ch22-p067 - 67. (a) Since the two charges in question are...

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67. (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 field vectors point in the opposite direction). Let the coordinate of the second particle be x' ( x' > 0). Then, the magnitude of the field 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' = 4 q 1 /4 πε 0 ( x' x ) 2 . We set the net field equal to zero: G EE E net =⇒ =
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This note was uploaded on 07/10/2010 for the course PHYS 241 taught by Professor Fr during the Fall '10 term at The Petroleum Institute.

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