A water particle of mass $10.0$ mg and having a charge of stays suspended in a room. What is the magnitude of electric field in the room ? what is its direction?

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Solution

Data:
Mass of water drop $(m) = 10 m g = 10^{- 5} k g$

Charge, $q = 1.5 \times 10^{- 6} C$

As the water drop is balanced in the air, we can say that the gravitational force of the water drop is balanced by the electric force on the water drop.

Let the electric field $= E$
Electric force = weight

$E q = m g$

$E = \frac{m g}{q}$
$E = (10^{- 5}) \times \frac{9.8}{1.5 \times 10^{- 6}}$

$E = 65.33 N / C$

Directed vertically upwards.

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A water particle of mass 10.0 mg and having a charge of stays suspended in a room. What is the magnitude of electric field in the room ? what is its direction?

A water particle of mass $10.0$ mg and having a charge of stays suspended in a room. What is the magnitude of electric field in the room ? what is its direction?