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# Applied Elec & Magnetism ECE 2317

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This 50 page Class Notes was uploaded by Karolann Wiegand on Saturday September 19, 2015. The Class Notes belongs to ECE 2317 at University of Houston taught by David Jackson in Fall. Since its upload, it has received 84 views. For similar materials see /class/208292/ece-2317-university-of-houston in Electrical Engineering at University of Houston.

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Date Created: 09/19/15

Electromagnetics Workshop Solutions Boundary Conditions Problem 1 The uniform elds shown in the gure below are near a dielectricdielectric boundarybut on opposite sides of it Which con gurations are correct Assume that the boundary is charge ee and that 52 gt 1 m m n I Solution a Normal D is not continuous a this con guration is not correct b Normal D is not continuous a this con guration is not correct c Normal D is continuous D1r gt D2 gt 51E gt 52E E1r E2r gt 51 gt 52 4 this con guration is not correct d Tangential E is not continuous 4 this con guration is not correct e This con guration is correct 0 Tangential E is not continuous a this con guration is not correct Electromagnetics Workshop Solutions Boundary Conditions Problem 2 A silver coated sphere of radius 5 cm carries a total charge of 12 nC uniformly distributed on its surface in free space Calculate a IDI on the surface of the sphere b D external to the sphere Solution Q 3nC 3nC 2 D 38197 C a ps 4an 5cm2 n25x10quot n m Q A3nCAN0955A bD W1 NZ rNV Zr nCmz Electromagnetics Workshop Solutions Boundary Conditions Problem 3 Match the following descriptions with the gures shown below Fields are near the interface but on opposite sides of the boundary E2 E1 D1 1 1 J 2 2 D 1V Vi 2 a Medium 1 and medium 2 are dielectrics with 81 gt 82 b Medium 1 and medium 2 are dielectrics with 81 lt 82 c Impossible d Impossible e There is a positive surface charge on the boundary between two dielectrics f Medium 2 is a perfect conductor Solution 0 0 ii f iii b iV a V d Vi e Electromagnetics Workshop Solutions Boundary Conditions Problem 4 Match the following descriptions with the gures shown The elds in the gures are very near the interface but on opposite sides of the boundary Descriptions can be used once more than once or not at all a Medium 1 and medium 2 are dielectrics with 81 gt 82 Medium 1 and medium 2 are dielectrics with 81 lt 82 Impossible 383 Possible but description is not provided e There is a positive surface charge on the boundary between two different dielectrics f There is a negative surface charge on the boundary between two different dielectrics Medium 2 is a perfect conductor 3 h There is a surface current on the boundary between the dielectrics D1 E1 1 1 2 l 1 O 2 2 D2 0 2 ii iii D1 1 1 DZ 2 2 V El 1 1 2 2 39 E2 0 viii 1 1 ID2 1 Di gt 2 2 DI 2 39 D2 0 X xi xii Electromagnetics Workshop Solutions Boundary Conditions Solution 0 b ii g iii 0 iV a V d Vi e Vii Viii ix 0 K d Xi f Xii c 006 Electromagnetics Workshop Solutions Boundary Conditions Problem 5 Given the potential d3 20 cos for p lt 2 and q 50 cos l20pcos for p gt 2 a Find E in both regions b Show that the boundary conditions are satis ed at the surface p 2 for all if g so for pgt2 and ggrgo for plt2 Find 5 S olution For cylindrical coordinates E9 V p 62 Az A 62 A 6 p 6 62 E9 20cos A pi20psin A5 20cos A 20sin A p lt 2 p E9 cos 50120p2 sin 