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709_Physics ProblemsTechnical Physics

# 709_Physics ProblemsTechnical Physics - Chapter 24 P24.72...

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Chapter 24 49 P24.72 Consider the gaussian surface described in the solution to problem 71. (a) For x d > 2 , dq dV Adx CAx dx = = = ρ ρ 2 E A = = = F H G I K J F H G I K J z z z d dq EA CA x dx CA d d 1 1 3 8 0 0 2 0 2 0 3 E Cd = 3 0 24 or E i E i = > = − < − Cd x d Cd x d 3 0 3 0 24 2 24 2 ± ; ± for for (b) For < < d x d 2 2 E A = = = z z z d dq CA x dx CAx x 1 3 0 0 2 0 3 0 E i E i = > = − < Cx x Cx x 3 0 3 0 3 0 3 0 ± ; ± for for P24.73 (a) A point mass m creates a gravitational acceleration g r = − Gm r 2 ± at a distance r . The flux of this field through a sphere is g A = − = − z d Gm r r Gm 2 2 4 4 π π e j . Since the r has divided out, we can visualize the field as unbroken field lines. The same flux would go through any other closed surface around the mass. If there are several or no masses inside a closed surface, each creates field to make its own contribution to the net flux according to g A = − z d Gm 4 π in . (b) Take a spherical gaussian surface of radius r . The field is inward so g A = °= − z d g r g r
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