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When studying Gauss Law, I learned the following two things: $$\oint \vec E \cdot d \vec A= Q/ \epsilon_0$$

  1. That when using the equation above, the $Q$ has to be the amount charge "inside" the Gauss surface, because the $\vec E \cdot d \vec A$ caused by charges that are "outside" of the Gauss surface cancel out

  2. That $E$ (roughly) means the density of lines(electric lines), $ \vec E \cdot d \vec A$ means the number of lines that are crossing a tiny segment, and that hence if we integrate $ \vec E \cdot d \vec A$, we get $Q/ \epsilon_0$, which is the number of lines. (That, in other words, Gauss Law is just a law stating a simple fact that (density)*(area)=(total number)

With those two things in mind, I looked at how the Gauss Law was used to find the electric field of a line charge. ( http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html ). But then I was confused: knowing that $E$ represents, the density of lines, and that the Gauss Law is just another way of saying that (density)*(area)=(total number), I presumably thought that $\vec E$ should refer only to the electric field caused by the charge "inside" the Gauss surface. (and not the net $\vec E$). However, I found that when using Gauss Law to find the electric field, the net$\vec E$ was considered, not the $\vec E$ caused only by charges inside the Gaussian surface. What is wrong here? Is there something that I am missing?

I found that Electric field and Gauss law this post asks a similar question, but it didn't seem to provide a satisfying answer, so I'm asking it here.

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  • $\begingroup$ $E$ is just the electric field, no more, no less. It is a quantity (a vector) that is defined at every point in space, and Gauss' Law says that its integral over a closed surface is proportional to the charge enclosed by that surface. Every time you see $E$ you should assume that it refers to the total electric field unless otherwise indicated. $\endgroup$ – Javier Aug 31 '17 at 13:40
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Well, trying to face it without calculus is really hard haha. I see you are very confused: there is not ANY "in" or "out" there.

When you calculate $\iint_S \vec{E}\cdot d\vec{S}$, the integral is the sum of that quantity ALONG the surface. So you don't compute any field inside or outside, but in the very surface. What you are doing there is summing the product $E\cdot dA\cdot\cos({\vec{E},\vec{n})}$ (shortened as the scalar product above) all the surface long. In other words, it means "checking that product, skipping to the next point of the surface, summing that product to the previous one, skipping again, and so on".

We could talk about this law for hours, but I think that's your main problem. I hope that helps you to rethink everything again. I'm omitting so many important ideas that might correspond to another question, especially

  • What field flow lines really mean.
  • What the flux actually is
  • If it is actually useful to think about it in terms of flow lines.
  • Whether that flux tells you about the electric field or not.

And so many others. Hope I helped anyways.

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  • $\begingroup$ Thank you! I guess I'll have to learn more math! (In fact, my math professor says we'll learn the double integral thing by the end of the semester(yeah!)) Once again, thank you! $\endgroup$ – Danny Han Sep 7 '17 at 14:50
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This is my own answer to the question above. Although it kind of feels weird to question and answer my own answer, since a co-founder of Stack Exchange says that doing so enables others to check the validity of my answer and/or suggest an even better answer, I will be writing this. (https://stackoverflow.blog/2011/07/01/its-ok-to-ask-and-answer-your-own-questions/)

First off, we will divide $\vec E$ into two types: $\vec E_{in}$ and $\vec E_{out}$, where $\vec E_{in}$ stands for electric field made from charges that are inside the Gaussian surface, and $\vec E_{out}$ stands for electric field made from charges that are outside the Gaussian surface. As a side note, therefore, by the superposition principle, $\vec E_{net} = \vec E_{in} + \vec E_{out}$.

It is NOT WRONG to say that when applying the Gauss Law, $\vec E_{inside}$ can be used to represent the $\vec E$. In other words, $\oint \vec E_{in} \cdot d \vec A = Q/{\epsilon_0}$ is valid. (Note that we have used $\vec E_{in}$). But, this has one problem

  1. Even if we solve the integral above, what we will obtain will be $\vec E_{in}$ not $\vec E_{net}$ (because we didn`t consider $\vec E_{out}$)

So then, how do we find $\vec E_{net}$? We add zeroes. From $$\oint \vec E_{in} \cdot d \vec A = Q_{in}/{\epsilon_0}$$ $$\oint \vec E_{in} \cdot d \vec A + 0 = Q_{in}/{\epsilon_0} + 0$$ $$\oint \vec E_{in} \cdot d \vec A + \oint \vec E_{out} \cdot d \vec A = Q_{in}/{\epsilon_0}+0$$ Therefore, since $\oint \vec E_{in} \cdot d \vec A + \oint \vec E_{out} \cdot d \vec A =\oint \vec E_{net} \cdot d \vec A$, (because $\vec E_{net} = \vec E_{in} + \vec E_{out}$) we can conclude that $$\oint \vec E_{net} \cdot d \vec A = Q_{in}/{\epsilon_0}$$

In the derivation above, we utilized the fact that $\oint \vec E_{out} \cdot d \vec A=0$. But there's something that I must confess. Since I didn't learn multivariable calculus yet, I'm not 100% sure if $\oint \vec E_{in} \cdot d \vec A + \oint \vec E_{out} \cdot d \vec A =\oint \vec E_{net} \cdot d \vec A$ is true. (I just thought that it would be true because of intuition. Please check if it is true)

My answer may be incorrect (I just started learning this). Please tell me if my answer is incorrect, or if there is a better answer.

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  • $\begingroup$ That is correct; integration is, we say, a "linear operation", meaning that $f(x +y)=f(x)+f(y)$. The problem that $E_\text{out}$ is not obtained by this method (because any of its field lines that come in eventually go out by this very definition of not connecting to a charge in the volume), is irreducible. We would say it's a problem of "boundary conditions" of the volume you're working with, or if it's an infinite volume, the boundary conditions "at infinity." There are nontrivial vacuum solutions for the electromagnetic fields that can always be superimposed. $\endgroup$ – CR Drost Aug 31 '17 at 13:37

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