Uniform Circular Motion Inside a Sphere of Charge

AI Thread Summary
The discussion focuses on calculating the speed of a tau particle in uniform circular motion inside a uranium nucleus, treating the nucleus as a uniformly charged sphere. Initial calculations using the total charge of the nucleus led to an incorrect speed, prompting a reevaluation of the charge within the tau's orbital radius. The correct approach involves determining the proportion of the nucleus's charge that is enclosed within the tau's orbit, which significantly affects the force calculations. Participants highlight the importance of understanding electric fields inside a uniform spherical shell, noting that such shells produce no net field within. The conversation emphasizes the need for a solid grasp of physics principles, particularly when applying calculus to these concepts.
Medeiros

Homework Statement


"*Question 44: Uniform Circular Motion Inside Sphere of Charge
The tau particle is a negatively charged particle similar to the electron, but of much larger mass - its mass is 3.18 x 10-27 kg, about 3480 times the mass of the electron and about twice the mass of a proton or neutron. Nuclear material is transparent to the tau; thus the tau can orbit around inside a nucleus, under the influence of attraction of the nuclear charge. Suppose the tau is in a circular orbit of radius 2.9 x 10-15 m inside a uranium nucleus. Treat the nucleus as a sphere of radius 7.4 x 10-15 m with charge 92e uniformly distributed throughout its volume. Find the speed of the orbital motion of the tau. Note that the charge of the tau is the same as that of the electron."

tau;
mass= 3.18 x 10-27 kg
circular orbit of radius= 2.9 x 10-15 m
charge= 1e- (1.602E-19)

u nucleus;
radius= 7.4 x 10-15 m
charge= 92 e (92* 1.602E-19C)

Homework Equations



1) F=|q1| × |q2| × k / r2
(k=9E9)

2) F=mv2 / r

The Attempt at a Solution



What I did, which is the wrong answer, is

Found the force using Eq. 1:
(1.602E-19 * 92* 1.602E-19 * 9E9) / (2.9 x 10-15)2

=2526.73 Nrearranged that F to = mv2 /r so that I can find V, which is

V = √F×r / m
(mass of tau, same 'r' as above)
=7.68E7 m/s

However, the correct answer was something like 1.118E7 or E11

I think that I need to use the radius of the nucleus to find how much charge is outside the tau's orbital radius? I haven't seen anything like that yet and wouldn't know where to start if this is the case

Thanks in advance
 
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Medeiros said:
find how much charge is outside the tau's orbital radius?
Yes. You are told it is uniform, and you know the two radii. What proportion of the charge is inside the orbital radius?
 
haruspex said:
Yes. You are told it is uniform, and you know the two radii. What proportion of the charge is inside the orbital radius?

I just tried that and I used the volume of a sphere 4/3 pi r^3
I did the ratio between the volume of the sphere from the orbital radius of the tau to the volume of the sphere from the actual nucleus. V1/V2 = 0.0602

Then I did 0.0602 * charge of nucleus to get the charge that is inside the nucleus. I used that as the new q2 for equation 1 and got a force of F = 152.15 N

Used both formulas as before and the new velocity is 1.178 E7 m/s

However, what about the charge that is now outside the nucleus and producing a force away from the center of the nucleus, so that the net charge is F to nucleus - F outside nucleus?
 
Medeiros said:
what about the charge that is now outside the nucleus
Do you mean in the nucleus but outside the orbit?
What is the field (gravitational or electric) inside a uniform spherical shell?
 
haruspex said:
Do you mean in the nucleus but outside the orbit?
Yep that's what I meant

What is the field (gravitational or electric) inside a uniform spherical shell?
Well since it has mass it has a negligible gravitational field and also an electric field because it has charge?
 
Medeiros said:
Yep that's what I meant

Well since it has mass it has a negligible gravitational field and also an electric field because it has charge?
No. Seems to be a significant gap in your knowledge of the subject. Take a look at http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html. It discusses gravitation, but applies equally to electric fields.
 
haruspex said:
No. Seems to be a significant gap in your knowledge of the subject. Take a look at http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html. It discusses gravitation, but applies equally to electric fields.
I've started learning calculus only a couple of weeks ago so I couldn't follow that

I think that I didn't pay attention to the force of gravity on that enough:
since g=GM/r^2, & r^2 = very small value, then that means g is significantly strong in that and i need to take it into account?
 
Medeiros said:
I've started learning calculus only a couple of weeks ago so I couldn't follow that
Proving the result, through calculus, is quite hard. Most students only learn that the result is true, not how to prove it:
A uniformly charged spherical shell produces no net field inside itself.
A uniform gravitational spherical shell produces no net field inside itself.
And further, outside the shell the field is the same as though all the charge (mass) were concentrated at the centre of the sphere.
 
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