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

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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  0 e − r r 1 r 2 where q is the magnitude of the electronic charge, and α -1 = a 0 /2, a 0 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 =− 0 Expand this in spherical coordinates: 1 r 2 r r 2 ∂  r 1 r 2 sin ∂  sin ∂  ∂ 1 r 2 sin 2 2 ∂  2 =− 0 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 =− 0 Evaluate the equation explicitly: 1 r 2 r r 2 r [ q 4  0 e − r r 1 1 2 r ] =− 0 q 4  0 1 r 2 r r 2 r [ e − r r ] r 2 r [ 2 e − r ] =− 0 q 4  0 1 r 2 r r 2 [ 1 r r e − r e − r r 1 r ] 2 2 r 2 e − r =− 0

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q 4  0 1 r 2 r − r e − r r 2 e − r r 1 r 2 2 r 2 e − r =− 0 q 4  0 − r 2 e − r 3 2 e − r q 4   0 1 r 2 r r 2 e − r r 1 r =− 0 = 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 3 8 e − r if r > 0 At r ≈ 0, we have
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Homework2 - Homework 2 Answers 95.657 Fall 2011...

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