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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 ?
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Solution
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.
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