45 physics212lecture3slide20 physics212lecture3slide

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Unformatted text preview: ΦA stays same dΦB stays same BB “The flux through dA would increase because the density of the lines would increase as you move the charge closer to it. Likewise as the charge is moved away from dB there will be less lines passing through dB. “ “The same number of field lines will pass through the shell surface regardless of where the charge is placed.” 50 40 30 20 10 0 41 Physics 212 Lecture 3, Slide 19 Physics 212 Lecture 3, Slide Think of it this way: 1 2 The total flux is the same in both cases (just the total number of lines) The flux through the right (left) hemisphere is smaller (bigger) for case 2. 45 Physics 212 Lecture 3, Slide 20 Physics 212 Lecture 3, Slide Return of the Point Charges BB Case 1 Case 2 -q +q +2q -q Compare Φ1, the flux through the sphere in Case 1, to Φ2, the flux through the sphere in Case 2 A) Φ1 < Φ2 B) Φ1 = Φ2 C) Φ1 > Φ2 The flux through any CLOSED SURFACE is equal to the CHARGE ENCLOSED ! The distribution of the charge matters not ! VERIFY: Count the lines leaving the sphere (20) Physics 212 Lecture 3, Slide 21 Physics 212 Lecture 3, Slide Things to notice about Gauss Law r r Qenclosed Φ S = Ñ E ×dA = ∫ ε0 closed surface If Qenclosed is the same, the flux has to If be the same, which means that the be which integral must yield the same result for any s...
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This document was uploaded on 02/13/2014.

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