Homework2 - Homework 2 Answers, 95.657 Fall 2011,...

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Unformatted text preview: Homework 2 Answers, 95.657 Fall 2011, Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Problem 1 Jackson 1.5 The time-averaged potential of a neutral hydrogen atom is given by = q 4 e r r 1 r 2 where q is the magnitude of the electronic charge, and -1 = a /2, a being the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give this potential and interpret your result physically. SOLUTION: The Poisson equation links charge densities and the electric scalar potential that they create. We use it here to find the charge density. We must perform a straight-forward differentiation in spherical coordinates. 2 = Expand this in spherical coordinates: 1 r 2 r r 2 r 1 r 2 sin sin 1 r 2 sin 2 2 2 = The potential is spherically symmetric, so that the potential depends only on the radial coordinate - the partial derivatives of the potential are all zero, except for the one with respect to the radial component. 1 r 2 r r 2 r = Evaluate the equation explicitly: 1 r 2 r r 2 r [ q 4 e r r 1 1 2 r ] = q 4 1 r 2 r r 2 r [ e r r ] r 2 r [ 2 e r ] = q 4 1 r 2 r r 2 [ 1 r r e r e r r 1 r ] 2 2 r 2 e r = q 4 1 r 2 r r e r r 2 e r r 1 r 2 2 r 2 e r = q 4 r 2 e r 3 2 e r q 4 1 r 2 r r 2 e r r 1 r = = q 3 8 e r e r q 4 1 r 2 r r 2 r 1 r Now we must be careful because 1/ r blows up at the origin. Split the last term into two cases: = q 3 8 e r e r q 4 1 r 2 r r 2 r 1 r if r 0 = q 3 8 e r e r q 4 1 r 2 r r 2 r 1 r if r > 0 Away from the origin, 1/ r does not blow up and the derivatives can be evaluated normally. The last term ends up equating to zero, so that our equations now becomes: = q 3 8 e r e r q 4 1 r 2 r r 2 r 1 r if r 0 = q...
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Homework2 - Homework 2 Answers, 95.657 Fall 2011,...

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