A charged water droplet hangs from ceiling of a room which has a uniform vertically upwards electrical field of strength 4.00 × 105 NC−1 . This field levitates and prevents a droplet of mass 1.0 ×10−4kg from falling. Can you find the net charge on the droplet?
2.45 × 10–9 C
The forces acting on the droplet are
(i) Electrical force q →E upwards
(ii) Force of gravity m→g downwards
So, if in order to prevent the drop from falling, the two external forces should balance each other out.
q→E=m→gq(4.00×108Kg−ms−2C−1)=(1.00×10−4kg)×9.8m−s−2q=2.45×10−9C