# A Proton is orbiting a metal ball

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 $$\mu s$$.

What is the charge on the ball?

2. Homework 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 $$F_{cp} = F_c$$

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

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

Kurdt
Staff Emeritus
Gold Member
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.

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

Kurdt
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
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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.

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
Staff Emeritus
Gold Member
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.

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

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

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

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

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
Homework Helper

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

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

Kurdt
Staff Emeritus
Gold Member
Its not q^2. There are two different charges.

Dick
Homework Helper
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.

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