Lab201_Feb 10

Lab201_Feb 10 - aq1 1 21 1 q E ( x) 1 E ( x) 2 Part . . ....

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Unformatted text preview: aq1 1 21 1 q E ( x) 1 E ( x) 2 Part . . . I: x 2, 1.2 191 ( x a) q. 1 3 ( (x a )) xa q. 2 3 (x a) E (x Ex Ex t ) 1 ) 2() o( a l E t o a l ( x) = 0 . 9 3 1 1 . 3 5 2 . 1 7 8 4 . 1 6 1 . 3 1 0 . 2 7 9 . 7 2 3 1 0 . 7 6 5 3 . 5 6 1 . 4 9 0 . 4 8 0 . 4 8 1 . 4 9 3 . 5 6 1 0 . 7 6 5 9 . 7 2 3 1 0 . 2 7 Questions: 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. E (0=0 o t) a l 2 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 3) 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. 3 (Plot for problem 5) Part II: 1. (Plot for Question 3) 4 2. 3. X component of electric field (Plot for Question 5) 5 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. 6 Plot for Question 7) 8. Plot for Question 8) ...
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