question_answer

A charged oil drop is suspended in uniform field of \[3\times {{10}^{4}}V{{m}^{-1}},\]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 = 10\[m{{s}^{-2}}\])

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