Water drop in electric field of the Earth

AI Thread Summary
A small water drop of 0.6 mm in diameter is suspended in Earth's electric field of 140 N/C, prompting a discussion on calculating the extra electrons on the drop. Participants emphasize the importance of converting diameter from mm to meters and calculating the drop's volume using the formula v=1/6πd³. The weight of the drop is determined by multiplying its volume by the density of water and gravitational acceleration. The electric force acting on the drop is equated to its weight to find the charge, and then the number of excess electrons is calculated using the charge value. The conversation also touches on related homework questions, with students seeking clarification on concepts and calculations.
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Homework Statement


A small drop of water measuring 0.6 mm in diameter, hangs suspended above the ground due to the Electric field of the Earth (140 N/C). How many extra electrons are on the drop of water?


Homework Equations


i know k=8.99 x 10^9
abs(f)=K((q)/(r^2))

The Attempt at a Solution


I don't even know where to start given the radius being .3mm and the E being 140 N/C
I know it always helps to draw a diagram, and i have that done but it's getting me absolutely nowhere. Please help, it would be greatly appreciated
 
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Your diagram should have two forces. One is the weight of the drop and the other is the electric force. What must be true about these forces?
 
Ok, I just got it and the answer is correct!(= we should be taking the same class and working on the same homwork.
calculate the weight of the water drop first:v=1/6pi*d*d*d
density of water is 1000, so weight of water drop=v*1000*9.8
that's just the F you need for the question, then F=qE (E is a given number)
finally, q=ne(here e is a constatnt, 1.606e-19), n is the answer!
 
221 PU by any chance? if so do u know how to work the first question with the 2 charges and then one being placed on the origin making it zero. It wants us to find the distance from from the origin to the charge on the right. hopefully that makes sense

thanks
 
disque said:
221 PU by any chance? if so do u know how to work the first question with the 2 charges and then one being placed on the origin making it zero. It wants us to find the distance from from the origin to the charge on the right. hopefully that makes sense

thanks

Yes, I am doing 221 now!
For #1, you just need to make an equation shows the force between the left charge and the origin = the force between the right charge and the origin.
Use that F equation and plug in numbers, that's it.
 
I'm in 221 as well and am stuck on the same problem. I've tried to answer it with the advice you have given but it is telling me the answer is wrong. If I followed your steps where am I messing up?

Thanks
 
Oh wow, so many people from 221? Anyhow, amslater: which question are you talking about? disque's initial question, or his second?
 
amslater88 said:
I'm in 221 as well and am stuck on the same problem. I've tried to answer it with the advice you have given but it is telling me the answer is wrong. If I followed your steps where am I messing up?

Thanks


You mean the water drop one?
I already gave all the euqations you would use for the question. But I am wondering if you converted mm unit to m.( the diameter they give you is in mm, but you have to convert it to m and calculate the volume)
 
the water one
 
  • #10
my answer 4.942e19 and that's wrong
 
  • #11
amslater:you need to convert mm to m
 
  • #12
is the answer to the 10th power?
 
  • #13
amslater88 said:
the water one

"A small drop of water measuring 0.6 mm in diameter, hangs suspended above the ground due to the Electric field of the Earth (140 N/C). How many extra electrons are on the drop of water? "

Tha's my original question and finally I got 4.927e10 correctly.
 
  • #14
I got it thanks a bunch!
 
  • #15
cynthiayyf said:
Ok, I just got it and the answer is correct!(= we should be taking the same class and working on the same homework.
calculate the weight of the water drop first:v=1/6pi*d*d*d
density of water is 1000, so weight of water drop=v*1000*9.8
that's just the F you need for the question, then F=qE (E is a given number)
finally, q=ne(here e is a constant, 1.606e-19), n is the answer!

The only thing i couldn't figure out about this is the volume of the rain drop and i still don't quite get it. How does it end up being d3/(6*pi) ?
 
  • #16
So hey another 221 student. Is question #2 very similar to question #5? I gather from the question that since the particle is moving horizontally the electric field is opposite the force of gravity, is that true? Also, when it says the electric force is equal to the weight, does it mean equal to the force of gravity? Any help would be much appreciated.
 
  • #17
For #2 you need to convert the mass to kg. Then, calculate weight by multiplying by 9.8, which is your Force. Finally, plug the Force calculated and E (given) into F=Eq. Solve q and that's it.

Any suggestions for #4? I answered the same question (#47) in the book correctly, but when I apply the same method to the values online I don't get the right answer.
 
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