7
$\begingroup$

''electric field always undergoes a discontinuity when you cross a surface charge $\sigma$'' GRIFFITHS

In the derivation; Suppose we draw a wafer-thin Gaussian Pillbox, extended just barely over the edge in each direction. Gauss law states that:

$$\int_{S} E \cdot A = Q_\text{enc}/ \epsilon$$

and so $$E_{\perp\;\text{above}} - E_{\perp\;\text{below} } = \sigma/ \epsilon . $$

My question is why not $2A$? $$\int_{S} E \cdot A = 2EA$$ because the top area of pillbox and the bottom area of pillbox, just as because the 2 parts of the flux...

SO.. WHY NOT : $$E_{\perp\;\text{above}}- E_{\perp\;\text{below} } = \sigma/ 2\epsilon ~? $$

And why there is tangencial component of electric field; not just perpendicular to the surface, which can be seen as flat just looking very close to the surface.

Improve this question
$\endgroup$

5 Answers 5

Reset to default
6
$\begingroup$

Immediately above the surface, write $$ \mathbf{E}=E_{\text{above}}\widehat{\mathbf{n}}+\mathbf{E}_\text{tangential,above} $$ and immediately below the surface, write $$ \mathbf{E}=E_{\text{below}}\widehat{\mathbf{n}}+\mathbf{E}_{\text{tangential,below}}, $$ where $\widehat{\mathbf{n}}$ is the unit surface normal pointing in the above direction, and both tangential components, by definitition, are orthogonal to $\widehat{\mathbf{n}}$. In the limit where the "pillbox" gets arbitrarily small, we can treat $E_{\text{above}}$ and $E_{\text{below}}$ as approximately constant in the pillbox, so that the flux is given by $$ A(E_{\text{above}}-E_{\text{below}})+O(h), $$ where $h$ is the height of the pillbox (this is Big $O$ notation). The minus sign arises because the unit normal to the surface below is $-\widehat{\mathbf{n}}$, not just $\widehat{\mathbf{n}}$. By Gauss's Law, this is equal to $\frac{A\sigma}{\varepsilon _0}$, and hence, in the limit $$ \boxed{E_{\text{above}}-E_{\text{below}}=\frac{\sigma}{\varepsilon _0}}. $$ Obviously, techincally speaking more careful limiting arguments are needed, but hopefully it is clear that this argument can be made completely precise. Let me know if you have any more questions.

To be fair, this is essentially exactly what Ron was getting at, but hopefully the increased detail made this more clear.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Nice derivation. I understood.Now, just 1 question: why there exist the tangencial component? remembering on flat capaitors... there are just perpendicular.. same any flat surface.. So if i can look to the surface very very close i can regard the surface as flat surface so there isnt tangencial component. did you understand my question? @JonathanGleason thanks $\endgroup$
    – AlexandreH
    Commented May 2, 2012 at 20:28
  • 2
    $\begingroup$ "Why there exist the tangencial component?" -- Well, why not? Unless I know that the tangential component is 0, I can't just assume it is, even though that wouldn't affect this result. I am pretty sure Griffiths does show that the tangential component has to be continuous, i.e., in the notation above, $\mathbf{E}_{\text{tangential,above}}=\mathbf{E}_{\text{\tangential,below}}$. Thus, if the surface is a conductor (which by definition means the electric field inside it must be $0$), we do require that the tangential component is $0$. This is not necessarily true in general however. $\endgroup$
    – Jonathan Gleason
    Commented May 2, 2012 at 20:51
  • $\begingroup$ Yes, just a perfect conductor doesnt have eletric field inside. But how about just the surface, the membrane.. without close the surface to form a volume. one surface not flat and not closed. the eletric field doesnt to have be perpendicular in the region where the surface canbe regarded as a flat surface? Well, I think You didnt understand me =/ $\endgroup$
    – AlexandreH
    Commented May 4, 2012 at 2:12
  • $\begingroup$ @AlexandreH For the surface of conductors, you're correct that the tangential component is zero. But the derivation Jonathan gave above is also valid for surfaces which have a charge $\sigma$ but are NOT conductors, in which case there could be a tangential component. $\endgroup$
    – Jahan Claes
    Commented Dec 10, 2015 at 5:48
2
$\begingroup$

Because one value of E, the inner one, multiplies the inner A, and the outer E multiplies the outer A, and you then subtract the two.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Yes, i see. But and the tangencial component?? $\endgroup$
    – AlexandreH
    Commented May 1, 2012 at 23:41
  • $\begingroup$ The box can be made arbitrarily flat, so the tangential component contributes no flux. $\endgroup$
    – Mark Eichenlaub
    Commented May 2, 2012 at 0:56
  • $\begingroup$ Not about the tangencial flux.. But about why? Wht there exisist tangencial component? if the surface can be seen as a flat.. $\endgroup$
    – AlexandreH
    Commented May 2, 2012 at 1:53
  • $\begingroup$ Any tangential component is not discontinuous across the charged surface. $\endgroup$
    – Ron Maimon
    Commented May 2, 2012 at 2:05
  • $\begingroup$ @user8479: The lack of tangential component would mean the surface is equipotential. There is no such indication in what you describe. $\endgroup$
    – Siyuan Ren
    Commented May 2, 2012 at 12:07
2
$\begingroup$

The tangential component due to the locally flat piece of surface, is indeed zero, at the surface. But, the total electric field is the field due to the locally flat patch, plus the rest of the surface. Hence, the total tangential component need not be zero.

Improve this answer
$\endgroup$
0
$\begingroup$

there will be no tangential component of E^ in the case of the charged surface in electrostatics. If so then this would result in a force on the free electrons which would cause a drift hence a current which is not desired in electrostatics.

Improve this answer
$\endgroup$
2
  • $\begingroup$ This is incorrect. The continuity of the tangential component of the electric field follows from the boundary-condition version of Faraday's law (or equivalently $\nabla\times\mathbf E=0$ in eletrostatics), and not from free-electrons considerations. $\endgroup$
    – Emilio Pisanty
    Commented Sep 30, 2015 at 14:01
  • $\begingroup$ @EmilioPisanty I don't understand why this is incorrect. Wouldn't the free-electron argument indeed show that the tangential component of the electric field must be zero? I am not saying that this argument shows why the tangential component is continuous in general because it does not do that. $\endgroup$
    – JosephSanders
    Commented Sep 16, 2022 at 21:02
0
$\begingroup$

Maybe this will help:

http://physweb.bgu.ac.il/ARCHIVE/Courses/2011B/Electro1/extra/Discontinuity_in_an_Electric_Field.pdf

I think the issue is the way you define: $$E_{above}; E_{below}$$

Take a look at the link and note that, for example: $$E_{above} = \vec{E}_2\cdot\hat{n}$$

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.