# emsol13 - Electromagnetic Theory I Solution Set 13 Due: 28...

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Unformatted text preview: Electromagnetic Theory I Solution Set 13 Due: 28 November 2011 49. Consider the ideal circular parallel plate capacitor of radius a and plate separation d a , hooked up to a straight current-carrying wire on the axis as pictured in the figure to J, Problem 6.14. The current in the wire varies harmonically, I ( t ) = I cos t = Re I e it . In this problem we neglect the effect of fringing fields, which means that the fields within the capacitor are assumed to be those between infinite parallel plates, which discontinuously drop to zero at the edge of the capacitor. This exercise will give you experience with the use of complex fields in solving physical problems. a) In the approximation described above we may make the ansatz that the (complex) fields within the capacitor have the form E = zf ( ) e it , B = zg ( ) e it From the (complex) Maxwell equations determine g in terms of f and show that f ( ) satisfies the n = 0 Bessel equation, whose solution is f ( ) = AJ ( ). Solution : To write the Maxwell equations we evaluate the curls of the above forms (dropping the e it factors: E = zf ( ) = f ( ) B = z ( g ) z ( g ( )) = z (2 g ( ) + g ( )) Then the sourceless Maxwell equations read f ( ) = ig ( ) , (2 g ( ) + g ( )) = if (1) The first equation determines g = f /i , which substituted in the first equation gives 2 f f f = 2 f f + f + 2 f = 0 (2) Changing variables to x = , we recognize the last equation as the Bessel equa- tion of order = 0, with general solution a linear combination of J ( ) and N ( ). However, since there are no sources between the plates, N , which is singular as 0, is not allowed. Thus we concude that f = AJ ( ) , g = A i J = A i J 1 ( ) (3) 1 b) Now consider the low frequency (quasi-static) limit. Expand the electric and magnetic fields within the capacitor up to quadratic order in . Determine A up to this order by demanding charge conservation, I = iQ , where Q is the (complex) charge on the top plate obtained by identifying the surface charge density = n E in terms of the electric field between the plates. (Here we are neglecting the field outside the plates.) Solution : From the properties of Bessel fundtions J ( x ) = 1 x 2 / 4 + O ( x 4 ) and J 1 ( x ) = ( x/ 2)(1 x 2 / 8 + O ( x 4 )). Then E = A z parenleftbigg 1 2 2 4 + O ( 4 ) parenrightbigg B = iA 2 ( z ) parenleftbigg 1 2 2 8 + O ( 4 ) parenrightbigg Q = 2 A integraldisplay a d parenleftbigg 1 2 2 4 + O...
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## This note was uploaded on 01/08/2012 for the course PHY 6346 taught by Professor Staff during the Spring '08 term at University of Florida.

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emsol13 - Electromagnetic Theory I Solution Set 13 Due: 28...

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