Two charges attract each other with a force of 1.9 n. what will be the force if the distance between them is reduced to one-sixth of its original value?

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asked Jun 8, 2019 111k views

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Gradosevic · Answer 1 · 2019-06-14T19:21:54+0000

From Couloumbs's law we have: F = q1q2/r^2 So 1.9 = q1q2/r^2 ----------(i) When the distance is one-sixth we have. F = q1q2/(r/6)^2 This can be written as F = 36 x q1q2/r^2 ----------(2) F = 36 x 1.9. Recall 1.9 = q1q2/r^2 from (1) So F = 68.4N

Two charges attract each other with a force of 1.9 n. what will be the force if the distance between them is reduced to one-sixth of its original value?

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