question_answer

A charged oil drop is suspended in unifor m field of \[3\times {{10}^{4}}\,V\text{/}m\]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\[g=10\,m\text{/}{{s}^{2}}\])

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       \[\therefore \]  \[qE=mg\] \[\Rightarrow \]               \[q=\frac{mg}{E}\]    \[=\frac{9.9\times {{10}^{-15}}\times 10}{3\times {{10}^{4}}}\]    \[=3.3\times {{10}^{-18}}C\]