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.
 Fall '08
 SPRUNGER
 Physics, Charge

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