Question

A water particle of mass 10.0 mg and with a charge of 1.50 × 10⁻⁶ C stays suspended in a room. What is the magnitude of electric field in the room? What is its direction?

Answer 1

Mass of the particle, $m=10\mathrm{mg}=10\times {10}^{-5}\mathrm{kg}$
Charge on the particle, $q=1.5\times {10}^{-6}\mathrm{C}$
Let the magnitude of the electric field be E.
The particle stays suspended. Therefore,
Downward gravitational force = Upward electric force
That is, mg = qE
$$\begin{gathered} \Rightarrow E=\frac{mg}{q}=\frac{10\times {10}^{-5}\times 10}{1.5\times {10}^{-6}} \\ =\frac{1000}{15}=66.7\mathrm{N}/\mathrm{C} \end{gathered}$$
The direction of the electric field will be upward to balance the downward gravitational force.