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Charge distribution with charge density

  1. Mar 6, 2009 #1
    The following is a worked example in my notes im having difficulty with:

    A charge distribution with charge density [itex]\rho \neq 0[/itex] exist in the half space [itex]V:z>0[/itex]
    We can view this as a system with an earthed plate at the z=0 plane. So at z=0 we have the Dirichlet boundary condition [itex]\varphi(x,y,0)=0[/itex]

    We know that the Greens' function [itex]G_D(\vec{r},\vec{r'}) = \frac{1}{4 \pi \epsilon_0} \frac{1}{|\vec{r}-\vec{r'}} + f_D(\vec{r},\vec{r'})[/itex] i.e. it satisfies the Poisson equation (so essentially we're letting [itex]\varphi = G_D(\vec{r},\vec{r'})[/itex]

    Our Dirichlet boundary condition tells us that [itex]G_D(\vec{r},(x',y',0))=0[/itex]

    [itex]\Rightarrow f_D(\vec{r},(x',y',0))=-\frac{1}{4 \pi \epsilon_0} \frac{1}{\sqrt{(x-x')^2+(y-y')^2+z^2}}[/itex]

    this is generalised to give

    [itex]\Rightarrow f_D(\vec{r},(x',y',0))=-\frac{1}{4 \pi \epsilon_0} \frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z+z')^2}}[/itex]

    and we note that [itex]\nabla^2 f_D(\vec{r},\vec{r'})=0[/itex] i.e. it satisfies Laplace's equation and so G still satisfies Poisson's equation.

    However it then states

    [itex]\varphi(\vec{r})=\frac{1}{4 \pi \epsilon_0} \int_V dV' \rho(\vec{r'}) \left[ \frac{1}{|\vec{r}-\vec{r'}|} - \frac{1}{|\vec{r}-\vec{r'_m}|} \right] [/itex]
    where [itex]\vec{r'}=(x',y',z'), \vec{r'_m}=(x',y',-z')[/itex] (*)

    but the formula given earlier for calculating the potential from Dirichlet boundary conditions was:

    [itex]\varphi(\vec{r})= \int_V dV' G_D(\vec{r},\vec{r'}) \rho(\vec{r'}) - \epsilon_0 \int_S dS' \frac{\partial{G_D(\vec{r},\vec{r'})}}{\partial{n}} \varphi(\vec{r'})[/itex]
    and so it appears that we've neglected the second term in the formula when we've written down the potential above at (*). Is this because the formula given for phi has a phi in the second term and so we can't apply it or something???
     
  2. jcsd
  3. Mar 7, 2009 #2

    tiny-tim

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    Hi latentcorpse! :smile:

    In ∫SdS'…φ, if S is the z=0 plane (and a hemisphere at infinity?), then isn't φ 0 over S?
     
  4. Mar 7, 2009 #3
    Re: Electrostatics

    sorry could u perhaps give a bit more explanation please - i can't really see what you mean?
     
  5. Mar 7, 2009 #4

    tiny-tim

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    perhaps i'm misunderstanding the question :confused:

    i meant, if S is the surface with the Dirichlet boundary condition mentioned in the question,
    then φ = 0 all over S, so the second integral is ∫S0 dS' = 0 ?
     
  6. Mar 7, 2009 #5
    Re: Electrostatics

    lol. good point.
     
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