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Going through Jackson's book (electrodynamics, method of images)

  1. Feb 12, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I would like to
    1)Find the electric field in all the space in the following set up: There's an electric charge q in a region with dielectric constant ##\varepsilon _1## which is at a distance d from a plane that separates the space into 2 regions, namely the one of the charge and another region with dielectric constant ##\varepsilon _2##.
    2)Find the polarization charge density over the plane. (Jackson gives out the answer to that task though he does not give the details which is awesome for me since I have to train hard).

    2. Relevant equations
    To get the electric field in all the space I will use the method of images (just like Jackson's do in order to get ##\Phi## in all the space)
    Page 155 and 156 in the 3rd edition of Jackson's book.
    The main idea I believe is first to get the potential in all the space and then get ##\vec E## via ##\vec E = \nabla \Phi##.
    3. The attempt at a solution
    I followed Jackson's book up to where he found out the potential in the region of the electric charge. It's ##\Phi (\rho, \phi , z )= \frac{1}{4\pi \varepsilon _1 } \left [ \frac{q}{\sqrt{\rho ^2 + (d-z)^2}} + \frac {q'}{\sqrt{\rho ^2 + (d+z)^2}} \right ]## where q' is a pseudo-charge situated in the region with dielectric constant ##\varepsilon _2## at a distance d from the plane (well it's the conventional method of images).
    And then I'm lost when he derives the potential in the region with ##\varepsilon _2##.
    He says that since there's no charge in that region, the potential must be a solution to Laplace equation without singularities. So far so good. It's just ##\varepsilon _2 \nabla \cdot \vec E =0##, just as he himself wrote a few lines before.
    But then he says that "Clearly the simplest assumption is that for z<0 the potential is equivalent to that of a charge q'' at the position A of the actual charge q: ##\Phi = \frac{q''}{4\pi \varepsilon _2 \sqrt{\rho ^2 + (d-z)^2}}##, for z<0.""
    ???
    I don't understand why he isn't summing up the potential due to q and q' like in the first region. He now uses another pseudo charge which is located at the same spot as the real charge. It's like he's not using the method of images anymore but he somehow thinks the problem as an "effective charge" right on the spot of the real charge. I don't understand why this work. I know that since the solution to Laplace equation is unique here, if this method works then it works and it leads to the only solution but I just don't understand the strategy here.
    Thanks for any clarification.
     
  2. jcsd
  3. Feb 13, 2013 #2

    TSny

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    As you say, the uniqueness theorem guarantees that if “by hook or by crook” you come up with a potential function that satisfies Poisson's equation for the given charge distribution throughout all regions and also satisfies the required boundary conditions, then you have the solution. Jackson is just making (highly) educated guesses to construct a solution. Basically, he’s saying “let’s try this and see if we can make it work”.

    For z > 0 he assumes the potential is that produced by the given charge q and an image charge q’ of magnitude yet to be determined.

    For z < 0 he assumes the potential is that of a single point charge of magnitude q’’ (yet to be determined) located at the same position as q and no other charges anywhere. So, he is guessing that the potential for z < 0 is that due to just one charge q’’ located in the region z > 0. These choices for the potential in the two regions clearly produce a potential function that satisfies Laplace’s equation throughout both regions except where it shouldn’t – namely, at the singularity of the given point charge q. He then shows that the boundary conditions at z = 0 can be satisfied with appropriate choices of q’ and q’’. So, he has found the solution to the problem.

    Note that you would definitely not want to include the image charge q’ in the construction of the potential for z < 0 since the solution for that region must satisfy Laplace’s equation without any singularities.

    I agree that the guess for the form of the potential for z < 0 is a little surprising. It’s somewhat unintuitive to me that the electric field lines in this region are that of a point charge located at the same position as q. I would have guessed that if the field for z < 0 is that of a point charge, then the location of that point charge would not be the same as the location of q. I would have guessed that the location of q’’ would depend on the relative magnitudes of the dielectric constants.

    However, from a mathematical point of view, you can see that putting the charge q’’ at the location of q is what makes it possible to satisfy the boundary conditions over the entire plane z = 0.
     
  4. Feb 13, 2013 #3

    fluidistic

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    Thank you very much, very neat explanation.
    I still haven't figured out why the boundary conditions imply the conditions on the charges, namely that q-q'=q'',etc. but I will think about it -alone-, I won't ask help for this. :smile:

    So now I've rewrote the potential in all the space. Do you think that using the formula ##\vec E = - \vec \nabla \Phi## is the way to go to get the electric field in all the space? Of course I must take the gradient in cylindrical coordinates.
    In fact since the potential has absolutely no dependence on the cylindrical coordinate phi, I could have used 2 dimensional polar coordinates to solve the problem, right?
    The gradient of Phi the potential is the vector ##\left ( \frac{\partial \Phi }{\partial \rho} , 0 , \frac{\partial \Phi }{\partial z} \right )##. Of course I must calculate this explicitly (I've done it and it does not look that beautiful, and it's a bit long to write in latex!).
    And it looks like evaluating the electric field in z=0 will help me to get the polarization in the dielectric at z=0 which in turn helps me to get the polarization-surface-charge density, ##\sigma##, what they ask me to find.
     
  5. Feb 14, 2013 #4

    TSny

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    Yes, you can get E that way. Or, you can just write down E directly from knowing that in each region E is the field of point charges. For example, in region 2 E is the field of point charge q'' located in region 1 at x = d and ρ=0. You also have to throw in a factor of 1/ε2 to account for the dielectric material in region 2. In region 1 you have a superposition of the point charge fields of q and q' with a factor of 1/ε1. But, using the gradient of the potential will get the same result.
    Yes, that's right. The field at z = 0 has already essentially been determined when satisfying the boundary conditions at z = 0 to determine q' and q''.
     
  6. Feb 14, 2013 #5

    fluidistic

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    Thank you very much!
     
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