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Unformatted text preview: PHYS809 Fall 2011 First takehome exam October 12 th 14 th , 2011 Please write your name on each answer sheet. Since this is a take home exam, there are a few rules that you must follow. You cannot discuss or communicate about the exam with any one by any means. You cannot look for or use solutions to the problems on the internet. You cannot use printed solutions to Jackson problems. You can use your textbook, notes, and the online course notes. You can use Mathematica or similar software, but you must say when you have used it. The problems do not have equal weight. 1. (Basically Jackson problems 1.12 and 1.13) (a) By using Greens theorem or otherwise, prove Greens reciprocation theorem: if is the potential due to a volume charge density in volume V and surface charge density on the conducting surface S bounding V, and is the potential due to another charge distribution and , then . V S V S dV dS dV dS + = + (b) A point charge q is located between two parallel grounded conducting planes, as shown in the figure. Use Green's reciprocity theorem to find the net charge induced on each plane in terms of q and the distances a and b . (a) Greens second theorem is ( ) ( ) 2 2 , V S dV d  =  S where and are scalars fields in a volume V bounded by surface S . If and are the electrostatic potentials resulting from charge distributions and in V and surface charge densities and on S respectively, then Greens second theorem gives ( ) ( ) 1 , V S dV dS  =  E E n where n is the unit vector normal to S pointing out of V . Now if the surface S is a conductor (so that the electric field is zero inside the conductor) . =  n E Hence ( ) ( ) , V S dV dS  = which can be rearranged to . V S V S dV dS dV dS + = + Note that since the conductors are equipotential surfaces, this can be written as , V V dV QV dV Q V + = + where the sums are over the surfaces....
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This note was uploaded on 12/02/2011 for the course PHYS 809 taught by Professor Macdonald during the Fall '11 term at University of Delaware.
 Fall '11
 MacDonald

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