50 l20p2 A p gt 2 At p 2 the tangential components of the electric eld should be equal 20sin sin 50 120p2 pZ The tangential electric eld components are equal for all at the surface p 2 At p 2 the normal components of the electric ux density should be equal 20 50 120 2 COS 8r cos p p The normal electric ux density components are equal for all at the surface p 2 if g 4 Electromagnetics Workshop Solutions Boundary Conditions Problem 7 A cylindrical capacitor as shown carries a total charge of Q and 7Q on the inner and the outer conductors respectively The capacitor is not connected to the ground or to any voltage source A dielectric washer with inner and outer diameters matching the dimensions of the cylindrical capacitor is placed x meters inside the capacitor as shown in the gure below a Find the electric eld in the capacitor in terms of Q b Find the voltage between the two conductors l X Hlt h X gtl 17IIIIIIIIA o WIIIIIM Solution aQp1xpohexEp2p p lp 27r 0p 1 50 p0 glx 0hix p1 1x 0hix Q EF 27rp 1x 0 hex b cDab Lba Q d Q 2 p 1x 0hix P 2 1x 0hix Electromagnetics Workshop Solutions Boundary Conditions Problem 8 A parallelplate structure with a plate separation d is lled With a material that contains a volume charge density pV az Cm3 and has permittivity g Both plates are held at ground potential 0 V a Using Poisson s equation determine the potential between plates b Determine the surface charge density on the top plate c Determine the surface charge density on the bottom plate AZ I I Solution a Poisson s equation pV 13qu 12 6d OlZZ 3 V2 2 C1 g 62 g 62 25 3 3 z ClzC2 00C20 a q 3 612 V d03C10d2 Z 652 Z g b For top plate pip D i s V i 3zz d2 adz 2 3 Cm l c For bottom plate 20p 5 0 am A A a pf Dz 8 V z E3zZ d2 20 pfm39 2 Cmz Electromagnetics Workshop Solutions Electric Flux Problem 1 Given the eleCLI lC ux density D3 50057 A 717p quot 29p cosZ gt A CmZ Find the net electIic ux in the upward 2 directionthrough the annulus shown inthe gure below agpgb X 1 jD ds ds2pdpd S lIzjli9pcosz pdpd 3b3 fagcosz d I a I w37rb37a3 C Electromagnetics Workshop Solutions Electric Flux Problem 2 A charge Q nC is at the origin of a spherical coordinate system Calculate the electric ux crossing the portion of a spherical shell described by 0 S 639 S g in the outward sense Solution r 4727quot2 In2 D4m2QD Q Total ux 2 2 1 iDrrz sin6d6d 0 0 0 2 M risinadadgj j Igsin d dg 47rr 0 0 47139 Til ag P Q nC Electrom agnetics Workshop Solutions Gauss s Law Problem 1 Consider two long concentric circular cylindrical shells On the inner shell of radius 1 In there is a uniform surface charge density equal to 5 Cmz On the outer shell of radius 2 In there is a uniform surface charge density equal to 5 Cmz Find the electric field in all regions rd 3345 0M haw 59quot g of yd dz f610d d2 00 00 Qlt f Afggffof 0 2 ES l0312f139l thl Us 7 ft f Electromagnetics Workshop Solutions Gauss s Law Problem 2 A point charge of strength q C is located at the center ofa spherical cloud of nonuniform charge density pV 27 ra Crng The radius of the spherical cloud is a Find the vector electric eld E at all points in space including the region inside the charge density and outside the charge density f l39rr O4K af 149 ngrrw39VQd de 37 W V DO DrLlrra 3gt r M9dV49d a Y z v D wr 2041 9V4 Qar Wgrg quotigt77 Electromagnetics Workshop Solutions Gauss s Law Problem 3 An in nite planar slab of charge exists from x 0 m to x 2 m and has volume charge density pV x l Cm3 a Sketch the charge