MIT6_641s09_sol_exam2009

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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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6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 Final - Solutions - Spring 2009 Prof. Markus Zahn MIT OpenCourseWare Problem 1 Figure 1: A diagram of a sheet of surface charge at y = 0 between two grounded perfect conductors at y = b and y = a (Image by MIT OpenCourseWare). A sheet of surface charge with surface charge distribution σ s ( x, y = 0) = σ 0 sin kx is placed at y = 0, parallel and between two parallel grounded perfect conductors at zero potential at y = b and y = a . The regions above and below the potential sheet have dielectric permittivities of 2 and 1 . Neglect fringing field effects. A Question: What are the electric potential solutions in the regions 0 y a and b y 0 ? Solution: A sinh k ( y a ) sin kx 0 < y < a Φ( x, y ) = B sinh k ( y + b ) sin kx b < y < 0 Φ( x, y = 0 ) = Φ( x, y = 0 + ) ⇒ − A sinh ka = B sinh kb Φ E y ( x, y = 0 + ) = = Ak cosh k ( y a ) sin kx = Ak cosh ka sin kx y =0+ y =0 + ∂y Φ E y ( x, y = 0 ) = = Bk cosh k ( y + b ) sin kx = Bk cosh kb sin kx y =0 y =0 ∂y 1
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C Final - Solutions - Spring 2009 6.641, Spring 2009 σ s ( x, y = 0) = σ 0 sin kx = 2 E y ( x, y = 0 + ) 1 E y ( x, y = 0 ) = 2 Ak cosh ka sin kx + 1 Bk cosh kb sin kx σ 0 = 2 Ak cosh ka + 1 Bk cosh kb A sinh ka A sinh ka cosh kb B = sinh kb ⇒ − 2 Ak cosh ka 1 k sinh kb = σ 0 ⇒ − Ak 2 cosh ka + 1 sinh ka cosh kb = σ 0 sinh kb σ 0 sinh kb A = k [ 2 cosh ka sinh kb + 1 sinh ka cosh kb ] A sinh ka σ 0 sinh ka B = = sinh kb k [ 2 cosh ka sinh kb + 1 sinh ka cosh kb ] Φ( x, y ) = σ 0 sinh kb sinh k ( y a ) sin kx A sinh k ( y a ) sin kx = k [ 2 cosh ka sinh kb + 1 sinh ka cosh kb ] 0 < y < a σ 0 sinh ka sinh k ( y + b ) sin kx B sinh k ( y + b ) sin kx = k [ 2 cosh ka sinh kb + 1 sinh ka cosh kb ] b < y < 0 B Question: What are the electric field distributions in the regions 0 < y < a and b < y < 0 ? Solution: Φ Φ 0 < y < a E = −� Φ = i x + i y = Ak cosh kx sinh k ( y a ) i x + sin kx cosh k ( y a ) i y ∂x ∂y b < y < 0 E = Bk [cos kx sinh k ( y + b ) i x + sin kx cosh k ( y + b ) i y ] Question: What are the free surface charge distributions at y = b and y = a ?
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