MIT6_641s09_sol_exam2008

# MIT6_641s09_sol_exam2008 - MIT OpenCourseWare...

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

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 .

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

View Full Document
C 6.641 — Electromagnetic Fields, Forces, and Motion Spring 2008 Final- Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 1 A Question: What are the electric potential solutions in the regions 0 y a and b y 0 ? Solution: V 0 cos kx sinh k ( y a ) 0 x a Φ( x, y ) = sinh ka V 0 cos kx sinh k ( y + b ) b x 0 sinh kb B Question: What are the electric field distributions in the regions 0 < y < a and b < y < 0 ? Solution: Φ Φ E = −� Φ = i x + i y ∂x ∂y V 0 k sin kx sinh k ( y a ) i x + cos kx cosh k ( y a ) i y 0 < x < a E = sinh ka V 0 k sin kx sinh k ( y + b ) i x cos kx cosh k ( y + b ) i y b < x < 0 sinh kb Question: What are the free surface charge distributions at y = b, y = 0 , and y = a ? Solution: ε 0 V 0 k σ f ( x, y = b ) = ε 0 E y ( x, y = b ) = cos kx sinh kb ε 0 V 0 k σ f ( x, y = a ) = ε 0 E y ( x, y = a ) = cos kx sinh ka σ f ( x, y = 0) = ε 0 [ E y ( x, y = 0 + ) E y ( x, y = 0 )] V 0 k V 0 k = ε 0 cosh ka + cosh kb cos kx sinh ka sinh kb = ε 0 V 0 k cos kx [coth ka + coth kb ] 1
+ + + + + Spring 2008 Final Exam 6.641, Spring 2008 D Question: What are the x and y components of force per unit area on a wavelength of width 2 π on the y = 0 electric potential sheet? k Hint: π k π k k 2 π cos 2 kxdx = 1 2 π k k + 2 π Solution: i) cos kx sin kxdx = 0 π k Force 1 = σ f ( x, y = 0) [ E y ( x, y = 0 + ) + E y ( x, y = 0 )] area 2 y 1 = ε 0 [ E y ( x, y = 0 + ) E y ( x, y = 0 )] [ E y ( x, y = 0 + ) + E y ( x, y = 0 )] 2 1 2 2 = ε 0 [ E y ( x, y = 0 + )] [ E y ( x, y = 0 )] 2 1 2 = ε 0 ( V 0 k cos kx ) coth 2 ka coth 2 kb 2 Force = σ f ( x, y = 0) E x ( x, y = 0 + ) = σ f ( x, y = 0) E x ( x, y = 0 ) area x = ε 0 V 0 k cos kx [coth ka + coth kb ] V 0 k sin kx = ε 0 V 0 2 k 2 sin kx cos kx [coth ka + coth kb ] π k Force k Force = dx area 2 π y π k area y π k k 1 ε 0 V 0 2 k 2 coth 2 ka coth 2 kb cos 2 kxdx = 2 π 2 π k 1 2 = ε 0

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern