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. $15N/C,$ vertically upwards
• B. $15N/C,$ vertically dowards
• C. $65.3N/C,$ vertically upwards
• D. $65.3N/C,$ 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.,
$F=qE=mg$
$\Rightarrow$ $E=\frac{mg}{q}$
$=\frac{{10}^{-5}\times 9.8}{1.5\times {10}^{-6}}$
$=65.3N/C$
The direction of electric field must be vertically upwards, so that upward force due to the field balances the weight.