Method of images& half a sphere

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SUMMARY

The discussion focuses on solving a potential problem in electrostatics involving a half-sphere configuration. The user proposes placing a charge of -q beneath the z=0 plane to cancel the potential in that region. They further explore the placement of additional charges, q' and -q', symmetrically around the z=0 plane to achieve a zero potential on the surface of the sphere. The solution emphasizes the importance of symmetry in charge distribution to maintain equilibrium in electrostatic potential.

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  • Knowledge of symmetry in physics
  • Ability to apply Pythagorean theorem in three-dimensional space
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Students and professionals in physics, particularly those studying electrostatics, as well as educators looking for examples of charge configurations and potential calculations.

Ellyl
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Homework Statement


[PLAIN]http://img532.imageshack.us/img532/3751/39842434.png
Find the potential in all space for this configuration.

Homework Equations


..basic electrostatics stuff, I guess.


The Attempt at a Solution


I can easily cancel the whole z=0 plane by placing a charge of -q under it, of course, but I have no idea what I should do about the half-sphere. I naturally went for placing a charge in the center of the sphere to cancel the other two, and I attempted this by calculating the potential on the top of the sphere ([PLAIN]http://img15.imageshack.us/img15/6324/kqr.gif, if pythagoras was right) and negating that with another charge, but this naturally messes the z=0 plane. Is it some other configuration of charges that's symmetrical for the z=0 plane? I guess it has to be, but I have no idea what such configuration would also negate the potential of the sphere.
 
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..no, wait, I think I actually got this. Once I place the -q charge under the plane, the problem becomes equivalent to this:
[PLAIN]http://img189.imageshack.us/img189/6430/spherel.png
, and then I just need to place charges q' and -q' symmetrically to the z=0 plane so that the potential on the sphere equals zero. Right?
 
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