Question

If a water particle of mass $10mg$ 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 downwards
• C. $65.3N/C$, vertically upwards
• D. $65.3N/C$, vertically downwards

Answer 1

The correct option is C $65.3N/C$, vertically upwards

$\text{Step 1: Balance forces on the drop[Ref fig]}$

The direction of the electric field must be vertically upwards so that the upward force due to the field balances the weight as shown in figure.

The drop will remain suspended in the room if,

$F_e=mg$, where $F_e$ is the force due to electric field on drop.

$\Rightarrow qE=mg$

$\Rightarrow E=\frac{mg}{q}$ $....\left(1\right)$

$\text{Step 2: Substituting all the values}$

Given: $m=10mg={10}^{-5}kg$, $q=1.5\times {10}^{-6}C$, $g=9.8m/s^2$

Equation $\left(1\right)\Rightarrow$ $E=\frac{{10}^{-5}\times 9.8}{1.5\times {10}^{-6}}=65.3N/C$

$\therefore E=65.3N/C$, vertically upwards

Hence, Option C is correct.