distribution as a function of x Label appropriate values on the vertical aXis b Find the electric eld in all regions State all assumptions and show all work I pvx l Cm 3 b CLane is add so no E 704554 slab Bads I QM 011 y mm A A x 2 on A 331 Dxgt gtlt a 39 Dgtlt X 2 E 0 lg xlt0 2 DeE 75 gtltzf x A X X A 12 ET Y x Ly7 0ltxlt2 60 E O 1 1xgt2 W Electromagnetics Workshop Solutions Gauss s Law Problem 4 A very long cylinder of uniform constant volume charge density pV K 4 CmS has a radius of p 4 m Surrounding this charge is a very long cylinder of uniform surface charge density p K8 Cmz having a radius of p 8 n1 K 4 and K 8 are arbitrary constants Find the electric field vector in all regions of space C Fmd Ihe electric eld vector in all three regions ofspace j E d 0 L S plt4 m 4ltpltlej ka gt mr zrk A of r8 Z fw tgdwayfff pd dfa n 21L l O ZW39DL KL JiPL LE 4 9 Er 39 f3 4 M 26 pLSImI L lt Elt 8 EM 2pr D K 4100th 5 Ef Fl LI llque pr1lt4rcvm1 p Ks CIml Em g gt 6 Electromagnetics Workshop Solutions Gauss s Law Problem 6 An in nite slab of charge of volume charge density pVcoszsinz and thickness 7139 is centered about the zaXis as shown below Find the electric eld at all points in space this includes the regions both inside and outside the slab z72392 coszsinz 20 J pV z 7r2 For the cosine varying charge 7r2 Di idA DZi idA 2DZA A J cosz dz Amp Atomquot 42 2Dzsinz 2 3 E z lso zZIr2 lso zS Ir2 Jquot Dzi idA Dzi 2DZA AT cos 2 dz Amp Aboutquot 4 2DZ sinz 2 sin 2 2sinz 3 EZL Sz 4 so 2 2 For the sine varying charge ame14 DZi idA 2DZAAH2sinz dz 0 Amp Atom 42 E2 0 lleIr2 jjpzi2dADzAAT sinz dz Amp 42 DZ coszz cosz 3 EZ COSZ Sz 7 so 2 2 L ZZ so 2 Using superposition E2 i z S so 2 sin cosz ltZlt g 2 2 Electromagnetics Workshop Solutions Gauss s Law Problem 7 A line source of constant charge density pi0 that exists along the zaXis is surrounded by an electrically neutral metal cylinder of thickness b a and a cylinder with uniform charge density pV0 ofthickness d c Note that d gt c gt b gt a a Find the electric eld in all regions b What must the value of pV0 be so that the electric eld is zero for p gt d c Now that the metal cylinder is grounded Modify the answers in part a to obtain the electric eld in all regions Solution pie2 forplta Q foraltpltb a E221 pmg forbltpltc Vm l39goJ p0pvoirp2 cz forcltpltd p0pvo dZ CZ forpgtd pro 3 b pm ml Cm pie2 forplta Q foraltpltb c E 1 Q forbltpltc Vm 2780 pv0 p2 Cz forcltpltd pvt612 cz forpgtd Electromagnetics Workshop Solutions Gauss s Law Problem 8 E Given an electric ux density vectorQ yzi 3yx2 x22 Cm2 l nd the value of the total charge Q contained within the cube with one vertex at 000 and the diagonally opposite vertex at 111 The point coordinates are measured in meters Fquot What is the value of the net electric ux out of the cube c A point charge q C is added at 1000 Now what is the value for the net electric ux out of the cube d The point charge in c is moved to l2l2l2 Now what is the value for the net electric ux out of the cube Solution 11 11 3 1 a QmQ dSI3xdxdz x2dxdy C S 00 00 2 3 6 11 b wz C 11 c wz C 11 1 7 d c W 62 3 Electromagnetics Workshop Solutions Capacitance Stored Energy Image Theory Problem 1 A cylindrical capacitor with inner radius a and outer radius b is lled with an inhomogeneous dielectric having 8 80k p Where k is a constant Calculate the capacitance per unit length of the capacitor S olution 013513ij dz C 27 skA A P E d dz Z 3 E Vm p pppp pt 0 Z gok oIN q3j39 pi WPM V bZ 39 Ok