Electric Charge of Water droplet falling

AI Thread Summary
A charged water droplet with a radius of 0.02 mm remains stationary in an electric field of 100 N/C. The charge on the droplet is calculated using the formula q = ma/E, resulting in a charge of approximately 8.22 x 10^-3 Coulombs. To determine how many electrons the droplet has gained or lost compared to its neutral state, the charge is divided by the elementary charge of an electron, yielding about 5.13 x 10^16 electrons lost. The term "neutral state" refers to the droplet being uncharged. The calculations confirm that the droplet's charge is sufficient to account for the loss of this many electrons.
hemetite
Messages
50
Reaction score
0

Homework Statement



A charged water droplet of radius 0.02 mm remains stationary in air. If the electric field of the Earth is 100 N/C downward,what is the charge on the water droplet? By comparison with its neutral state, how many electrons must the water droplet have gained or lost? [ Density of water is 103 kg m-3 ]

Homework Equations



E=F/q , F=ma , p=m/v

The Attempt at a Solution



I got the first answer ...which i have used

q = ma/E which mass of water droplets is derive from the density formula of a sphere and the density formula...

I don't quite understand by the second part?
"By comparison with its neutral state, how many electrons must the water droplet have gained or lost?"

what it is meant by neutral state?
 
Physics news on Phys.org
hemetite said:
I don't quite understand by the second part?
"By comparison with its neutral state, how many electrons must the water droplet have gained or lost?"

what it is meant by neutral state?
Uncharged.
Work out how many Coulombs of charge are on the drop.
Then from knowing the charge on the electron, how many electrons this represents.
 
thanks..here is my attempt...

from answer of first part q=8.22 x 10^-3..

there at drop the number of electrons lost = (8.22 x 10 ^ -3) / (1.6021765 x 10 ^ -19)

= 5.13 x 10 ^ 16

i have strong feeling that this is the answer...since when it is falling...the charge of the drop is strong enough to dispel the calculated amount of electrons...
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top