A) \[15\text{ }N/C,\]vertically upwards
B) \[15\text{ }N/C,\]vertically dowards
C) \[65.3\text{ }N/C,\]vertically upwards
D) \[65.3\text{ }N/C,\]vertically downwards
Correct Answer: C
Solution :
Given, \[m=10.0,\,mg=10\times {{10}^{-6}}kg={{10}^{-5}}kg\] \[q=1.5\times {{10}^{-6}}C.\] \[E=?\] As the drop stays suspended in the room, force (F) due to electric field must be balancing the weight of the drop i.e., \[F=qE=mg\] \[\Rightarrow \] \[E=\frac{mg}{q}\] \[=\frac{{{10}^{-5}}\times 9.8}{1.5\times {{10}^{-6}}}\] \[=65.3\,N/C\] The direction of electric field must be vertically upwards, so that upward force due to the field balances the weight.
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