ch21-p030 - and sign of q 2 : q 1 q 2 4 o r 2 = F q 2 = 1.5...

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30. Since the graph crosses zero, q 1 must be positive-valued: q 1 = +8.00 e . We note that it crosses zero at r = 0.40 m. Now the asymptotic value of the force yields the magnitude
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Unformatted text preview: and sign of q 2 : q 1 q 2 4 o r 2 = F q 2 = 1.5 x 10-25 k q 1 r 2 = 2.086 10 18 C = 13 e ....
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