ElectrostaticsHard
Question
A charged oil drop is suspended in a uniform field of 3 × 104 V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10-15 kg and g = 10 m/s2)
Options
A.3.3 × 10-18C
B.3.2 × 10-18C
C.1.6 × 10-18C
D.4.8 × 10-18C
Solution
Since ball is hanging in equilibrium, force by gravity is balanced by electric force.
qE = mg
⇒
⇒
∴ q = 3.3 × 10-18C
qE = mg
⇒
⇒
∴ q = 3.3 × 10-18C
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