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Induced Surface Current in Conducting Plane, Dropped Charge

  1. Mar 28, 2016 #1
    1. The problem statement, all variables and given/known data
    a) A charge q is released a distance d above a grounded infinite conducting plane. It's non relativistic velocity is v. Find the induced surface current density on the plane.
    b) Show that the above current density produces a vanishing magnetic force on the charge.
    2. Relevant equations
    Magnetic Boundary Conditions: B_above-B_below = K_f x n.
    Magnetic Force on Charge: F=qvBsinΘ

    3. The attempt at a solution
    I'm not at all sure what to do for part a) but for b) Ampere's Law applied to a rectangle passing through the plane implies that the field is parallel to the plane so the cross product gives 0.
     
  2. jcsd
  3. Mar 28, 2016 #2

    TSny

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    If the B field is parallel to the plane at the location of the charge, why would the cross product of v and B yield zero?
     
  4. Mar 28, 2016 #3
    Right, they would be perpendicular so one of the magnitudes needs to zero.
     
  5. Mar 28, 2016 #4

    TSny

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    Symmetry in the system should help in determining the magnitude of B along the line of fall of the charge.
     
  6. Mar 29, 2016 #5
    Which symmetry are you referring to?
     
  7. Mar 29, 2016 #6

    TSny

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    If the system is rotated about a certain axis, there is no change in the induced surface-current distribution. So, there can be no change in the B field pattern if the system is rotated about this axis. That should tell you something about B for points on the axis of symmetry.
     
  8. Mar 29, 2016 #7
    So azimuthal symmetry about the line of fall of the charge would make sense. I'm just not seeing the connection between that and zero magnetic field on the line. Is it an Amperian loop?
     
  9. Mar 29, 2016 #8

    TSny

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    Question (b) only asks you to show that the magnetic force on the charge is zero. Azimuthal symmetry alone implies that even if B is nonzero on the axis of symmetry, it must point in a certain direction. From this direction you should be able to conclude that the force is zero.

    If you want to show that B is actually zero on the axis of symmetry, you can use the Biot-Savart law along with the symmetric distribution of the current density.
     
  10. Mar 29, 2016 #9
    Oh, thank you. I understand where I was going wrong now. The magnetic field will have to point along the line of fall so the force vanishes.
     
  11. Mar 29, 2016 #10

    TSny

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    Yes, that's right.
     
  12. Mar 29, 2016 #11

    TSny

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    You could also use the fact if there were a magnetic force, it would necessarily be perpendicular to the velocity (i.e., perpendicular to the symmetry axis). But this force sticking out in some direction would violate the rotational symmetry.
     
  13. Mar 29, 2016 #12
    So that's part b), could I get a hint how to begin part a)? The only idea I have is to somehow use magnetic boundary conditions.
     
  14. Mar 29, 2016 #13

    TSny

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    Can you see a way to get the surface charge density σ on the conducting plane?
     
  15. Mar 30, 2016 #14
    If the charge wasn't moving, yes.
    The logic would be that from Gauss' Law the electric field at the surface of the plane is σ/ε, which is equal to the normal derivative of the electric potential evaluated at the surface of the plane. So from the method of images the potential is the same as that due to two oppositely charged point charges.
    So I suppose my real difficulty with the question is that I'm not sure how the electric potential changes when the charge is accelerating.
     
  16. Mar 30, 2016 #15

    TSny

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    σ at a point of the plane will depend on the distance d of the point charge from the plane. Therefore dσ/dt will depend on the velocity v of the point charge. You should be able to relate dσ/dt to the current density K.
     
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