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Physics Question on Electric Field

The electric field required to keep a water drop of mass in just to remain suspended, when charged with one electron:

A

em/g\text{em/g}

B

mg/e\text{mg/e}

C

emg\text{emg}

D

mg\text{mg}

Answer

mg/e\text{mg/e}

Explanation

Solution

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=mgemg=qE=eE\,\,\,\Rightarrow \,\,\,E=\frac{mg}{e}