InterviewSolution
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InterviewSolution
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`n` small drops of same size are charged to `V` volts each .If they coalesce to from a single large drop, then its potential will be - |
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Answer» Let r be the radius of each small drop and R be the radius of bigger drop. As volume of bigger drop = volume of n small drops `:. (4)/(3) pi R^(3) = n xx (4)/(3) pi r^(3)` As capacity varies directly as the radius, `:.` Capacity of bigger drop becomes `n^(1//3)` times the capacity of each small drop. Also, potential of bigger drop `= ("total charge")/("capacity")` `:. V = (nq)/(4 pi in_(0) R) = (nq)/(4pi in_(0) n^(1//3) r) = n^(2//3) (q)/(4pi in_(0) r)` `= n^(2//3)` times the potential of each small drops. |
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