• # 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)  $4.8\times {{10}^{-18}}C$

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