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Electrostatics, method of images. infinite conducting plane at constant potential

  1. Sep 20, 2014 #1
    Hi there. The example to obtain the electric potential in the region z>0 by the method of images for a infinite grounded conducting plane, with a point charge q located at a distance d is a typical example of the application of the method of images.

    If we consider that the plane is located in the xy plane, at z=0, and the charge at z=d, we must fulfill the boundary conditions:

    1. ##V=0## when ##z=0##
    2. ##V \rightarrow 0## far from the charge, that is for ##x^2+y^2+z^2>>d^2##.

    So locating the image charge at z=-d one can get the desired potential for the region of space of interest, being:

    ##\displaystyle V(x,y,z)=\frac{1}{4\pi \epsilon_0}\left [ \frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right ]##

    Now, if the plane is not grounded, how should I modify this to set it, lets say at a potential ##V_0##?

    I thought that one posibility would be to use an other image distribution, an infinite charged plane parallel to the conducting plane, at a position ##-z_0## such that ##\frac {\sigma}{2\epsilon_0}z_0=V_0##, where ##\sigma## is the charge per unit area in the charged plane. I think that another possibility would be just to put the charges over the conducting plane. Now the induced charges will be, ofcourse infinite. And the electric field will be the sum of the images plus the real charge in the region z>0.

    So, what I should have would be something like:
    ##\displaystyle V'(x,y,z)=\frac{1}{4\pi \epsilon_0}\left [ \frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right ]+\frac{V_0}{z_0}(z_0-z)##

    The potential due to the infinite charged plane should grow with z, and at the same time it should meet the condition of V(x,y,z=0)=V_0. I'm having trouble with where I'm puting the origin of the potential with respect to the two planes I'm considering.

    Thanks in advance.
     
    Last edited: Sep 20, 2014
  2. jcsd
  3. Sep 21, 2014 #2

    Orodruin

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    Hint: You can shift the potential by a constant without affecting the partial differential equation it has to fulfill (the poisson equation).
     
  4. Sep 21, 2014 #3
    Wait, so you're just saying to add the potential [itex]V_0[/itex] to the expression for [itex]V[/itex], and it'll all be ok? That seems weird to me- mostly because it won't change the potential difference between the plate and the point charge, which is what I think Telemachus wants to see; what happens when the charge [itex]q[/itex] is above the plate but the plate isn't at zero potential.

    Or maybe I just didn't get your hint at all :confused:

    I've always had trouble with the whole image charges concept, so I'm quite interested in the answer to this question (don't think I can help- especially not at around 5 in the morning).
     
    Last edited: Sep 21, 2014
  5. Sep 21, 2014 #4

    Orodruin

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    Yes, this is exactly what I am saying. What makes you think the potential difference between the plate and the charge (which is infinite, but anyway ...) should change? The potential is defined up to an arbitrary constant so this still satisfies the Poisson equation. If you want to be more specific you will have to define the potential at infinity. Specifying it as anything other than V0 will just make your computations more involved while not really changing the field.
     
  6. Sep 21, 2014 #5
    Oh wow right, infinite potential at the point charge.. Quite stupid of me, sorry about that. Thanks for clarifying.
     
  7. Sep 21, 2014 #6
    I thought of that, but I don't think that it's correct. I think I must use the expression I've found, and considere both images in the region of space of interest, the one that generates the potential difference at z=0, and the mirror image for the point charge. The potential I obtain isn't the same, note that the potential due to the infinite charged plane increases linearly with z. So I've got two possible answers to get a solution that fulfills the boundary conditions, but the one I obtain by just adding a constant doesn't change anything from a physical point of view. The electric field (which we may considere a physical entity) wouldn't change, and in this case I think that isn't quite right.
     
    Last edited: Sep 21, 2014
  8. Sep 21, 2014 #7

    Orodruin

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    Yes, the two correspond to different solutions and the difference is the boundary conditions that you put at infinity. The linear solution requires an infinite potential at infinity and would correspond to a constant field far away from the point charge. The solution I proposed is what you get if you want the field to go to zero at infinity. So the question is, which is being asked for?

    Note that the boundary condition ##V \to 0## as ##r \to \infty## is impossible to satisfy in all directions since you have an infinite plane with a non-zero potential.

    Edit: Note that potentials are only defined up to a constant. I could therefore just redefine the potential at the plane to be zero without changing the physics assuming I take this into account in all the boundary conditions. This does not, and should not, change the physics.
     
  9. Sep 21, 2014 #8
    I haven't thought on the boundary conditions I should have, I just thought of the plane as being charged with respect to a reference, and I think that must imply that there are charges over the plane (so the force exerted over the real charge wouldn't be the same in both cases, just adding a constant won't change the force). The solution you proposed wouldn't meet the boundary contition at infinity neither, it would set it at ##V_0##.

    Anyway, I think your observation was fine.
     
  10. Sep 21, 2014 #9

    Orodruin

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    Assuming that the fields vanish is quite common in physical situations and is what more closely corresponds to the original question. If I had put a question like that to my students, it would have been my intended meaning. That being said, if there is nothing specified, then the boundary condition at infinity could also be a constant field. In fact, the free space solution also allows a linear potential, corresponding to a constant field in all of space.
     
  11. Sep 21, 2014 #10
    Yes, you are right, thank you Orodruin, I think this discussion has been enriching and clarifying. There are physical differences in both situations. For example, the induced charges in my case would be infinite, while if I've used your proposal, the induced charges would be the same in both cases. I think its interesting to mention it.
     
  12. Aug 17, 2015 #11
    so if I understood this whole discussion correctly, the only thing that would change when the infinite plate is grounded is the magnitude of the image charges?
    I was reviewing my E&M and got stuck on this problem about two infinite grounded planes with point charge in the middle. It's easy to find the an expression for force when they're grounded, but then it asked what would happen if the metal plates were not grounded and I wasn't sure, so I tried to google it and stumbled upon this forum.
     
  13. Aug 17, 2015 #12

    Orodruin

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    The only relevant thing for the field is the potential difference between the plates. Unless you put a potential difference between the plates, the situation will be exactly equivalent to that of the grounded plates.
     
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