# Homework Help: Metal ball charged with electron orbiting

1. Feb 7, 2012

### conov3

1. The problem statement, all variables and given/known data
An electron orbits a 3.0 cm diameter metal ball 5.0 mm above the surface. The orbital period of the electron is l2.0 micro seconds. What is the charge on the metal ball?

r=1.5cm or .015m
R=.02m (.015+.005)
Orbit=12*10^-6s
2. Relevant equations

E=(kq/r)R which gives electric field
k=9*10^9
Maybe F=qE?
Is this the only equation?

3. The attempt at a solution
(9*10^9*1.6*10^-19/.015)*.02 which gives E. not sure what to do next!

I fell like the time is involved with the problem.. but do not know where it takes part and cannot find another formula in my book.

2. Feb 7, 2012

### Staff: Mentor

You have an orbiting body; it's traveling in a circular orbit. What forces act on a body in circular motion?

3. Feb 7, 2012

### conov3

ω angular rotation
angular velocity and then we would have a period of one rotation?
ω=2∏/T?

4. Feb 7, 2012

### conov3

Also, v=a^2/r

5. Feb 7, 2012

### Staff: Mentor

Well, you're on the right track. It looks like you know how to find the angular velocity. What force is required to maintain circular motion? What is providing that force?

6. Feb 7, 2012

### conov3

? kind of confused with that question
Wouldnt that be angular acceleration?
so
a=v^2/r
I actually switched my a and v on the last post. I meant to give angular acceleration.
So from there, if I find my acceleration.. I can say F=ma? then ma=qE?

7. Feb 7, 2012

### conov3

Then find E=ma/q....

then q=Er/kR?

8. Feb 7, 2012

### Staff: Mentor

Yes, you can balance forces or accelerations -- they are proportionally related via the mass of the orbiting object.

No, it's not angular acceleration that you want. What acceleration is involved in circular motion?

9. Feb 7, 2012

### conov3

Tension?

10. Feb 7, 2012

### Staff: Mentor

Tension is a force in a connecting filament or string. No, you're looking for either centripetal acceleration or centripetal force (When an object is whirled around on the end of a string, the centripetal force is provided by tension in the string).

What are the formulas for centripetal force and centripetal acceleration?

Last edited: Feb 7, 2012
11. Feb 7, 2012

### conov3

Centripetal force=mv^2/r
Centripetal acc.=ω^2*r

12. Feb 7, 2012

### Staff: Mentor

Yup. You might want to note that centripetal acceleration is also ac = v2/r . It gets made into centripetal force by multiplying by the mass: f = ma. Similarly, centripetal force can be had as: Fc = mω2r.

So, suppose you consider the formula for centripetal force, Fc = mω2r. What is supplying it? (What's holding the electron in orbit?)

13. Feb 7, 2012

### conov3

Are you meaning the charge of the metal ball? thats keeping it in place

14. Feb 7, 2012

### Staff: Mentor

Yes, what force does the charged ball exert on the orbiting electron? What's the formula?

15. Feb 7, 2012

### conov3

E=(kq/r)R? Im not sure this is where I struggled in our first phys semester. I know that the centripetal acceleration of the electron would point towards the ball..
but E does not equal F so that's not the right equation right?

16. Feb 7, 2012

### Staff: Mentor

Your text or notes should have the equation for the electric force:

$F = k \frac{Q_1 Q_2}{r^2}$

17. Feb 7, 2012

### conov3

So from there I could say... mv^2/r=kqq/r^2

18. Feb 7, 2012

### conov3

so from there I could say mv^2/r=kQQ/r^2

19. Feb 7, 2012

### conov3

or do i use ω instead of v^2/r and it equals 12*10^-6

If this is correct, I got a final answer or q=3.644*10^-35C

Last edited: Feb 7, 2012
20. Feb 7, 2012

### Staff: Mentor

Keep in mind that there are two charges involved -- the charge of the ball and the charge on the electron. So write 'Qq' rather than 'QQ'.

You can use either formula for the centripetal acceleration. One requires you to find the angular velocity, the other the tangential speed of the electron. You've already shown that you can find the angular velocity quite easily... So your equation equating the centripetal force to the electrical force becomes:

$m_e \omega^2 r = k \frac{Q q_e}{r^2}$

Where me is the electron mass and qe the magnitude of the charge on the electron.