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 :
[a] In steady state, (i.e., at equilibrium), electric force on drop |
= weight of drop |
\[\therefore \] \[qE=mg\] |
\[\Rightarrow \] \[q=\frac{mg}{E}\] |
\[=\frac{9.9\times {{10}^{-15}}\times 10}{3\times {{10}^{4}}}\] |
\[=3.3\times {{10}^{-18}}C\] |
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