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)
N C –1
N C –1
Text Solution
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(b, c)
(i)
(initially) ∴ mg = f air
(ii) 
∴ QE = mg + f air = 2mg
∴ charge is –ve, so electric field 'E' is directed downwards.
& QE = 2 mg
∴ E = 
NC –1
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