Water drop in electric field of the Earth

In summary, a student was struggling with a physics homework problem involving a small drop of water suspended above the ground due to the Earth's electric field. Other students in the same class were also working on the same homework. The solution involved finding the weight of the water drop first, then using that value to calculate the electric force and finally solving for the number of extra electrons on the drop. The student also mentioned another question involving the electric field and the force of gravity, and asked for help with a different question.
  • #1
disque
29
0

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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  • #2
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?
 
  • #3
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!
 
  • #4
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
 
  • #5
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.
 
  • #6
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
 
  • #7
Oh wow, so many people from 221? Anyhow, amslater: which question are you talking about? disque's initial question, or his second?
 
  • #8
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)
 
  • #9
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.
 

Related to Water drop in electric field of the Earth

1. What is a water drop in the electric field of the Earth?

A water drop in the electric field of the Earth refers to the phenomenon of a charged water droplet being influenced by the Earth's electric field. This can occur when the water droplet comes into contact with a charged object or when it falls through the Earth's atmospheric electric field.

2. How does the Earth's electric field affect a water drop?

The Earth's electric field exerts a force on the charged particles within the water droplet, causing it to move in a certain direction. This can also cause the droplet to change shape or even split into smaller droplets.

3. What causes a water drop to become charged in the first place?

A water droplet can become charged through various processes such as friction, contact with a charged object, or exposure to ionizing radiation. The charge is typically carried by ions or electrons within the droplet.

4. Can a water drop in the electric field of the Earth affect its surroundings?

Yes, a charged water droplet can have an impact on its surroundings. For example, if the droplet is charged positively, it can attract nearby objects with a negative charge, causing them to move towards the droplet.

5. Is the electric field of the Earth constant or does it vary?

The electric field of the Earth is not constant and can vary based on factors such as location, time of day, and weather conditions. It can also be affected by human activities such as power plants and lightning strikes.

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