ch22_p65 - 65(a Since the two charges in question are of...

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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 = ¡ = 0 so that q x q xx 1 0 2 1 0 2 4 4 4 π π ε = ′ − b g . Thus, we obtain x' = 3 x = 3(2.0 mm) = 6.0 mm.
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