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A Proton is orbiting a metal ball

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1. Homework Statement

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. Homework 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]
 

Answers and Replies

Kurdt
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You're on the right lines, just keep going. One issue is your value for r?
 
452
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so would q_1 = q_2 in this case, and for the r, I should use the 1.9mm instead?
 
Kurdt
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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.
 
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so the radius of the metal ball would be .5 cm = 0.005m which is the value I used.
 
Kurdt
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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.
 
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Plus the distance of the proton above the surface.
to calculate the velocity needed for the centripetal force?
 
Kurdt
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I don't see why I would need the distance of the proton from the surface.
 
Kurdt
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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.
 
10
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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
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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.
 
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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
 
10
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I just saw your reply as I entered this post above, what am I doing wrong?
 
10
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radius would be r = 1/2(0.019 m) = 0.0095 m
 
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no wait, r = 1/2(1 cm) + 1.9 mm = 0.0069 m
 
10
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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
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Radius would be 0.005m+0.0019m.
 
10
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You're right, I corrected it and it comes to 0.0069 m. Do you know if this procedure I used is correct??
 
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the final answer should come to about -#*10^-12, but this is nowhere near that :(
 
Kurdt
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Its not q^2. There are two different charges.
 
Dick
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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
Good up to here. Now as Kurdt said, q1 is not equal to q2, you can't replace that with q^2. Set q1 equal to the charge of the proton and solve for q2, the charge of the sphere.
 
10
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F_cp = F_c
so F_cp=(kq_1q_2)/r^2

q_1 = 1.602*10^-19 C

q_2= (F_cp*r^2)/(k*q_1)
q_2= [(2.02*10^-16 N)(0.0069 m)^2]/[(8.99*10^9 Nm^2/C^2*1.602*10^-19 C)]
q_2= 6.68*10^-12 C

this looks right, thank you!!

do you know why the answer has to be negative value?
 
Dick
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The proton has a + charge. If it's orbiting the sphere, then it's attracted to the sphere. Opposite charges attract.
 
10
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That makes perfect sense...

thanks for all your help!
 

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