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''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, extendind just barely over the edge in each direction. Gauss law states that:

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

and so $E_{above}^{perp} - E_{below}^{perp} = \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_{above}^{perp} - E_{below}^{perp} = \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.

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2 Answers 2

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.

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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 –  AlexandreH May 2 '12 at 20:28
2  
"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. –  Jonathan Gleason May 2 '12 at 20:51
    
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 =/ –  AlexandreH May 4 '12 at 2:12

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.

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

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