Question

A water droplet of mass $10\text{mg}$ and having a charge of $1.50\times {10}^{-6}\text{C}$ stays suspended in a room. What is the magnitude of electric field and its direction in the room ? (Take $g=10{\text{ms}}^{-2}$)

• A. $30\text{N/C}$, downwards
• B. $66.7\text{N/C}$, upwards
• C. $132\text{N/C}$, upwards
• D. $66.7\text{N/C}$, downwards

Answer 1

The correct option is B $66.7\text{N/C}$, upwards

Given,
mass of water droplets, $m=10\text{mg}=10\times {10}^{-6}\text{kg}$

charge on water droplets,
$q=1.5\times {10}^{-6}\text{C}$

$\to$As the water particle remains suspended in the room, so net force on it must be zero.

$\to$As gravitational force is acting downwards $(-\hat{j})$. To compensate for that force, electrostatic force must point upwards, i.e., in $\left(\hat{j}\right)$ direction, as shown in the following figure.

$\to$Since the charge on water is positive, the electric field also must point in the upwards direction.


From force equilibrium,

$mg=qE$

$\Rightarrow E=\frac{mg}{q}$

Substituting the values,

$\Rightarrow E=\left[\frac{(10\times {10}^{-6})\times \left(10\right)}{(1.5\times {10}^{-6})}\right]$

$\Rightarrow E=66.67\text{N/C}$

$\therefore E\cong 66.7\text{N/C}$

Hence, option (b) is correct asnwer.