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Question

# 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?

A

4.00 × 10–9 C

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B

2.45 × 10–9 C

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C

5.40 × 10–9 C

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D

1.75 × 10–9 C

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Solution

## The correct option is B 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

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