Question

urgently required.
no unnecessary answers...otherwise I'll report the answer. ​

Answer 1

$Here: mass of water drop =m,=m, so, weight of water drop =mg=mg so, mg=mg= electrostatic force applied by the field mg=qE=eE⇒E=mge</p><p></p><p></p><p>$

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Answer 2

the electric field required to keep a water drop of mass m just to remain suspended, when charged with one electron of charge e (take g as acceleration due to gravity) is: answer is mg/e