Two same charges separated by the distance 'd' and force of attraction between them is 'f' . If one charge transfers its 50% to the other charge then find new force if 'd' remains constant .
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Answer:
New force is 3/4 times the f.
Explanation:
Let the 2 charges be q .
when 50% is lost from 1st charge then,
q1 = q/2 and q2 = q/2+q = 3q/4
F = kq^2/d^2 .....................(1)
F' = k 3q^2/d^2 = 3F/4 ....................(2)
Therefore, new force is 3/4 times of actual force
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