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 \[g=10m/{{s}^{2}}\])

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 :

Key Idea In steady state electric force on drop balances the weight of the drop. In steady state, electric force on drop                 = weight of drop                 \[\therefore \]  \[qE=mg\]                 \[\Rightarrow \]               \[q=\frac{mg}{E}\]                                 \[=\frac{9.9\times {{10}^{-15}}\times 10}{3\times {{10}^{4}}}\]                                 \[=3.3\times {{10}^{-18}}C\]