|Jan15-09, 12:19 PM||#1|
Electric Charge of Water droplet falling
1. The problem statement, all variables and given/known data
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 ]
2. Relevant equations
E=F/q , F=ma , p=m/v
3. 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 dont 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?
|Jan15-09, 12:23 PM||#2|
Work out how many Coulombs of charge are on the drop.
Then from knowing the charge on the electron, how many electrons this represents.
|Jan15-09, 12:34 PM||#3|
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...
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