- #1
latentcorpse
- 1,411
- 0
The following is a worked example in my notes I am having difficulty with:
A charge distribution with charge density \rho \neq 0 exist in the half space V:z>0
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 \varphi(x,y,0)=0
We know that the Greens' function G_D(\vec{r},\vec{r'}) = \frac{1}{4 \pi \epsilon_0} \frac{1}{|\vec{r}-\vec{r'}} + f_D(\vec{r},\vec{r'}) i.e. it satisfies the Poisson equation (so essentially we're letting \varphi = G_D(\vec{r},\vec{r'})
Our Dirichlet boundary condition tells us that G_D(\vec{r},(x',y',0))=0
\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}}
this is generalised to give
\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}}
and we note that \nabla^2 f_D(\vec{r},\vec{r'})=0 i.e. it satisfies Laplace's equation and so G still satisfies Poisson's equation.
However it then states
\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]
where \vec{r'}=(x',y',z'), \vec{r'_m}=(x',y',-z') (*)
but the formula given earlier for calculating the potential from Dirichlet boundary conditions was:
\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'})
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?
A charge distribution with charge density \rho \neq 0 exist in the half space V:z>0
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 \varphi(x,y,0)=0
We know that the Greens' function G_D(\vec{r},\vec{r'}) = \frac{1}{4 \pi \epsilon_0} \frac{1}{|\vec{r}-\vec{r'}} + f_D(\vec{r},\vec{r'}) i.e. it satisfies the Poisson equation (so essentially we're letting \varphi = G_D(\vec{r},\vec{r'})
Our Dirichlet boundary condition tells us that G_D(\vec{r},(x',y',0))=0
\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}}
this is generalised to give
\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}}
and we note that \nabla^2 f_D(\vec{r},\vec{r'})=0 i.e. it satisfies Laplace's equation and so G still satisfies Poisson's equation.
However it then states
\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]
where \vec{r'}=(x',y',z'), \vec{r'_m}=(x',y',-z') (*)
but the formula given earlier for calculating the potential from Dirichlet boundary conditions was:
\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'})
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?
…