MIT6_641s09_sol_exam2008

MIT6_641s09_sol_exam2008 - 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 . 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 ? Solution: V cos kx sinh k ( y a ) x a ( x, y ) = sinh ka V cos kx sinh k ( y + b ) b x sinh kb B Question: What are the electric field distributions in the regions 0 < y < a and b < y < ? Solution: E = = i x + i y x y V k sin kx sinh k ( y a ) i x + cos kx cosh k ( y a ) i y 0 < x < a E = sinh ka V k sin kx sinh k ( y + b ) i x cos kx cosh k ( y + b ) i y b < x < sinh kb Question: What are the free surface charge distributions at y = b, y = 0 , and y = a ? Solution: V k f ( x, y = b ) = E y ( x, y = b ) = cos kx sinh kb V k f ( x, y = a ) = E y ( x, y = a ) = cos kx sinh ka f ( x, y = 0) = [ E y ( x, y = 0 + ) E y ( x, y = 0 )] V k V k = cosh ka + cosh kb cos kx sinh ka sinh kb = V 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 on the y = 0 electric potential sheet? k Hint: k k k 2 cos 2 kxdx = 1 2 k k + 2 Solution: i) cos kx sin kxdx = 0 k Force 1 = f ( x, y = 0) [ E y ( x, y = 0 + ) + E y ( x, y = 0 )] area 2 y 1 = [ E y ( x, y = 0 + ) E y ( x, y = 0 )] [ E y ( x, y = 0 + ) + E y ( x, y = 0 )] 2 1 2 2 = [ E y ( x, y = 0 + )] [ E y ( x, y = 0 )] 2 1 2 = ( V k cos kx ) coth 2 ka coth 2 kb 2 Force = f ( x, y = 0) E x ( x, y = 0 + ) =...
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MIT6_641s09_sol_exam2008 - MIT OpenCourseWare...

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