# Physics Problem

1. Dec 15, 2003

### mustang

Problem 22.
given: g=9.814m/s^2.
An object with a net charge of 40 uC is placed in a uniform electric field of 572 N/C, directed vertically.
What is the mass of this object if it floats in this electric field? In units of kg.
Note: What formula would you use?

2. Dec 15, 2003

### himanshu121

Use Equilibrium Condition, Relation b/w Field and Force

3. Dec 19, 2003

### mustang

I need more help. What does "b" and "w" mean?

4. Dec 20, 2003

### HallsofIvy

"b" and "w" don't mean anything. humanshu121 was using b/w to mean "between".

5. Dec 20, 2003

### mustang

I having a hard time solving problem since I don't know what formula I should use and the values to "sub" in.

6. Dec 20, 2003

### gnome

What is the formula for the force experienced by a charge q in an electric field of magnitude E?

7. Dec 20, 2003

### mustang

Would that formula be Electric field strength=Coulomb constant* charge producing the field(q)/(distance)^2.

8. Dec 20, 2003

### gnome

No -- that formula describes the strength of the field PRODUCED by a charge q.

You want the force EXPERIENCED by a charge q under the influence of a field from some other source.

9. Dec 20, 2003

### mustang

Is the formula electric force=Coulomb constant*(charge1)(charge2)/(distance)^2

10. Dec 20, 2003

### gnome

Not exactly, but you could find the right one "hidden" in there (because in there you have included the expression that you wrote up above for the field produced by a charge).

11. Dec 20, 2003

### mustang

I'm comfused?

12. Dec 20, 2003

### dodger

You have two forces,

1.weight from the mass acting towardds the earth
2.an electric force from the electric field, the key equation here is that: the force on a charged particle is its charge times the field -> F = qE

Note: Coulomb law is that just mentioned without the charge of the particle in question. Coulmbs law is with q1 and q2 is between two paticles, in this case the electric field is probably produced by large metal plates with millions of particles in them - point is how the field is made is not important.

13. Dec 20, 2003

### gnome

If the field E1 produced by a point charge q1 is
E1 = keq1/r2

and the force exerted on point charge q2 by point charge q1 is
F12 = keq1q2/r2

just substitute the formula up above for E to get
F12 = q2E1

Here, you aren't dealing with 2 point charges. Instead you have 1 point charge in a uniform field from some unspecified source. But as dodger said, the formula above can be generalized, so the force on a charge q from any uniform field is
F = qE

Now you just have to find the conditions so that force exactly offsets the weight of the particle -- basically a geometry question.

14. Dec 20, 2003

### dodger

This subject should not be a confusing thing, a field is produced by charged stuff and this exerts a force on other charged stuff, the size of this force (as given by the equation i said) tells you how big a force acts on the charged thing. Stop and think about it - if you have a charged particle in an electric field and you double its charge the effect on it goes the same - the force on it doubles. The same effect would happen if the field was doubled - the force on it would double.

15. Dec 20, 2003

### gnome

Right.

And forget about my unnecessary "geometry" comment.

There's really no geometry to deal with here. One force (the weight) acts straight down and the other (the electrical force) acts straight up (do you see why?).

If the object floats, these two forces must be equal, right? Simple algebra. Just solve for m.

16. Dec 20, 2003

### mustang

So F=qE is bascially F=ma, Right.

17. Dec 20, 2003

### gnome

Right; well, F=qE=F=mg.

I assume you meant g, right?

18. Dec 20, 2003

### mustang

So the equation to solve this problem is qE=mg, where g=9.81m/s^2, E=572N/C, and q= 40*10^-6.

19. Dec 20, 2003

### gnome

Correct.

20. Dec 20, 2003

Thanks!!!!

Thank you!