# What will the electric field be at the surface?

• I

## Main Question or Discussion Point

The electric field due to a dipole distribution in volume $V'$ can be viewed as electric field due to a volume charge distribution in $V'$ plus electric field due to a surface charge distribution in boundary of $V'$.

$\displaystyle\mathbf{E}=\int_{V'} \dfrac{\rho (\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^3} dV' + \oint_{S'} \dfrac{\sigma (\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^3} dS'=\mathbf{E}^{V}+\mathbf{E}^{S}$

$\mathbf{E}^{S}$ is discontinuous at the surface $S'$.

(1) Does this mean that $\mathbf{E}^{S}$ is undefined at the surface $S'$? (Consequently $\mathbf{E}$ will be undefined at the surface $S'$)

(2) If no, then what is $\mathbf{E}^{S}$ at the surface $S'$?

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Henryk
Gold Member
There is something wrong with your equation. The LHS, that is the electric field is a vector, while the RHS, is the integrated volume and surface charge densities, both scalar quantities.
However, the question you are asking:
Does this mean that ESES\mathbf{E}^{S} is undefined at the surface S′
is a valid one.
The answer to this is, there is no such thing as surface charge density !!.
The true surface charge, that is a finite amount of charge distributed over a zero-thickness surface would require an infinite charge density at the surface and that is physically impossible.
The concept of a 'surface charge density' is a mathematical approximation dealing with situation that the with of a charge layer is much smaller than the dimension of the system in consideration.
Take any metal, for example. Any 'reasonable' external field is screened within a fraction of an atomic layer. If a dimension of that metal surface are even as small as 1 um (0.001 mm), this is still much larger than the screening length (approximately 0.0000001 mm).
This approximation does lead to the electric field at the surface not being defined. That is, the limit at one direction being different that that at the other. However, it doesn't affect the values of electric field at some distance from the surface.
In those rare cases when you need to consider the electric field distribution on atomic scale, the surface charge density isn't a valid approximation anyway.

There is something wrong with your equation. The LHS, that is the electric field is a vector, while the RHS, is the integrated volume and surface charge densities, both scalar quantities.
$\dfrac{\rho (\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^3}=\rho\ \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}$

there is no such thing as surface charge density
Then in the equation

$\displaystyle\mathbf{E}=\int_{V'} \rho\ \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV' + \oint_{S'} \sigma\ \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dS'=\mathbf{E}^{V}+\mathbf{E}^{S} \tag1$

what does the second term imply?

The true surface charge, that is a finite amount of charge distributed over a zero-thickness surface would require an infinite charge density at the surface and that is physically impossible. The concept of a 'surface charge density' is a mathematical approximation dealing with situation that the width of a charge layer is much smaller than the dimension of the system in consideration.
So does the second term in equation (1) refer to a volume charge at surface $S'$ whose width is much smaller than the dimension of the system in consideration (i.e. $V'$) "?

PeroK
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Gold Member
2018 Award
So does the second term in equation (1) refer to a volume charge at surface $S'$ whose width is much smaller than the dimension of the system in consideration (i.e. $V'$) "?
This is the same issue as in your previous thread. These equations involve a model of surface charges densities (2D) and volume charge densities (3D).

Neither of these are an exact physical model. They are both approximations that allow calculus to be used.

If you want to analyse either of these more closely, then you at need to move to a model where the charges are discrete. This may be useful to explain polarisation, say, in the first place. But, when you zoom out to the macroscopic scale, there are so many tiny charges and they are so close together, that you can model the situation as though the charges were spread continuously across a volume and/or a surface.

If you have studied classical mechanics, you will have seen the same idea of studying a body as a 1D string or wire, a 2D surface or plate, or a 3D volume of continuous mass distribution. Despite the fact that macroscopic objects are a collection of atoms, that are themselves a collection of more fundamental particles.

This is the same issue as in your previous thread. These equations involve a model of surface charges densities (2D) and volume charge densities (3D).

Neither of these are an exact physical model. They are both approximations that allow calculus to be used.

If you want to analyse either of these more closely, then you at need to move to a model where the charges are discrete. This may be useful to explain polarisation, say, in the first place. But, when you zoom out to the macroscopic scale, there are so many tiny charges and they are so close together, that you can model the situation as though the charges were spread continuously across a volume and/or a surface.

If you have studied classical mechanics, you will have seen the same idea of studying a body as a 1D string or wire, a 2D surface or plate, or a 3D volume of continuous mass distribution. Despite the fact that macroscopic objects are a collection of atoms, that are themselves a collection of more fundamental particles.
I understand and do agree that surface and volume charge densities are in no way an exact physical model. However in classical electrostatics, as approximations of the exact physical model, we use surface and volume charge densities so that we can use calculus.

So in classical electrostatics, by using the methods of calculus, I get equation (1).

So my question is, if we use classical electrostatics and calculus, shall we have to accept that $\mathbf{E}^S$ is undefined at the surface $S′$ and consequently $\mathbf{E}$ is undefined at the surface $S′$?

PeroK
Homework Helper
Gold Member
2018 Award
I understand and do agree that surface and volume charge densities are in no way an exact physical model. However in classical electrostatics, as approximations of the exact physical model, we use surface and volume charge densities so that we can use calculus.

So in classical electrostatics, by using the methods of calculus, I get equation (1).

So my question is, if we use classical electrostatics and calculus, shall we have to accept that $\mathbf{E}^S$ is undefined at the surface $S′$ and consequently $\mathbf{E}$ is undefined at the surface $S′$?
Not necessarily. You could define the field at the surface to be what is most appropriate. For example, the average of the fields on either side.

If you look at the force on the surface itself then this would determine a suitable value for the field at the surface.

• Mike400
The average of the field on either side is zero $\left[ \dfrac{\sigma}{2\epsilon_0} (\mathbf{\hat{n}})+\dfrac{\sigma}{2\epsilon_0} (-\mathbf{\hat{n}})=0 \right]$. So can we conclude $\mathbf{E}^S=0$ at surface?

PeroK
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2018 Award
The average of the field on either side is zero $\left[ \dfrac{\sigma}{2\epsilon_0} (\mathbf{\hat{n}})+\dfrac{\sigma}{2\epsilon_0} (-\mathbf{\hat{n}})=0 \right]$. So can we conclude $\mathbf{E}^S=0$ at surface?
The field on the surface will depend on the charges elesewhere. It won't be normally be zero.

The field just outside a point on the the surface is $\vec{E_+} = \vec{E} + \frac{\sigma}{\epsilon_0} \hat{n}$ and the field just below the surface is $\vec{E-} = \vec{E} - \frac{\sigma}{\epsilon_0} \hat{n}$, where $\vec{E}$ is the field on the surface and $\vec{E_{\pm}}$ are the discontinuous field values.

Effectively, therefore, the field on the surface is $\vec{E} = \frac12(\vec{E_+} + \vec{E_-})$

• vanhees71