P23_066

# P23_066 - = ⇒ E = E so that q 1 4 πε x 2 = 4 q 1 4 πε...

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66. (a) Since the two charges in question are of the same sign, the point x =2 . 0 mm should be located in between them (so that the Feld vectors point in the opposite direction). Let the coordinate of the second particle be x 0 ( x 0 > 0). Then, the magnitude of the Feld due to the charge q 1 evaluated at x is given by E = q 1 / 4 πε 0 x 2 , while that due to the second charge 4 q 1 is E 0 =4 q 1 / 4 πε 0 ( x 0 x ) 2 . We set the net Feld equal to zero: ~ E net =0
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Unformatted text preview: = ⇒ E = E so that q 1 4 πε x 2 = 4 q 1 4 πε ( x − x ) 2 . Thus, we obtain x = 3 x = 3(2 . 0 mm) = 6 . 0 mm. (b) In this case, with the second charge now positive, the electric Feld vectors produced by both charges are in the negative x direction, when evaluated at x = 2 . 0 mm. Therefore, the net Feld points in the negative x direction....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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