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\[=9.9\times {{10}^{-15}}kg\]and\[=10m/{{s}^{2}}\])

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, electric force on drop = weight of drop \[\therefore \]  \[qE=mg\] \[\Rightarrow \]               \[q=\frac{mg}{E}\]                          \[=\frac{9.9\times {{10}^{-15}}\times 10}{3\times {{10}^{4}}}\]                 \[=3.3\times {{10}^{-18}}C\]