A) \[15N/C\], vertically upwards
B) \[15N/C\], vertically downwards
C) \[65.3N/C\], vertically upwards
D) \[65.3N/C\], 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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