Zirgok Q Zirgok C F I M b a m Electromagnetics Workshop Solutions Capacitance Stored Energy Image Theory Problem 2 a Determine the capacitance of a parallelplate capacitor lled with two dielectrics as shown in Figure 1 below b A parallelplate capacitor has two layers of dielectrics as shown in Figure 2 How is the potential difference Vapplied between the plates diVided across the dielectrics lb Figure 1 Figure 2 Solution a The two capacitances are in parallel Therefore C C1 C2 C1 sla7lc and C2gzm 3 C csl 82b82 b The two capacitances are in series VV1V2 and KC2 29 V2 C1 81 a Therefore V1 Vi and V2 Vi 1a52b slagzb Eleetron agnetaes Workshop Soluh ons Capacitance Stored Energy Image Theory Prnhlem 3 A parallelrplate eapaertor ls shown below The top plate ls at avoltage of mm whlle the bottom plate ls assunreolto be at avoltage of mV wrth y gt eapaeltor b Uslng your answer to part a ndthe eleetrre fleldlnslde the eapaertor rh hr an rt partb plates Also ndthe total eharge Q C on the top plate rh hr eapaertor A Mi place area Snllltinn d 2Vrno ltxlth Vr h 4 La Erin V2 Cn Ver k Q plA em 9 01 Cey Electromagnetics Workshop Solutions Capacitance Stored Energy Image Theory Problem 4 a A pair of 200mm long concentric cylindrical conductors of radii 50 and 100mm is lled with a dielectric g 1080 A voltage is applied between the conductors to establish an 6 electric eld E Vm between the cylinders Calculate the energy stored p b Determine the capacitance and the applied voltage between the two cylinders S olution a Stored energy is given by WE jsE 261V V 227 1 5 2 712 WE JJ 102 palpaly alzmx02x2 39x1012 gtltln2 2 0 0 05 p 2 WE 3856 J b C 2nd ln2 ln2 C0161 nF WEZiQVZiCTZEV ZWE 2 2x385 2 2 C 0161gtlt10 V 0692 Mv Electromagnetics Workshop Solutions Capacitance Stored Energy Image Theory Problem 5 Use the image theory to nd the capacitance per unit length of the cylinderandplane system shown below S olution 2 1 Electric eld due to p Ep M p561 so 27ml sop Electric eld due to its image E i L so p Zh so Zh p Potential difference between the cylinder and its image 2 a 1 1 2 a 2h a V IL 77 dpzim7 a so p 2h p so a The potential difference between the positive cylinder and the conducting plane is V 2 therefore the capacitance per unit length of the actual system is C pl Zita5 I V2 gm 2h a so a Cl 27rso ln 2h a a Electrom agneucs Workshop Solua ons Capacitance Stored Energy Image Theory Emblem 5 a L F m below b FigureB n a FxgmeA Fxgme 3 5 O 241 12 Snlutinn the xraxxs ahgned wnh the honzomal dAmEnsxon d This is an arbnxary ehoxee th nus assumpuon the eld due to the hue on the le xs Electromagnetics Workshop Solutions Capacitance Stored Energy Image Theory distance between the lines This choice allows us to simplify the eld solution for the line on the left to L Vm Zirgox Similarly the eld solution for the line on the right is E L Vm x 2 3980 x d pl d7 1 1 pl 61 61 a pl d a d J dx ln ln ln V 27239s x x d 2723950 a 61 61 723980 a 0 a pl 7239s 0 ziz o KG LIHVJ a lnd a Its a a 0 Fm The eld solution is exactly the same for the single line above an in nite ground plane However the ground plane is 612 away from the line so the voltage is half of what is applied to the twowire transmission line to have the same elds Hence the capacitance in this case is twice that of the twowire transmission line 2750 C a Fm I he Electromagnetics Workshop Solutions Dielectrics Dielectric Breakdown Problem 1 Fused silica SiOz has a relative permittivity of approximately 