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 \[\therefore \] \[qE=mg\] \[\Rightarrow \] \[q=\frac{mg}{E}\] \[=\frac{9.9\times {{10}^{-15}}\times 10}{3\times {{10}^{4}}}\] \[=3.3\times {{10}^{-18}}C\]You need to login to perform this action.
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