Q. 374.1( 7 Votes )

A water particle of mass 10.0 mg and having 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 :

Mass of the water particle : m= 10.0 mg = 10× 10-6 kg
Charge of the particle: q = 1.5× 10-6 C

The suspended particle will be under the influence of gravity.
Hence , gravity will act in downward direction as shown in the figure.
Formula used:
We know that,
Fe = qE
Where F is the electrostatic force, q is the charge and E is the electric field produced by the charge.
But, Force due to gravity : is FG= mg
Here, m is the mass of the particle and g is the acceleration due to gravity.
Now, for the particle to stay suspended in the room, the downward gravitational force must be equal and opposite to the electric force.


Thus, the magnitude of electric field due to the charged water molecule suspended in the room is 65.33 NC-1 and it is in upwards direction opposite to the gravitational force.

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