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PHYS809_11F_first_exam_solutions

# PHYS809_11F_first_exam_solutions - October 12th 14th 2011...

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PHYS809 Fall 2011 First take-home 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 Green’s theorem or otherwise, prove Green’s reciprocation theorem: if Φ is the potential due to a volume charge density ϱ in volume V and surface charge density σ on the conducting surface S bounding V, and Φ is the potential due to another charge distribution ϱ and σ , then . V S V S dV dS dV dS ρ σ ρ σ Φ + Φ = Φ + Φ (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) Green’s second theorem is ( ) ( ) 2 2 , V S dV d φ φ φ φ φ φ φ φ - = - S where φ and φ are scalars fields in a volume V bounded by surface S . If φ and φ are the electrostatic potentials resulting from charge distributions ρ and ρ in V and surface charge densities σ and σ on S respectively, then Green’s second theorem gives ( ) ( ) 0 1 , V S dV dS φρ φ ρ φ φ ε - - = - - 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)

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0 . σ ε = - n E Hence ( ) ( ) , V S dV dS φρ φ ρ φσ φ σ - - = - which can be re-arranged to . V S V S dV dS dV dS ρφ φ σ ρ φ φσ + = + Note that since the conductors are equipotential surfaces, this can be written as , V V dV QV dV Q V ρφ ρ φ + = + where the sums are over the surfaces.
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PHYS809_11F_first_exam_solutions - October 12th 14th 2011...

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