38 The density of fused silica is 22 gcm3 There are 2144352X1024 molecules per kg of fused silica a If an electric eld of 10 Vm is applied to fused silica what is the dipole moment of each water molecule Assume that all of the molecules in the volume have the same dipole moment b Assuming that the charge q in the dipole formed by a single fused silica molecule is equal to four times the charge on a proton qw 1602 gtlt10 19 C Calculate the effective displacement distance d between the positive and negative charges inside the fused silica molecule when the electric field inside the fused silica is 10 Vm S olution x P V 1280E8 l50E Assuming all of the molecules in the volume have the same dipole moment we can simplify to g l g EAV Np s lsOE 3 p1 r 0 AV N P g 1g0 EAV N We can arbitrarily choose AV 1 m3 Hence we need to find N the number of molecules in AV 1m3 24 6 3 N W 223g 10 011 amp m 47176gtlt1027moleculesm3 1 kg cm 1 m 1000 g If p and E are aligned we can also write lpl 28 lp W 5255x10 Cm d1 IIquot W e 820lgtlt103924 m q 461 Electromagnetic Workshop Solutions Dielectrics Dielectric Breakdown Problem 2 dielectric strength anywhere in the material a In a coaxial capacitor lledwith insulatingmatelial ofpeimittivity e at what value of p is I u pal a max lm m b What is the breakdown voltage if a 1 cm 17 2 cm and the dielectric material is mica with e 5 and Em zoo MVm Solution Ft a From Gauss s Law we can easily show that E r 2 7r 5 p Hence E isamaximumwhen pa b s m L 5 m b Vignede ZirelnEa lznrmlnm39 EBR 7 2 pLZ EBR IZ SDH um E lzuo Vm ln2 EM HlnZ200X1051X102ln2m1386MV Electromagnetics Workshop Solutions Dielectrics Dielectric Breakdown Problem 3 A metal sphere of radius a is in air The dielectric breakdown of the air is E Vm a What is the maximum total charge me that can be placed on the sphere before the air will break down b Assume that this charge is place on the sphere what is the potential on the surface of the sphere in terms of E5 Solution a By Gauss s Law E QL KZ gt a 47230quot Breakdown occurs where eld is a max ie at surface r a Q E m Vm 5 47139s a2 0 So Qmax 47r50a2E5 C 2 b Note E 2E5 a 2Vm r gt a r r a if Edr J Erdr jig QM QM 4 g a2E 2 aEE Vm 471780 47178061 47178061 Electromagnetics Workshop Solutions Poisson s Equation Problem 1 A parallelplate capacitor is shown below The top plate is at a voltage of V2 V while the bottom plate is assumed to be at a voltage of V V with V2 gt V a Obtain the solution for the potential function inside the capacitor b Find the electric eld inside the capacitor A m2 plate area r pvx71Cm3 I x 3 939 Solution x x2 x2 3 2 7 77 a V2 d pv 1 x d3 2 Cqgtx 2 6 CxD dx 08 08 dx 05 0 r 2 3 x3x x CxD 0 r x0 QDVIC0D3DV1 2 3 Z xzh q3V23h h Ch113CMM 650 h 680 2 3 2 So x3x x Vi Vl 3h h xV1 v 6508 h 650 Z 2 b E Vq6 2xx M3h h iVm0ltxlth 6x 2808 h 6808 Electromagnetics Workshop Solutions Poisson s Equation Problem 2 Consider the coaxial cylinders of length L The outer conductor radius b is at a potential of V and the inner conductor radius a is at zero Volts Between the two conductors is a dielectric with permittivitys This dielectric region is also lled with a volume charge density pV C7 Cm3 foraltpltb p where C0 is a constant Find the electric eld in the region between the conductors Solution V2q3ampii p63 C03 a 63 C0 g p p 6p 8p 6p 6p 8 3p gpclja gg 6p 8 6p 8 p C0 3q3p 7pCllnpC2 V s V0C b a Apply boundary conditions C Vi0 b a EV C0 g C005 Vm Electromagnetics Workshop Solutions Poisson s Equation