309-2008-Solutions9

309-2008-Solutions9 - ECE 309 Electromagnetic Fields...

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Unformatted text preview: ECE 309 Electromagnetic Fields University of Virginia Fall 2008 Homework # 9 Solutions 1. From either Gausss Law or Laplaces equation, the field between the inner and outer con- ductors of a cylindrical capacitor is, ~ E = V r ln( b/a ) r where V is the voltage across the capacitor, a is the inner radius and b is the outer radius. The displacement current density is thus, ~ J d = ~ E t ~ J d = j V r ln( b/a ) where the second relation above holds for phasor representations of ~ J d and V . The total current flowing across a cylindrical surface of radius r is thus, I d = j V ln( b/a ) Z 2 Z rddz r = j 2 V ln( b/a ) which is obviously independent of r . The capacitance of the coaxial capacitor is given by, C = Q V = V = 2 ln( b/a ) where is the charge per unit length on the coaxial capacitor. Thus, the charging current for the capacitor is given (from circuit theory) by, I c = C dV dt I c = jCV = j 2 V ln( b/a ) which is the same as the displacement current found above. 2. Given the electric field, ~ E = xA ( x + y ) + yB ( x- y ) cos t is in a source-free medium, we know that it is divergenceless, ~ ~ E = 0 . Thus, ~ ~ E = E x x + E y y + E z z = A- B = 0 Thus, the relation between the coefficients is, A = B 3. To find the ~ H-field, lets first express the electric field in phasor form, ~ E = yE sin x a e- j z z ( * ) From Faradays Law, ~ ~ E =- j ~ H, ~ H = j ~ ~ E so that ~ H = j h- x E y z + z E y x i = j h j z E sin x a e- j z z x + a E cos x a e- j z z i...
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This note was uploaded on 10/16/2010 for the course ECE 309 taught by Professor Weikle during the Spring '08 term at UVA.

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309-2008-Solutions9 - ECE 309 Electromagnetic Fields...

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