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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 timeaveraged 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 α1 = 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 straightforward 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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This note was uploaded on 02/13/2012 for the course PHYSICS 95.657 taught by Professor Staff during the Fall '11 term at UMass Lowell.
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