Problem%20Set%202%20Solutions

# Problem%20Set%202%20Solutions - 7. To exploit the symmetry...

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7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the shape of a cube, of edge length d , with a proton of charge 19 1.6 10 C q =+ × situated at the inside center of the cube. The cube has six faces, and we expect an equal amount of flux through each face. The total amount of flux is Φ net = q / ε 0 , and we conclude that the flux through the square is one-sixth of that. Thus, 19 92 12 2 2 0 1.6 10 C 3.01 10 N m C. 66 ( 8 . 8 5 1 0 C N m ) q × Φ= = = × ×⋅

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12. Eq. 23-6 (Gauss’ law) gives ε ο Φ = q enc . (a) Thus, the value 52 2.0 10 N m /C Φ= × for small r leads to 12 2 2 5 2 6 6 central 0 (8.85 10 C /N m )(2.0 10 N m /C) 1.77 10 C 1.8 10 C q ε −− = × × = × ≈ × . (b) The next value that Φ takes is 4.0 10 N m /C Φ=− × , which implies 6 enc 3.54 10 C. q =− × But we have already accounted for some of that charge in part (a), so the result for part (b) is q A = q enc q central = – 5.3 × 10 6 C. (c) Finally, the large r value for Φ is 6.0 10 N m /C × , which implies 6 total enc 5.31 10 C. q Considering what we have already found, then the result is total enc central 8.9 . A qq q C µ = +
14. None of the constant terms will result in a nonzero contribution to the flux (see Eq. 23-4 and Eq. 23-7), so we focus on the x dependent term only. In Si units, we have E non-constant = 3 x i ^ . The face of the cube located at x = 0 (in the yz plane) has area A = 4 m 2 (and it “faces” the + i ^ direction) and has a “contribution” to the flux equal to E non-constant A = (3)(0)(4) = 0.

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## This note was uploaded on 05/04/2010 for the course PHYS 11b taught by Professor Mobius during the Spring '09 term at Harvard.

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Problem%20Set%202%20Solutions - 7. To exploit the symmetry...

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