02/2010 for the course ECE 6.641 taught by Professor Zahn during the Spring '09 term at MIT.

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MIT6_641s09_sol_quiz2006_2 - MIT OpenCourseWare...

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