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Unformatted text preview: Physics 505 Fall 2007 Homework Assignment #1 — Solutions Textbook problems: Ch. 1: 1.5, 1.7, 1.11, 1.12 1.5 The timeaveraged potential of a neutral hydrogen atom is given by Φ = q 4 π e αr r 1 + αr 2 where q is the magnitude of the electronic charge, and α 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. We may obtain the charge distribution by computing ρ = ∇ 2 Φ. However, since Φ blows up as r → 0, we must be a bit careful. We first consider r > ρ = ∇ 2 Φ = q 4 π 1 r 2 ∂ ∂r r 2 ∂ ∂r e αr 1 r + α 2 = q 4 π 1 r 2 ∂ ∂r e αr 1 + αr + α 2 r 2 2 = qα 3 8 π e αr For r ≈ 0, on the other hand, we may expand Φ = q 4 π 1 r α 2 + ··· ≈ q 4 π r This is the potential of a point charge q at the origin. Hence the complete charge distribution can be written as ρ = qδ 3 ( r ) qα 3 8 π e αr The first term corresponds to the proton charge, and the second to the negatively charged electron cloud in the 1 s orbital around the proton. We can additionally verify that the hydrogen atom is indeed neutral Q = Z ρd 3 x = q qα 3 8 π Z ∞ e αr 4 πr 2 dr = q q 2 Γ(3) = 0 1.7 Two long, cylindrical conductors of radii a 1 and a 2 are parallel and separated by a distance d , which is large compared with either radius. Show that the capacitance per unit length is given approximately by C = π ln d a 1 where a is the geometrical mean of the two radii. This is essentially a twodimensional electrostatic problem. We choose the geom etry to be a 1 a 2 d x To calculate the capacitance, we use C = Q/ ΔΦ where C is the capacitance per unit length, Q ( Q ) is the charge per unit length on the first (second) conductor, and ΔΦ = Φ 1 Φ 2 is the potential difference between the conductors....
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 Fall '08
 Stephens,P
 Charge, Magnetism, Work, Electric charge, R1 R2, Gauss’ Law, R. Gauss, Ebottom Etop Etop

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