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Question
A charged oil drop weighing 1.6 x 10-15 N is found to remain suspended in a uniform electric field of intensity 2 x 103 Nc-1. Find the charge on the drop.
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Solution
Weight of the charged oil drop = `1.6 xx 10^-15`N.
Electric field = `2 xx 10^3 "NC"^-1`
∴ F = mg = qE
∴ q = `("mg")/"E" = (1.6 xx 10^-15)/(2 xx 10^3)`C
q = `0.8 xx 10^-18`C
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