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Electric Field and the Speed of a Proton

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle with a charge of -60.0 nC is placed at the center of a nonconducting spherical shell of inner radius 20.0 cm and outer radius 34.0 cm. The spherical shell carries charge with a uniform density of -2.26 µC/m3. A proton moves in a circular orbit just outside the spherical shell. Calculate the speed of the proton.

    2. Relevant equations
    We will need:
    [tex]E = F/q_{o}[/tex]
    [tex]\Phi=EA=q_{inside}/\epsilon_{o}[/tex] (no integral is needed because we know the electric field will be constant at the surface of the sphere and we know the surface area of a sphere).
    [tex] F = ma_{c}[/tex]

    3. The attempt at a solution

    Solving for E I get:

    E = [tex]q_{inside}/(\epsilon_{o}A)[/tex]

    [tex]q_{o}[/tex] is just the inner charge (-60.0nC) + the outer charge [4/3*pi*charge density*(0.34^3-0.20^3).

    Plugging in for E I get:


    [tex]q_{o} = q_{inside}[/tex] because the spherical surface should act as a point charge right?

    Therefore after some algebra and substitution for the centripetal acceleration I get:

    [tex] v = \sqrt{q^{2}/(4\pi\epsilon_{0}rm})[/tex]
    Where r = .34 and m is the mass of a proton.

    When I plug in all the values I get a speed on the order of [tex]10^{12}m/s[/tex]
    Which is faster than the speed of light if I am not mistaking...that being roughly [tex]3 x 10^{8}m/s[/tex]

    What am I doing wrong?
  2. jcsd
  3. Aug 31, 2008 #2
    My problem lies with the above. [tex]q_{o} = q_{inside}[/tex]. That is a false statement. The [tex]q_{o}[/tex] is actually the charge of the proton, not the charge of the entire charge configuration. With that adjustment I get an answer on the order of 10^5 which is must more realistic and was the correct answer.
  4. Aug 31, 2008 #3


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    Homework Helper

    Yup that's right. Good that you figured it out by yourself.
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