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Unformatted text preview: Further, since the net field points more strongly leftward for the small positive x (where it is very close to q 2 ) then we conclude that q 2 is the negativevalued charge. Thus, q 1 is a positivevalued charge. We write each charge as a multiple of some positive number ξ (not determined at this point). Since the problem states the absolute value of their ratio, and we have already inferred their signs, we have q 1 = 4 ξ and q 2 = −ξ . Using Eq. 223 for the individual fields, we find E net = E 1 + E 2 = 4 ξ 4 πε o ( L + x ) 2 – ξ 4 πε o x 2 for points along the positive x axis. Setting E net = 0 at x = 20 cm (see graph) immediately leads to L = 20 cm....
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 Spring '10
 BB
 Physics, Charge

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