Lab201_UPDATE

# Lab201_UPDATE - 5. a) No, it's 2 at the origin. b) Because...

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Part I 2 .11 .9 , 2 .1 . . . E to ta l x () E 1x ()E 2( -0 .931 -1 .353 -2 .178 -4 .16 -1 .3 -10 .27 9 .723 10 .765 3 .56 1 .49 0 .408 -0 .408 -1 .49 -3 .56 -10 .765 -9 .723 10 .27 (Plot for problem 3) E to ta l x ( ) x 2 1 0 1 2 10 50 0 50 Questions:

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1. If you don't, you'd end up dividing by 0. 2. Because x-a is r, and so the one in the numerator cancels out the one in the denominator. The reason for this is to get the right sign. 3. a) It does equal 0 b) Because it is exactly between two equal charges. c) Yes d) It would be in stable equilibrium because the net force along the x axis at the origin is 0. . (Plot for problem 5)

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Unformatted text preview: 5. a) No, it's 2 at the origin. b) Because not only is the positive charge repelling but the negative is attracting in the same direction. Thus the forces combine rather than cancel. c) No d) It wouldn't be under stable equilibrium because it would be under a net force. Part 2: 1. 2. 3. X component of electric field 4. 5. 6. Because at the x-axis, the field line is perfectly vertical. Thus there is no horizontal component to the field line. 7. 8....
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## Lab201_UPDATE - 5. a) No, it's 2 at the origin. b) Because...

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