Force on a charge inside a hollow charged sphere

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mheslep
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Homework Statement


What is the force on a point charge q1 contained inside a hollow conductive sphere (or fine grid) with charge q2 and radius R?


Homework Equations


Coulomb F=k q1q2/r^2
Gauss Law/method of closed surface: Integral over closed surface of Enormal da = E times the area of the surface = Flux = must equal the enclosed charge / eps0.
F=qE

The Attempt at a Solution



Am aware of familiar prediction that an _empty_ hollow sphere has no internal E field, which follows from applying gauss's law to any virtual surface internal to the sphere with no enclosed charge, therefore E must be zero everywhere. In this case w/ contained q1 there is some field. However, applying gauss's closed surface again seams to give just the solution for E from a point charge q: E=kq1/r^2 (r distance from q1) only, with no E contribution from the surrounding sphere. Thus no force on q1??? Ive done some computer sim. (2D only) by surrounding a pt charge w/ 2 dozen like pt charges arranged in a circle and then doing N - body Coulomb - there's plenty of resulting force on the internal charge excepting the center of the circle.

A little help please?

Falstaff
 

Answers and Replies

  • #2
Dick
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I admire your resourcefulness at trying to simulate the situation with discrete point charges. If you follow that route then you should be able to show that as the number of point charges gets larger and larger then the interior force gets smaller and smaller. In the limit of a continuous distribution it will go to zero. Really!
 
  • #3
mheslep
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Yes with an _empty_ interior you don't need many circular charges to see the E field approaching zero. With an internal charge, though, the E field can not be zero everywhere. Gauss's law: I can close a surface around a charge, there must be non zero E. For that matter, let the internal pt charge grow a bit to the point where it becomes a smaller sphere so that you then have a concentric sphere capacitor. In that case I know the E between the inner and outer spheres - some ~ constant radial E.
 
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Dick
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The internal charge doesn't produce a force on itself. Or do I misunderstand you?
 
  • #5
mheslep
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Ah, ok. You're stating there is an E field but only due to the internal charge - a normal pt charge radial E field w/ no contribution from the surrounding sphere, thus no force? Makes sense. What about the concentric spheres case then. Isn't that an example of the same thing?
 
  • #6
Dick
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Exactly. But what about the concentric spheres case? I don't quite get the picture?
 
  • #7
mheslep
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My point: Isnt the concentric spheres case merely an a specific instance of the general case of a charge contained inside a sphere? Ah, (again). Im thinking now that yes it is, but again as before there is no net force on the smaller sphere.

Bottom line is I was trying to tinker w/ the Paul Trap concept and use all positive rotating poles instead of the rotating dipole. Works great in the 2D sim, Im containing 100kev ions no problem.
 
  • #8
Dick
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Ah, you are editing your posts! Sure there is E field between the spheres, hence capacitance, but still no force on the internal sphere.
 
  • #9
Dick
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Wow, Paul Trap! Guess I'd better look that one up.
 
  • #10
mheslep
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  • #11
Dick
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Thanks! So this isn't just pure theory.
 
  • #12
mheslep
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Paul Traps have been around for some years apparently. Its ability to hold an ion I believe is the basis for some of our best atomic measurements. I was curious about using something like it to confine at higher densities and energies. The trick is converting the rotating dipole to all positive poles which seems to generate a significant virtual potential well for my 2D sim.
 
  • #13
Meir Achuz
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Homework Statement


What is the force on a point charge q1 contained inside a hollow conductive sphere (or fine grid) with charge q2 and radius R?
Falstaff
The charge q2 on the outer conducting sphere does not affect the field inside the conductor, so the force on q1 is the same as it would be for a grounded conductor. Use an image charge outside the sphere to solve the problem.
 
  • #14
Dick
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The charge q2 on the outer conducting sphere does not affect the field inside the conductor, so the force on q1 is the same as it would be for a grounded conductor. Use an image charge outside the sphere to solve the problem.
Good point. If the internal charge can cause the sphere charge to rearrange so it's no longer uniform, you do get internal field. And need to use the image concept to compute the force.
 

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