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# Note for ECE 2317 with Professor Jackson at UH stored energy

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This 7 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 277 views.

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Date Created: 02/06/15
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 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 a ib J T i 7 M Figure 1 Figure 2 S olution a The two capacitances are in parallel Therefore C C1 C2 ac ab c Cslj and C252 1 3 C csl 82b82 b The two capacitances are in series VV1V2 and 5gi V2 C1 8161 Therefore V1Vi and V2Vi 1a52b slagzb Eleetron agnetaes Workshop Soluu ons Capacitance Stored Energy Image Theory Prnhlem 3 A parallelrplate eapaettor ts shown below The top plate ts at ayoltage of mm whlle the bottom plate ts assurneolto be at ayoltage of mV wtth y gt y a Solve the Laplaee equataon to obtatn the solutton forthe potentaal funetaon tnstole the eapaettor b Ustng your answer to part a ndthe eleetrte fleldlnslde the eapaettor e Ustng your answer to part b oleternune the eharge densmes on the top and bottom plates Also ndthe total eharge Q C on the top plate d Ustng your answer from part e oleternune the eapaettanee othe parallelrplate eapaettor A Mi plate urea 11 z Snlutinn db V26 0 a a x0 VC oer3134 V xh qgtyzc hVC Vt So XVV 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 2 227 1 105 1 WE JJ J 2 palpaly alzmx02x2 39x1012 gtltln2 2 0 0 05 2 WE 3856 J b ZIrgL 27rgtlt10gtlt8854gtlt103912 gtlt02 ln2 ln2 C0161 nF WE 1QVlcw V W L8Sg 2 2 C 0161gtlt10 V0692 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 p5 27ml amp so 27ml so p Psa Psa sop 2h so2h p Potential difference between the cylinder and its image 2 a 1 1 2 a 2h a V I p5 7 dp p5 ln J 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 P1 27ml Electric field due to p E 0 Electric field due to its image E C 7 I V2 gm 2h a so a Cl 27rso Zh a In a Eleetrorn agnetaes Workshop Soluh ons Capacitance Stored Energy Image Theory Prnhlem 5 a Flnd the eapaertanee per unrtlength of the twowrre txansmlsslon llne shown m Flgure A below b Uslng the result m part a ndthe eapaertanee per unrt length ofa slngle transrnrssron llne above an rn nrte gonna plane as shown m Flgure B Forboth problems assume 1 III a Frgme A Frgme 3 i en O 211 12 T Snlntln The eenter ofthe llne on the lelt ls assurneolto be the ongln of our global coordlnate system wrth the xraxls allgned wth the honzontal dlmenslon a Thls ls an arbrtrary ehoree Wrth ths assurnptaon the fleld olue to the hue on the lelt ls 7 74 Vrn P 27rer Recogmzlng that we only neeol the eornplete fleld solutron for a slngle path between the two wrres we ean make the eonvenrent ehoree ofsolvlng for the total elds along the shortest 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 E L Vm x Zirgox Similarly the eld solution for the line on the right is pi E Vm x 2750 x d din d3 pl 1 dx pl lnda ln a j pl lnda V 2723950 a x x d 2723950 a 61 61 723980 a CI zgz pl g 80 Fm KG lm 61 61 In d a 723980 a a 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 C 2 Fm I ln d a a

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