Why do we ignore the contribution to a surface integral from the point r=0?

[Mike400]

I know that electric field of an infinite sheet of charge in $x$-$y$ plane is $\vec{E}=\frac{\sigma}{2 \epsilon_0} (\hat{k})$. From this fact, how can we conclude that electric field very close to the surface point of a $\sigma$ distribution is approximately $\vec{E}=\frac{\sigma}{2 \epsilon_0} (\hat{n})$ where $\hat{n}$ is normal to surface point.

I think I have got the answer. Please check its validity.

Consider a $\sigma$ distribution and a point $P$ infinitely close to it. Then:

$\displaystyle\mathbf{H}(P)_1=\int_{S'} \dfrac{\sigma}{r^2}(\hat{\mathbf{r}})\ dS'$

Now increase the linear dimensions of the system infinitely. Then:

$\displaystyle\mathbf{H}(P)_{\infty}=\lim\limits_{n \to \infty} \int_{S'} \dfrac{\sigma}{(nr)^2} \left( \dfrac{n\mathbf{r}}{nr} \right) n^2 dS' =\int_{S'} \dfrac{\sigma}{r^2}(\hat{\mathbf{r}})\ dS'=\mathbf{H}(P)_1$

Now after the increment, $\sigma$ distribution is an infinite sheet near point $P$. Point $P$ is also at a finite distance from $\sigma$ distribution (infinite sheet). Therefore:

$\mathbf{H}(P)_1=\mathbf{H}(P)_{\infty}=\dfrac{\sigma}{2 \epsilon_0} (\hat{\mathbf{n}})$

 

[Charles Link]

I think I have got the answer. Please check its validity.

Consider a σσ distribution and a point PP infinitely close to it. Then:

H(P)1=∫S′σr2(^r) dS′H(P)1=∫S′σr2(r^) dS′

Now increase the linear dimensions of the system infinitely. Then:

$\displaystyle\mathbf{H}(P)_{\infty}=\lim\limits_{n \to \infty} \int_{S'} \dfrac{\sigma}{(nr)^2} \left( \dfrac{n\mathbf{r}}{nr} \right) n^2 dS' =\int_{S'} \dfrac{\sigma}{r^2}(\hat{\mathbf{r}})\ dS'=\mathbf{H}(P)_1$

Now after the increment, σσ distribution is an infinite sheet near point PP. Point PP is also at a finite distance from σσ distribution (infinite sheet). Therefore:

H(P)1=H(P)∞=σ2ϵ0(^n)H(P)1=H(P)∞=σ2ϵ0(n^)

The equation actually reads $\vec{H}(\vec{r})=\int \frac{\sigma_m (\vec{r}')}{4 \pi \mu_o r^3} (\vec{r}-\vec{r}') dS'$ where $r=|\vec{r}-\vec{r}'|$.
I'm defining $\vec{M}$ so that $\vec{B}=\mu_o \vec{H}+\vec{M}$. $\\$ $\vec{H}=\frac{\sigma_m}{2 \mu_o}$.

 

[Mike400]

Now is there a similar way to show that $\displaystyle\psi=\int_{S'} \dfrac{\sigma}{r} dS'$ does not blow up near the surface?

 

[Charles Link]

The easiest way I know of is that $\psi=\pm \int H \cdot dr$. At $r=0$, the $H$ field remains finite. Like $E$ it is conservative, (when there are no currents in conductors) and $\nabla \times H=0$. Thereby, the equation $\psi=\int H \cdot dr$ can be used.

 

[Mike400]

The easiest way I know of is that $\psi=\pm \int H \cdot dr$. At $r=0$, the $H$ field remains finite. Like $E$ it is conservative, (when there are no currents in conductors) and $\nabla \times H=0$. Thereby, the equation $\psi=\int H \cdot dr$ can be used.

That is what I thought too.

 

[Mike400]

Is there a way to show:

$(1)\displaystyle\int_{S'} \dfrac{\sigma}{r} dS'$ is continuous all over space

$(2)\displaystyle\int_{S'} \dfrac{\sigma}{r^2}\ \hat{\mathbf{r}}\ dS'$ is continuous everywhere except at $S'$

 

[Charles Link]

Is there a way to show:

$(1)\displaystyle\int_{S'} \dfrac{\sigma}{r} dS'$ is continuous all over space

$(2)\displaystyle\int_{S'} \dfrac{\sigma}{r^2}\ \hat{\mathbf{r}}\ dS'$ is continuous everywhere except at $S'$

By Gauss' law, $E_{1 \, ||}-E_{2 \, ||}=\frac{\sigma}{\epsilon_o}$. and similarly for $H$. (parallel to $\hat{n}$). Meanwhile, the perpendicular $E_1$ and $E_2$ are equal. (parallel to the surface).

 

[Mike400]

By Gauss' law, $E_{1 \, ||}-E_{2 \, ||}=\frac{\sigma}{\epsilon_o}$. and similarly for $H$. (parallel to $\hat{n}$). Meanwhile, the perpendicular $E_1$ and $E_2$ are equal. (parallel to the surface).

How shall we show field is continuous at other points?