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Uniform Circular Motion Inside a Sphere of Charge

  1. Aug 10, 2017 #1
    1. The problem statement, all variables and given/known data
    "*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)


    2. Relevant equations

    1) F=|q1| × |q2| × k / r2
    (k=9E9)

    2) F=mv2 / r


    3. 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 N


    rearranged 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
     
  2. jcsd
  3. Aug 10, 2017 #2

    haruspex

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    Yes. You are told it is uniform, and you know the two radii. What proportion of the charge is inside the orbital radius?
     
  4. Aug 10, 2017 #3
    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?
     
  5. Aug 10, 2017 #4

    haruspex

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    Do you mean in the nucleus but outside the orbit?
    What is the field (gravitational or electric) inside a uniform spherical shell?
     
  6. Aug 11, 2017 #5
    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?
     
  7. Aug 11, 2017 #6

    haruspex

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    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.
     
  8. Aug 15, 2017 #7
    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?
     
  9. Aug 15, 2017 #8

    haruspex

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    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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