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                           [AIEEE 2004] \[=9.9\times {{10}^{-15}}kg\text{ }and\text{ }g=10\text{ }m/{{s}^{2}}\])

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 :

[a] In steady state, (i.e., at equilibrium), electric force on drop

= weight of drop

\[\therefore \]      \[qE=mg\]

\[\Rightarrow \]   \[q=\frac{mg}{E}\]

\[=\frac{9.9\times {{10}^{-15}}\times 10}{3\times {{10}^{4}}}\]

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