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Classical electrodynamics

  1. Sep 15, 2008 #1
    On the Z=0 plane the charge distribution is of the form
    [tex]\rho[/tex]s=[tex]\rho[/tex]0 sin( [tex]\alpha[/tex] x )sin( [tex]\beta[/tex] y )
    find the potential everywhere, assuming that [tex]\phi[/tex](z[tex]\rightarrow[/tex]±[tex]\infty[/tex])=0

    according to the answer, we should look for a potential of the form
    A sin( [tex]\alpha[/tex] x )sin( [tex]\beta[/tex] y )f(z)
    (due to the form of the charge distribution)

    in order to satisfy the condition [tex]\phi[/tex](z[tex]\rightarrow[/tex]±[tex]\infty[/tex])=0

    we conclude that the potential has the following form:
    A sin( [tex]\alpha[/tex] x )sin( [tex]\beta[/tex] y )exp(-z) , z>0
    A sin( [tex]\alpha[/tex] x )sin( [tex]\beta[/tex] y )exp(z) , z<0

    this potential clearly satisfies the boudary condition, but does f(z) have to be exponential? how can we be sure it isn't some other function that decays at infinity?

    and another question - in which cases is it possible to deduce that the potential has the same form as the charge distribution? is it always true when the charge distribution is infinite and is a product of separate functions of each variable?

  2. jcsd
  3. Sep 15, 2008 #2


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    There was a theorem for this, I think it went like: if the potential is given on the boundary and (... something else, about the charge distribution ...) , then the solution is (in particular) unique. So if you find any potential that solves the Poisson equation, you can be sure it is the solution. I forgot the name of the theorem, but it is in Griffith's EM book for a fact.
  4. Sep 15, 2008 #3


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    Yes, it is called the existence and uniqueness theorem and can be found in Griffith's Introduction to Electrodynamics. It guarantees that there is only one unique solution, so given that you found a solution that works, you can rest assured it is the only possible solution.

    I think the fact that the potential included a term that is proportional to the charge distribution is a result of the second derivative of a sin or cos function being -sin or - cos respectfully. This means that when you plug the potential into the Poisson's Equation Equation , if the potential is seperable (i.e. V(x,y,z)=g(x)h(y)f(z)), you clearly require that g(x) is proportional to sin(alphax) and h(y) is proportional to sin(betay). In general, I think it is very difficult to specify under which conditions you can expect the potential to be proportional to the charge distribution, but my guess is that anytime the charge distribution is a seperable, harmonic function, and the potential is seperable then it will occur. I think there ar other charge distributions as well though, for which the potential is proportional to the charge distribution.
    Last edited: Sep 15, 2008
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