ElectrostaticsHard

Question

An oil drop has a charge - 9.6 × 10-19 C and mass 1.6 × 10-15 gm. When allowed to fall, due to air resistance force it attains a constant velocity. Then if a uniform electric field is to be applied vertically to make the oil drop ascend up with the same constant speed, which of the following are correct. (g = 10 ms-2) (Assume that the magnitude of resistance force is same in both the cases)

Options

A.The electric field is directed upward     
B.The electric field is directed downward
C.The intensity of electric field is 1/3 × 102 N C-1
D.The intensity of electric field is 1/6 × 105 N C-1

Solution

(initially)    ∴    mg = fair 
(finelly )    ∴    QE = mg + fair = 2mg
∴    charge is −ve, so electric field ′E′  is directed downwards.
& QE = 2 mg
∴    E = NC-1

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