Charge distribution with charge density

In summary, the conversation discusses a worked example involving a charge distribution with charge density in a half space, viewed as a system with an earthed plate at z=0. The Greens' function is used to satisfy the Poisson equation and the Dirichlet boundary condition. The formula for calculating the potential from the Dirichlet boundary condition is given, but it appears that the second term has been neglected in the potential formula at one point. There is a discussion about the second term and its potential effect on the formula.
  • #1
latentcorpse
1,444
0
The following is a worked example in my notes I am 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?
 
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  • #2
latentcorpse said:
… 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]…
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?

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?
 
  • #3


sorry could u perhaps give a bit more explanation please - i can't really see what you mean?
 
  • #4
latentcorpse said:
sorry could u perhaps give a bit more explanation please - i can't really see what you mean?

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 ?
 
  • #5


lol. good point.
 

1. What is charge distribution?

Charge distribution is the arrangement or spread of electric charge within a given space or object. It can be described in terms of charge density, which is the amount of electric charge per unit volume or unit area.

2. What is charge density?

Charge density is the measure of electric charge per unit volume or unit area. It is typically denoted by the symbol ρ and has units of coulombs per meter cubed (C/m3) or coulombs per meter squared (C/m2).

3. How is charge density related to charge distribution?

Charge density and charge distribution are closely related. The charge density at a given point is determined by the charge distribution in the surrounding area. For example, if there is a higher concentration of electric charge in a certain region, the charge density will be greater in that area.

4. What factors can affect charge distribution?

Several factors can influence charge distribution, including the shape and size of an object, the type of material it is made of, and the presence of external electric fields. Different materials have different abilities to distribute and hold charges, which can impact the overall charge distribution in a given space.

5. How is charge distribution with charge density used in scientific research?

Charge distribution and charge density are important concepts in many fields of science, including physics, chemistry, and engineering. They are used to study the behavior of electric fields and the interactions between charged particles. In research settings, they can also be used to design and optimize electronic devices, such as capacitors and transistors.

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