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,

\[m = 10  \text{mg } = 10 \times  {10}^{- 5} \] kg

Charge on the particle,

\[q = 1 . 5 \times  {10}^{- 6}   C\]

Let the magnitude of the electric field be E.
The particle stays suspended. Therefore,
Downward gravitational force = Upward electric force
That is, mg  = qE 

\[\Rightarrow   E = \frac{mg}{q} = \frac{10 \times {10}^{- 5} \times 10}{1 . 5 \times {10}^{- 6}}\] 

\[                 =   \frac{1000}{15} = 66 . 7  \] N/C

The direction of the electric field will be upward to balance the downward gravitational force.