Question

If a water particle of mass 10 mg and having a charge of $ 1.5\times {{10}^{-6}}C $ stays suspended in a room, then the magnitude and direction of electric field in the room is

• 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

Answer 1

Answer: (C)

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.