# Homework Help: Electrostatic field

1. Feb 15, 2010

### cerium

b]1. The problem statement, all variables and given/known data[/b]
A see-saw with a central pivot is made of insulating material. the left hand side supports a conducting sphere of charge Q= 5.0x 10-6 which experiances an electrostatic force from an idetical sphere 10cm below with +Q
the right end has anouther conducting sphere of charge +3Q which is 10cm above a sphere with charge +4Q. A block is placed on the right hand side of the see-saw to balance the system and keep the rod horizontal.

2. Relevant equations

3. The attempt at a solution#
My question is calculate the mass of the block required to balance the see-saw. I have no idea were to start

2. Feb 15, 2010

### tiny-tim

Hi cerium! Welcome to PF!

Show us how far you've got, and where you're stuck, and then we'll know how to help!

3. Feb 15, 2010

### cerium

Thank you for that hopefully I can get going now

4. Feb 15, 2010

### cerium

I have worked out the electrostatic force for both sides of the see-saw but Im unsure of how to proceed to get the mass of the block

5. Feb 15, 2010

### tiny-tim

ok, now you have the three forces (two electrostatic and one gravitational), find the moment (the torque ) of each force about the pivot, and decide what the mass needs to be for them to balance.

6. Feb 15, 2010

### cerium

Im having a mental block how can I work out gravitatinal force if I havent been given any masses,

7. Feb 15, 2010

### tiny-tim

uhh? the mass is what you have to find.

call it m.

8. Feb 16, 2010

### coz

Hi tiny-tim,
Can I just check, to find the electrostatic force on either side of the see saw would you use the equation:
F_el = 1/4pi$$\epsilon$$_0 x q_1xq_2/r_2

9. Feb 16, 2010

### tiny-tim

Hi coz!

(are you the same person as cerium?)
A little difficult to read (try using the X2 and X2 tags just above the Reply box ), but yes, that looks like Coulomb's law !

10. Feb 16, 2010

### coz

Hi, no I am not the same person as cerium, just have a similar question to work out. I am new to this, so thanks for the tips!

For the LHS I have
[F][/el] = [8.988x10][/9] N m2 C-2 x ([5.0x10][/-6] x [5.0x10][/-6])/(0.1m)2
= 22.47N

For the RHS I have
[F][/el] = [8.988x10][/9] N m2 C-2 x ([1.5x10][/-5] x [2.0x10][/-5])/(0.1m)2
= 269.64 N

I must have done something wrong though, as this shows that electrostatic force is greater on the RHS than the LHS, so why would the block be placed on the RHS to balance the see-saw?

(i hope all my calculations come out correctly)

11. Feb 16, 2010

### coz

Obviously I didn't use the tags correctly! How do they work then? Thanks again!

12. Feb 16, 2010

### cerium

Hi Coz I worked out the same as you and Im still not sure how to proceed Have you had any joy yet

13. Feb 16, 2010

### coz

Not yet, but I won't let it beat me! :)

14. Feb 16, 2010

### tiny-tim

Welcome to PF!

hmm … something to do with the direction of the electrostatic force?
Press the "QUOTE" button under this post, and you'll see lots of exciting tags!

15. Feb 16, 2010

### coz

Re: Welcome to PF!

Ok, so the charges are like and so they repel, and the RHS repels with greater force than the LHS, because the charge is higher there.

16. Feb 16, 2010

### coz

The only equations I know for gravitational force are;

Fgrav = -Gm1m2/r2
Fgrav (on m at r) = mg(r)

But both of these need the mass to work out the gravitational force. I could use F=ma to work out the mass, but this seems too easy! Also you mentioned torque earlier, so I don't think this is right!

Could I make the electrostatic force on th RHS equal the gravitational force and then rearrange Fgrav (on m at r) = mg(r)? Again this doesn't use torque, so I am assuming it is not right!

17. Feb 16, 2010

### tiny-tim

Hi coz!

(erm … you're using the X2 tag on the wrong bits! )
As I said to cerium … the mass is what you have to find.

call it m.

And your formula is for a general radius r from the Earth's centre.

When you're here (on the Earth's surface), just use F = mg.
?? you're just rambling … get some sleep! :zzz:

18. Feb 16, 2010

### coz

Sorry for rambling!

m=F/g = 27.486kg

This seems too simple.

19. Feb 16, 2010

### tiny-tim

Well, in this case all the distances happen to be the same, so it is that simple!

20. Feb 16, 2010

### coz

Thank you so much for all your help :)

21. Feb 17, 2010

### sand

Hi. I'm new but I wanted to clarify something in Coz's working. It looks like they used the electrostatic force of the LHS for the force in F=mg. But is that right? or should it be the difference between the LHS and the RHS? But that leads to me to wonder if you can just subtract forces like that?
Thanks

22. Feb 17, 2010

### tiny-tim

Welcome to PF!

Hi sand! Welcome to PF!

(i'm not sure what you're asking, but …)

technically, we weren't subtracting forces, we were subtracting moments of forces …

but the distances are all the same, so the moments are all proportional to the forces in this question.

23. Feb 17, 2010