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# Note for ECE 2317 with Professor Jackson at UH curl and faraday

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This 8 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 42 views.

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Date Created: 02/06/15

Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 1 Use Stokes Theorem to evaluate V X V dS S i Vxyz x26 yzexzfi zze yi and S is the hemisphere x2 y2 z2 4 z Z 0 oriented upward ii Vx y z xyz xyfi xzyzi and S consists ofthe top and the four sides but not the bottom of the cube with vertices i1 ilil oriented outward S olution i IltVgtltV dS mV dr 2fx2 y2 2d 22J7 4cos2 sin 4sin2 cos d l l 8 cos3 sin3 0 a M d 0 1 1 71 1 ii IltVgtltV dS mVdrxyzyildxJxyl alyJcyzy1 dxjxy dy0 s c 71 Z 1 71 x1 1 24 1 x71 Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 1 Use Stokes Theorem to evaluate V X V dS S i Vxyz x26 yzexzfi zze yi and S is the hemisphere x2 y2 z2 4 z Z 0 oriented upward ii Vx y z xyz xyfi xzyzi and S consists ofthe top and the four sides but not the bottom of the cube with vertices i1 ilil oriented outward S olution i IltVgtltV dS mV dr 2fx2 y2 2d 22J7 4cos2 sin 4sin2 cos d l l 8 cos3 sin3 0 a M d 0 1 1 71 1 ii IltVgtltV dS mVdrxyzyildxJxyl alyJcyzy1 dxjxy dy0 s c 71 Z 1 71 x1 1 24 1 x71 Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 2 Use Stokes Theorem to evaluate mV dr In each case C is oriented counterclockwise as C Viewed from above i Vxyz xy2 y 22 zx2 2 and C is the triangle with vertices 100 010 lad 000 ii Vxyzyz 2xz 7e yiandCisthe circle x2y2l6z5 S olution i ledrlVxV39 dSlVXV idS lVXVz dSliloM 2ydydxi ii UlvdrIVxV39 dSlVXV39idS WW ls lowlan S H pdpdg 807 Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 3 A x yX a Explain how you know that Exyz x2 y2 x2 y2 7 is conservative In other words its integrals of the form IE d are path independent C b Find a potential function G for the vector eld E given above such that E V S olution a Exyz sin cos DIH V X E 0 so it is path independent b E V E V i i3 l p p 13 13 C Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 4 A possible electric eld is given as E 2y 2x sin y 7 a Determine V X E and use it to decide whether integrals of the form IE 6 are C independent of path b If IE 616 is path independent nd a potential G such that E V C c Evaluate the line integral IEd where C is the path from 0 0 0 to 2 7r2 0 both in C cylindrical and rectangular coordinate S olution x y z VXE aax My 1362 22 20 IEd is path independent 2y 2xsiny 0 C 6d E Vqgt 3m y 6 22 6x By 62 a 2y 3 2yxCr 6x 6d y2xs1ny 3 cosy 2yxCy 6y 6 Z0 3 CZ 62 d3 cosy 2xyC If points are interpreted in cylindrical coordinates Z Z Ilt2y lt2xsinygt 7gt Idy sin ydy l cos2 m 1416 0 160 0 If points are interpreted in rectangular coordinates 7r2 7r2 Z J2yy0 dx J 2xsinyx2 dy 0 J 4sinydy 27rlm 7283 0 0 0 Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 5 Two students propose the following two different ux densities as solutions to an electrostatics problem Dp 80p2 A Dp sopz sin2 A 7380 pzcos2 A a Show that the volume charge densities that correspond to these two fields are identical b Choose one of the following equivalent statements and showing all work use it to determine which of the ux densities in part a is a valid electric field i mEd 0 C iiVgtltE0 iii E Vqgt c Assume that the potential associated with the charge distribution in part a depends on p alone that is we may write the potential as p Show that setting E V results in a simple differential equation for the potential Solve this equation for the unknown potential by integrating the equation using the valid field from parts a and b You may assume any convenient reference point VD pV VD1 pgopz3pgo Cm3 VD2iipgop2 sin2 g i3p3sopzcos2 3p80 sin2 3pg0 cos2 3ps0 CmS pap p62 1 l3 l i VXEl 66p 66 aaz0 p 3p2 0 0 9 pt I Bqu 66 662 6pzcos sin A 3pzcos2 A 2p2sin cos A 0 p2 sin2 0 3pzcos2 Dp gopz A is the valid field ad 16d am 3 2 M p3 E Vqgt z F q6p pa az VgtltEpi p Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 6 A disk of surface charge p50 Cmz with radius a is centered at the origin in the Xy plane a Find the potential due to this charge at point P00z Assume d 00 0 Validate this assumption by evaluating your answer in the limit as z gt oo b Find the electric eld at point P00z c Assume that the disk is in nitely large Find the electric eld at point P00z How does it compare with the electric eld due to an in nite planar charge distribution p50 Solution a 2 ps0 p dp d p50 pdp pm 2 2 p50 2 2 q V Z 47504 pr2zz 280 pr222 280 p 2 0 80 a 2 ia2zz z zgt0 b E V Ap50 if 28 Iazz2 2 zlt0 dz Apso IZI Z E V E250Z Iazz2 m c hm p50 1L 2 p50 1 HwZSO 2 612422 280 z Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 7 A slab of charge has a uniform charge density pm At x x1 d3 1 and Ex E There is no variation of E in the y andz directions Find the potential x within the slab y 1191 va Z x x1 9 Solution Ex x Since VDpV 3Eh or J39dExth39dx dx 80 El 80 I 3Exhx xlE1 for Jgltxltx2 80 Now E V dd pm D pm 3 EExx xlEl or idV 39 870x x1E1 dx Hence dx 7V c Jc12 E1c clq31 forx1ltxltx2 s

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