{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT6_641s09_sol_quiz2006_2

MIT6_641s09_sol_quiz2006_2 - MIT OpenCourseWare...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 0 z lim r →∞ E = E 0 i z System is in time independent steady state E = E 0 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 ( ) inside and outside the sphere in the time independent steady state? Solution: Φ( r = 0) finite Φ( r = R ) = Φ( r = R + ) 1
Image of page 2
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 0 z = E 0 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 0 r cos θ C = E 0 1 BC II Φ( r = R ) = Φ( r = R + ) AR = E 0 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 0 ∂r ∂r R 3 1 σ 2 2 D σ 2 2 σ 2 = = E 0 D E 0 + D E 0 R 3 σ 1 R 3 σ 1 R 3 σ 1 1 2 σ 2 1 σ 2 R 3 E 0 1 σ σ 1 2 σ 1 σ 2 D R 3 + R 3 = E 0 E 0 D = 2 σ 2 D = R 3 E 0 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 0 + E 0 = E 0 = E 0 = A σ 1 + 2 σ 2 σ 1 + 2 σ 2 σ 1 + 2 σ 2 3 σ 2 E 0 r cos θ r < R Φ = 2 σ 2 + σ 1 ( σ 1 σ 2 ) R 3 E 0 r (2 σ 2 + σ 1 ) r cos θ r > R 2 Φ 1 Φ E = −� Φ = i r + i θ ∂r r ∂θ 2 3 σ σ 2 2 + E σ 0 1 cos θi r sin
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern