Questions
A charged oil drop is suspended in a uniform field of 3x10 4 v/m so that it neither falls nor rises .The charge on the drop will be (take mass of the charge =9.9 x10-15 kg and g = 10m /s2
detailed solution
Correct option is A
In equilibrium position m g = g Eor q=mgE=9⋅9×10−15×103×104 =3⋅3×10−18CTalk to our academic expert!
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