Huang_sp09mt2_p234 - Problem 1 solution 1 1.1 Part a...

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1 Part a 1.1 solution The easiest way to calculate the potential is to take advantage of the effective spherical symmetry of infinitesimal charge elements - Z l E · dl = - Z l dl · Z Q kdq r 2 ˆ r = - Z Q kdq Z r dr · ˆ r r 2 = Z Q kdq r The apparent cylindrical symmetry of the problem encourages cylindrical coordinates, in which the charge element is dq = σρdρdθ , and r = p ρ 2 + a 2 k Z a 0 Z 2 π 0 σρdρdθ p ρ 2 + z 2 = k Z z 2 + a 2 z 2 πσdu u 1 2 = 2 πσku 1 2 | z 2 + a 2 z 2 = 2 πkσ p z 2 + a 2 - z 2 · = σ 2 ² k p z 2 + a 2 - z · The u substitution being u = z 2 + a 2 = du = zdz 1.2 rubric 3 points for recognizing cylindrical symmetry. 3 points for distinguishing r from ρ (!!), writing r correctly, writing dq correctly, and setting up the integral 2 for calculating correctly and reporting the potential as a scalar 2 part b E = -∇ V . Since there is only z dependence E = - ∂V ∂z ˆ z = σ 2 ² ± 1 - z z 2 + a 2 ˆ z 2.1 rubric 4 points for setting up the calculation correctly 3 points for recognizing the potential only has dependence z . (Some thought there was a dependence, but a parametrizes the radius of the disk, and is not a coordinate). 1 point for including vector direction of
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This note was uploaded on 02/09/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at University of California, Berkeley.

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Huang_sp09mt2_p234 - Problem 1 solution 1 1.1 Part a...

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