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 :


Given:
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.
Fe=FG

qE=mg


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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