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Question

A charged oil drop weighing 1.6×1015N is found to remain suspended in a uniform electric field of intensity 2×103NC1. Find the charge on the drop.

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Solution

Let, Charge on drop be q
Given:
Weight, w=1.6×1015
Intensity of electric field, E=2×103NC1
When charged oil drop is suspended in uniform electric field then its weight will act in vertically downward direction and force due to uniform electric field will act in upward direction.
qϵ=mg
q=mgϵ
=1.6×10152×103
=0.8×1018C
=8×1019C

625458_595807_ans_8f7d984e04784999b989bcfafe419a83.png

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