# Water drop in electric field of the Earth

1. Aug 28, 2009

### disque

1. The problem statement, all variables and given/known data
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?

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

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 28, 2009

### kuruman

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. Aug 29, 2009

### cynthiayyf

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. Aug 29, 2009

### disque

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. Aug 30, 2009

### cynthiayyf

Yes, Im 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. Aug 30, 2009

### amslater88

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. Aug 30, 2009

### ideasrule

Oh wow, so many people from 221? Anyhow, amslater: which question are you talking about? disque's initial question, or his second?

8. Aug 30, 2009

### cynthiayyf

You mean the water drop one?
I already gave all the euqations you would use for the question. But I am wondering if you coverted 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. Aug 30, 2009

### amslater88

the water one

10. Aug 30, 2009

### amslater88

my answer 4.942e19 and that's wrong

11. Aug 30, 2009

### fatsack

amslater:you need to convert mm to m

12. Aug 30, 2009

### amslater88

is the answer to the 10th power?

13. Aug 30, 2009

### cynthiayyf

"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. Aug 30, 2009

### amslater88

I got it thanks a bunch!!

15. Aug 31, 2009

### brandon.baker

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. Aug 31, 2009

### bowieknife50

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. Jan 13, 2010

### mlk7929

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 thats 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.