# A Proton is orbiting a metal ball

1. Apr 9, 2008

### cse63146

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

A proton orbits a 1.0-cm-diameter metal ball 1.90 mm above the surface. The orbital period is 1.50 $$\mu s$$.

What is the charge on the ball?

2. Relevant equations

$$F_{cp} = \frac{m v^2}{r}$$

$$F_c = \frac{k q_1 q_2}{r^2}$$

3. The attempt at a solution

Since the proton is orbiting the metal ball, there is obviously a centripetal force involved, but first I need to velocity.

$$v_{cp} = {\omega r} = \frac{2 \pi r}{T} = \frac{2 \pi (0.005)}{1.5*10^{-6}} = 2.1*10^4 m/s$$

I was wondering whether this statement is true $$F_{cp} = F_c$$

2. Apr 9, 2008

### Kurdt

Staff Emeritus
You're on the right lines, just keep going. One issue is your value for r?

3. Apr 9, 2008

### cse63146

so would q_1 = q_2 in this case, and for the r, I should use the 1.9mm instead?

4. Apr 9, 2008

### Kurdt

Staff Emeritus
You'll be solving for for one of the charges. The other will be the charge on the proton. The radius if measured from the centre of the metal sphere.

5. Apr 9, 2008

### cse63146

so the radius of the metal ball would be .5 cm = 0.005m which is the value I used.

6. Apr 9, 2008

### Kurdt

Staff Emeritus
Plus the distance of the proton above the surface.

7. Apr 9, 2008

### cse63146

to calculate the velocity needed for the centripetal force?

8. Apr 9, 2008

### Kurdt

Staff Emeritus
Yes.

9. Apr 9, 2008

### cse63146

I don't see why I would need the distance of the proton from the surface.

10. Apr 9, 2008

### Kurdt

Staff Emeritus
The r is the distance of the proton from the centre of rotation. The distance is the radius of the sphere plus the height above the surface.

11. Apr 12, 2008

### original1

so F_cp=(mv^2)/r

and F_cp=F_c
so F_cp=(kq_1q_2)/r^2
also q_1=q_2 --> F_cp=(kq^2)/r^2

Is this right?

12. Apr 12, 2008

### Kurdt

Staff Emeritus
If you're doing the same question you can't assume both charges are the same since it is one of the charges you are trying to find.

13. Apr 12, 2008

### original1

r = 1.0cm + 1.9 mm = 1.19 cm = 0.0119 m
m_proton = 1.67*10^-27 kg
T = 1.5 us = 1.5*10^-6 s

v_cp = omega*r = (2pi*r)/T = [2pi(0.0119 m]/(1.5*10^-6) = 4.98*10^4 m/s

F_cp = (mv^2)/r = (1.67*10^-27 kg*(4.98*10^4 m/s)^2)/0.0119 m = 3.49*10^-16 N

F_cp = F_c
so F_cp=(kq_1q_2)/r^2
F_cp=(kq^2)/r^2
3.49*10^-16 N = (k*q^2)/r^2
q^2 = [(3.49*10^-16 N)(r^2)]/(k)
q^2 = [(3.49*10^-16 N)((0.0119m)^2)]/(k)
q^2 = 4.937*10^-20 Nm^2 / 8.99*10^9 Nm^2/C^2
q^2 = 5.49*10^-15 C^2
q = 2.34*10^-15 C

Is this the correct procedure and answer? Someone plz help

14. Apr 12, 2008

### original1

I just saw your reply as I entered this post above, what am I doing wrong?

15. Apr 12, 2008

### original1

radius would be r = 1/2(0.019 m) = 0.0095 m

16. Apr 12, 2008

### original1

no wait, r = 1/2(1 cm) + 1.9 mm = 0.0069 m

17. Apr 12, 2008

### original1

new calculation:

r = r = 1/2(1 cm) + 1.9 mm = 0.0069 m
m_proton = 1.67*10^-27 kg
T = 1.5 us = 1.5*10^-6 s

v_cp = omega*r = (2pi*r)/T = [2pi(0.0069 m]/(1.5*10^-6) = 2.89*10^4 m/s

F_cp = (mv^2)/r = (1.67*10^-27 kg*(2.89*10^4 m/s)^2)/0.0069 m = 2.02*10^-16 N

F_cp = F_c
so F_cp=(kq_1q_2)/r^2
F_cp=(kq^2)/r^2
2.02*10^-16 N = (k*q^2)/r^2
q^2 = [(2.02*10^-16 N)(r^2)]/(k)
q^2 = [(2.02*10^-16 N)((0.0069m)^2)]/(k)
q^2 = 9.62*10^-21 Nm^2 / 8.99*10^9 Nm^2/C^2
q^2 = 1.07*10^-30 C^2
q = 1.03*10^-15 C

18. Apr 12, 2008

### Dick

19. Apr 12, 2008

### original1

You're right, I corrected it and it comes to 0.0069 m. Do you know if this procedure I used is correct??

20. Apr 12, 2008

### original1

the final answer should come to about -#*10^-12, but this is nowhere near that :(