question_answer

A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field\[100\text{ }V{{m}^{-1}}\]. If the mass of the drop is\[1.6\times {{10}^{-3}}g,\] the number of electrons carried by the drop is\[(g=10\text{ }in{{s}^{-2}})\]

A)  \[{{10}^{18}}\]                 

B)         \[{{10}^{15}}\]

C)  \[{{10}^{6}}\]                   

D)         \[{{10}^{9}}\]

E)  \[{{10}^{12}}\]