A water particle of mass 10.0 mg and having a charge of `1.50 xx 10^(-6)C` stays suspended in a room. What is the magnitude of electric field in the room ? What is its direction?

Explanation

To solve the problem, we need to find the magnitude of the electric field in the room that allows a water particle to remain suspended. The forces acting on the water particle are the gravitational force acting downward and the electrostatic force acting upward.

### Step-by-Step Solution:

1. **Convert the mass of the water particle to kilograms:**
The mass of the water particle is given as 10.0 mg. To convert this to kilograms:
\[
\text{Mass (m)} = 10.0 \, \text{mg} = 10.0 \times 10^{-3} \, \text{g} = 10.0 \times 10^{-6} \, \text{kg} = 1.0 \times 10^{-5} \, \text{kg}
...