Problem 3 An in nitely long cylinder of radius a is lled with a charge of uniform density pm Ifthe potential on the surface of the cylinder is V0 what is the potential within the cylinder Solution Poisson s equation VZQD amp 80 10 d2 pm p dp dp 80 p 3 p V0 p2 CllnpC2 480 To keep q 0 nite we must have C1 0 Apply the condition a V3 2 C2 V0M 4s 0 qgtpV04g p a2 p2 for p S a 0 Electromagnetics Workshop Solutions Divergence and Divergence Theorem Problem 1 Vector eld E is characterized by the following properties a E points along 1 b The magnitude of E is a function of only the distance from the origin c E vanishes at the origin d VE 6 everywhere Find an expression for E that satis es these properties Solution E frf l 6 l 6 VB75VZEr7grzfr6 6 3 5rzfr 6r2 3 r2fr I6r2d 3r2fr2r3C 3 fr 2r 2 r Since E vanishes at the origin 3 C 0 3 fr 2r 39 E2rr Vm Electromagnetics Workshop Solutions Divergence and Divergence Theorem Problem 2 a The electric eld in a free space region is given by E 3yz 2 4yzz y 23y i Vm What is the charge density in the region if any Use a volume integral to nd the total charge in the unit cube with one corner at the origin one at 100 another at 010 and another at 001 c Check your result in part b by using a flux integral to compute the charge in the cube b V Solution a VDpV 80 8yz3zzypy pV 50 8yz3zzy Cm3 b Q 33380 8yz3zzy dxdydz sol42gzz dz 0 0 0 0 55 7 c c Q mD dS go 3342 dxdzjjy dxdy50 Cl Q Electromagnetics Workshop Solutions Divergence and Divergence Theorem Problem 3 Find the net ux out ofthe surface shown if D cos 3 sin z i i By examining D explain why the ux through each of the surfaces and is zero ii Evaluate the ux through surfaces and and hence determine the net ux out of the closed surface shown b Determine the net charge enclosed in the surface shown 2 V Solution a i D D M 0 D DZZ0 0 ah pm ID dS illicos pd dz 00 ha ii JD dSIIsin dszl g ah N s a jD dSHzpdpd 00 71761211 71761211 2 4 Hence 1 2ah C 20201 b an 13 dS Qmm 2 QMW 2ah j h C Electromagnetics Workshop Solutions Divergence and Divergence Theorem Problem 4 a Show that the volume charge density corresponding to D lOsin2 3 p l cos2 2 is p independent of and z b Verify the divergence theorem for D with the cylindrical volume shown below Solution 1 6 16D 6D a pVVD pr Z p 6p p 13 62 pV iilOpsin2 g iipi cos2 gi p 6p p 13 62 p 10sin2 cosz p p py Cm3 hZ Z alo J II pdpd dz207rah hZO 0 p LHS jD dSjD dS j D dS battam RHSvaDdV 5de mp hZ 27 27 a 102 LHS j j 10sin2 pd dz 2 j j cosz pdpd rhZ 0 pg 0 0 p zh2 LHS 107rah107rah 207239ah Thus LHS RHS Electromagnetics Workshop Solutions Divergence and Divergence Theorem Problem 5 A static electric eld in free space is given by E i2r2 rsin639 r2 sin6cos Find the volume charge density that produced this electric eld Solution pv VQ50VE In spherical coordinates we have vgi23r2E 1E9sin6 E r 6r rs1n666 rs1n639 6 where Er2r2 E9rsin6 2 Eq r sm6cos Hence we have pvso8r2cos6 rsin Cm3 Electromagnetics Workshop Solutions Exam 1 Review Problem 1 An electric eld is de ned in rectangular coordinates as E x 3y2 Vm 1 Calculate the voltage drop VA where A is the point 200 and B is the point 020 in rectangular coordinate by integrating along a path C1 b path C2 2 Find the potential B when the reference is A l V B 1 a VAR ldLIEdLIxamp3Jc 3d VAR JJ 2 p2 cos sin 302 sin cos gtd Iii7202 sin cos gtd VA 2cos2 Z 4V 0 A B 0 2 b VA IEdLIEdLdexI3ydy B A 2 20 2 VABx 22 0 2 A BVAB3 B A VABl 4 3 V 3y2 7 264V Electromagnetics Workshop Solutions