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A charge oil drop of charge q is falling...

A charge oil drop of charge `q` is falling under gravity with termined velocity v in the absence of electric filed . A electric field can keep the oil drop stationary . If the drop acquires an additional charge , it moves up with velocity `3v` in that field . find the new charge on the drop.

Text Solution

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`mg =6pi eta r upsilon, Eq=mg`
`Eq^(1) = mg + 6pietar(3upsilon), Eq^(1) = Eq +wq+3Eq`
`Eq^(1) =4ERightarrow q^(1) =4q`
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