A charged oil drop weighing 1.6×10−15N is found to remain suspended in a uniform electric field of intensity 2×103NC−1. Find the charge on the drop.
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Let, Charge on drop be q Given:
Weight, w=1.6×10−15 Intensity of electric field, E=2×103NC−1 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 =1.6×10−152×103 =0.8×10−18C =8×10−19C