Electrostatics Question 671

Question: A charged oil drop is suspended in a uniform field of $ 3\times 10^{4}V/m $ so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge as $ 9.9\times {{10}^{-15}}kgandgas10m/s^{2} $

Options:

A) $ 3.3\times {{10}^{-18}}C $

B) $ 3.2\times {{10}^{-18}}C $

C) $ 1.6\times {{10}^{-18}}C $

D) $ 4.8\times {{10}^{-18}}C $

Show Answer

Answer:

Correct Answer: A

Solution:

[a] In equilibrium, $ qE=mg $

$ \Rightarrow q=3.3\times {{10}^{-18}}C $



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