question_answer

A charged oil drop is suspended in 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 \[=9.9\times {{10}^{-15}}kg\text{ }and\text{ }g=10\text{ }m/{{s}^{2}}\])

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

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

C)        \[1.6\times {{10}^{-18}}C\]    

D)        \[4.8\times {{10}^{-18}}C\]

Correct Answer: A

Solution :

In steady state, (i.e., at equilibrium), electric force on drop = weight of drop \[\therefore \]  \[qE=mg\] \[\Rightarrow \]               \[q=\frac{mg}{E}\]                 \[=\frac{9.9\times {{10}^{-15}}\times 10}{3\times {{10}^{4}}}\]                 \[=3.3\times {{10}^{-18}}C\]