MIT6_641s09_sol_exam2008

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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 .
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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
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+ + + + + 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
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