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Homework8

Homework8 - Homework 8 Answers 95.657 Electromagnetic...

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Homework 8 Answers, 95.657 Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Supplemental Problem C For the two cases shown below, (a) calculate all the monopole, dipole and quadrupole moments and write down the potential far away in terms of these moments, (b) sketch the equipotential lines and electric field lines, and (c) find the exact solution to the first case using Coulomb's law and take the far away limit to check part a. SOLUTION: Case 1 (a) We can use the spherical harmonics expansion derived in lecture to find the different moments. The charge density for case 1 is expressed as: = q a 2  r a  cos  −−−/ 2 −− 3 / 2 This can be checked by integrating over all space and making sure the total charge equals zero. Plugging this into the multipole moment definition after expanding the spherical harmonics: q lm = Y l m *  ' , ' r ' l  x ' d x ' q lm = 2 l 1 4 l m ! l m ! 0 0 2 0 e i m ' P l m cos ' r ' l  x ' r ' 2 sin ' dr ' d ' d ' q lm = q a l 2 l 1 4 l m ! l m ! P l m 0 0 2 e i m '  '  ' −− ' −/ 2 − ' 3 / 2 d ' q lm = q a l 2 l 1 4 l m ! l m ! P l m 0 [ 1 − 1 m −− 1 m i m i m ] q lm = q a l 2 l 1 4 l m ! l m ! P l m 0  1 − 1 m  1 i m The problem asks for the explicit forms of the monopole, dipole, and quadrupole moments. We can now find these easily by plugging in the appropriate l 's and m 's in the equation above.

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The monopole moment is: l = 0, m = 0: q 00 = 0 This makes sense because the monopole moment is just the total charge and the total charge in this case is zero. The dipole moments are: l = 1, m = -1: q 1, 1 = 0 l = 1, m = 0: q 1,0 = 0 l = 1, m = 1: q 1,1 = 0 This also makes sense. A close inspection of the geometry reveals that the symmetry would lead to no overall dipole moment. The quadrupole moments are: l = 2, m = -2: q 2, 2 = qa 2 15 2 l = 2, m = -1: q 2, 1 = 0 l = 2, m = 0: q 2,0 = 0 l = 2, m = 1: q 2,1 = 0 l = 2, m = 2: q 2,2 = qa 2 15 2 The potential in terms of these moments is:  r , , ≈ 1 4  0 l = 0 2 m =− l l 4 2 l 1 q l m Y lm  ,  r l 1  r , , ≈ 1 4  0 4 5 q 2, 2 Y 2, 2  ,  r 3 1 4  0 4 5 q 2,2 Y 2,2  ,  r 3  r , , ≈ 1 4  0 4 5 1 4 15 2 sin 2 1 r 3 [ q 2, 2 e i 2 q 2,2 e i 2 ]  r , , ≈ 3 qa 2 4  0 sin 2 cos 2  r 3
(b) It is hard to plot three dimensions and have it be visually meaningful. Let us instead plot the cross- section of three-dimensional space at the x - y plane. This corresponds to an angle θ = π /2, so that the potential becomes:  r , =/ 2, ≈ 3 q a 2 4  0 cos 2  r 3

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