Exam 1 Review Problem 6 Find the electric eld in the x y plane due to the pair of uniformly charged line sources shown in the gure Useful integral J LL13 u2 b2 7 u2 b2 4 Rp z392 l R p2 z 22 311 1 pimpZ 3 4717801 p2z2 1 ap1p3Z 2d Iibp1p3z 2d 4 l7 3 in 3 quotgquot of 262 of 262 A z Az 2 a AZ ZA IMP3 86121 pimp 3azLJlpp 3dz p2z3922 p2z3922 p2z3922 1 moi2 2 amp32 E 1 3612 4171 st 0 j b p2zy2i p2zy2i 1 471780 A a z 22p1L 3dz p2 ZVZ 2 2zpl l l Vm inXyplane g blwzE chf Electromagnetics Workshop Solutions Exam 1 Review Problem 3 A line source of constant charge density p20 exists along the z aXis It is surrounded by an electrically neutral metal cylinder of thickness b a and a cylinder with uniform charge ps Cmz with radius 0 c gt b gt a a Find the electric eld in all regions b What are the surface charge densities at p a and p b c Now that the neutral metal cylinder is grounded Find the electric eld in all regions p20 quot Solution plo forplta 0 foralt ltb a E 1 A p Vm 27803 pZOE forbltpltc pl0 ps0 270 for p gt c pzoL pro 2 pzoL pro 2 b Cm S Cm p5 ZiraL Zita p Zn39bL Zn39b plo forplta 0 foralt ltb c E 1 p Vm Zirsop Q forbltpltc p50 27mg for p gt c Electromagnetics Workshop Solutions Exam 1 Review Problem 4 A static electric eld in free space is given by E ltcos 3zsin zp Vm a Find the volume charge density that produced this electric eld b Find the value of the total charge Q contained within the cylinder formed by the planes p 2 m and z i3 m centered about the origin c What is the value of the net electric ux out of the cylinder d A point charge q 5 C is added at 5 0 0 in rectangular coordinate Now what is the value for the net electric ux out of the cylinder e The point charge in d is moved to 111 in rectangular coordinate Now what is the value for the net electric ux out of the cube Solution a pv VQ 50 VE pV so M mi Cm3 p b Q pdVlzf p pdpd dz Qsoilirjpz dpd dzgo6 ogo so84232 250 c c 11 Q 327250 C d 11 32717280 C e 1 32250 5 0 Electromagnetics Workshop Solutions Electric Potential Problem 1 A disk 0 S p S a z 0 0 S S 27239 carries a surface charge density p p50 p2 a2 Find 00Z in free space Assume oo 0 S olution q J39 p dS p50 27 0 p3 S 4780 4750 a2 I p2 22 dp d d3 222 dp 622 2z3a2 222 la2 22 Electromagnetics Workshop Solutions Electric Potential Problem 2 A line charge in free space with a constant charge density pl forms a quarter circle of radius a that lies in the x y plane with the center of curvature at the origin as shown Find the potential at a point P on the zaXis Assume that the potential is zero at the origin Z P Solution p a 7r2 61 p a 00z l 7 50 I czzz2 850xlczzz2 This is the potential along the zaXis with respect to in nity To change the reference to z 0 we can just subtract the potential at z 0 with respect to in nity pza 00z qgt000 88 V Electromagnetics Workshop Solutions Electric Potential Problem 3 A regular hexagon of charge with line charge density pl exists in the x y plane centered around the origin The hexagon is inscribed Within a circle of radius a a Find the potential due to this charge distribution along the zaxis assuming the reference is at infinity where the potential is zero b Repeat a assuming the reference is at the origin where the potential is 5 V S olution Since the hexagon is regular it can be split up into six equilateral triangles Hence the sides of ME Since the charge density is the triangles are of length a The height of each triangle is uniform we can deduce that the electric field