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 :
Because the drop neither falls nor rises, so it remains stationary, then For equilibrium position, the electric farce\[qE\]will be equal to the weight of the drop. So, \[qE=mg\] \[q=\frac{mg}{E}\] \[=\frac{(9.9\times {{10}^{-15}})\times 10}{3\times {{10}^{4}}}\] \[=3.3\times {{10}^{-18}}C\]
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