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A Proton is orbiting a metal ball

by cse63146
Tags: ball, metal, orbiting, proton
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cse63146
#1
Apr9-08, 02:07 PM
P: 452
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 [tex]\mu s[/tex].

What is the charge on the ball?

2. Relevant equations

[tex]F_{cp} = \frac{m v^2}{r}[/tex]

[tex]F_c = \frac{k q_1 q_2}{r^2}[/tex]

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.

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

I was wondering whether this statement is true [tex] F_{cp} = F_c[/tex]
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Kurdt
#2
Apr9-08, 04:40 PM
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You're on the right lines, just keep going. One issue is your value for r?
cse63146
#3
Apr9-08, 05:54 PM
P: 452
so would q_1 = q_2 in this case, and for the r, I should use the 1.9mm instead?

Kurdt
#4
Apr9-08, 06:06 PM
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A Proton is orbiting a metal ball

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.
cse63146
#5
Apr9-08, 06:29 PM
P: 452
so the radius of the metal ball would be .5 cm = 0.005m which is the value I used.
Kurdt
#6
Apr9-08, 06:33 PM
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Quote Quote by cse63146 View Post
so the radius of the metal ball would be .5 cm = 0.005m which is the value I used.
Plus the distance of the proton above the surface.
cse63146
#7
Apr9-08, 07:13 PM
P: 452
Quote Quote by Kurdt View Post
Plus the distance of the proton above the surface.
to calculate the velocity needed for the centripetal force?
Kurdt
#8
Apr9-08, 07:32 PM
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Quote Quote by cse63146 View Post
to calculate the velocity needed for the centripetal force?
Yes.
cse63146
#9
Apr9-08, 07:35 PM
P: 452
I don't see why I would need the distance of the proton from the surface.
Kurdt
#10
Apr9-08, 07:37 PM
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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.
original1
#11
Apr12-08, 12:25 PM
P: 10
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?
Kurdt
#12
Apr12-08, 12:34 PM
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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.
original1
#13
Apr12-08, 12:41 PM
P: 10
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
original1
#14
Apr12-08, 12:42 PM
P: 10
I just saw your reply as I entered this post above, what am I doing wrong?
original1
#15
Apr12-08, 12:45 PM
P: 10
radius would be r = 1/2(0.019 m) = 0.0095 m
original1
#16
Apr12-08, 12:46 PM
P: 10
no wait, r = 1/2(1 cm) + 1.9 mm = 0.0069 m
original1
#17
Apr12-08, 12:51 PM
P: 10
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
Dick
#18
Apr12-08, 01:04 PM
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Thanks
P: 25,242
Radius would be 0.005m+0.0019m.


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