and hence the potential will be equally affected by each of the six sides of the hexagon Hence using superposition aZ z6 pl 31 471396 0 az gig2 y392zz 3p m aZazz2 V a2xazz2 023 1n3C5 2 C5 27139 so 3pz 1 3 271780 n 0 z3ima2 W 5 V 27 50 3 a2a2zz Electromagnetics Workshop Solutions Electric Potential Problem 4 A hollow cylinder of surface charge p5 sin2 Cmz is centered at the origin a Find the potential due to this cylinder at point P00z Assume oo 0 Validate this assumption by evaluating your answer in the limit as z gt oo b Find the electric eld at point P00z by taking the gradient of the potential c Find the electric eld at point P00z using Coulomb s law and compare with the answer in part b Solution Z a h2 27 Sin2 ld ldzl zihZ dZ39 ziln zh2la2zh2Z 471804120aZz z392 4804x2azz z392 450 z h2quotazz h22 lim zh21 a2zh22 1 3 lim 2 a 120 2 z h21lazz h22 2 450 Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 1 Use Stokes Theorem to evaluate V X V dS S i Vxyz xzeyzx 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 27 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 xy2fiy227zx2 2 and C is the triangle with vertices 100 010 nd000 ii Vxyzyz 2xz 7e yi and C is the circle x2 y2 16 Z 5 Solution i EV drjvxv dsjVxV2dS vaz dS m0quot1 2ydydx C S S S ii UlvdrIVxV39 dSlVXV39idS WW 251042 H pdpdg 807 Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 3 y A 2 7 is conservative In other words a Explain how you know that Ex y z m 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 Solution 1 A A l A a Exyz s1n xcos y P P V x E 0 so it is path independent b E VqgtE v i i p p6 13 C Electromagnetics Workshop Solutions Gradient Curl Faraday s Law Problem 4 A possible electric eld is given as E 2y 2x sin y7 a Determine V X E and use it to decide whether integrals of the form IE 6 are independent of path C b If IE 616 is path independent nd a potential G such that E V C O V Evaluate the line integral IE 6 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 Isinydy l cos2 m 1416 0 160 0 proints are interpreted in rectangular coordinates Z Z 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 elds 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 eld i mEd 0 C iiVgtltE0 iii E Vqgt Assume that the potential associated with the charge distribution in part a depends on p O V 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 eld from parts a and b You may assume any convenient reference point VD pV VD1 pgopz3pgo Cm3 l 6 l 6 VD s 2sin2 2 p6pp op p62 1 l3 l i VXEl 66p 66 aaz0 p 3p2 0 0 p350pz cos2 3p50 sin2 3ps0 cos2 3p50 CmS 9 MP 2 Bqu 66 662 6pzcos sin A 3pzcos2 A 2p2sin cos A 0 p2 sin2 0 3pzcos2 Dp 80 is the valid eld MD A 1 MD A MD A 6 E Vq3 z 2 2 2 q C F Flap WM 02 J p p p 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 V 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 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 O V S olution a Px p dp d pxquot p dp p I quot 4 Hm z lm z lpz z 000 0 0 hm ampVazzz lzl0 z gtoo 250 Cili39czzz2 z zgt0 pxo 28 0 Iaz z2 zV 8 iVa2zz 2 zlt0 dz Apso M Z E V 250z l m c hm p50 1L 2 pail HwZSO 2 612422 280 z 25 z Electromagnetics Workshop 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 and 2 directions Find the potential x within the slab x Solution Ex x Since VDpV 3Eh or IdEXhIdx dx 50 E1 50 x1 3ExpV x xlE1 for x1ltxltx2 80 Now E V dd pm I pm 3 EExx xlEl or idV 39 80x x1E1 dx Hence x V x Jcl2 E1x g 1 forx1ltxltx2 80

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