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 $
BETA
