Gauss' law states that the net outward normal electric flux through a closed surface is equal to $q_{total, inside}/\epsilon_0$. However, I'm a bit confused of why the presence of an external charge doesn't influence the solution. I always hear as a response - "any field lines entering the enclosed surface, will also leave it", which is fine. However, the field is inversely proportional to $r^2$, therefore, it will "exit" with a different magnitude and thus, with a different flux. So how they possibly cancel each other out? And if possible, I really look forward to a simple explanation, without involving too much math. I just want to get the concept. I'm sorry for such a newbie question.

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    $\begingroup$ I think the best intuitive,nonmathematical way to understand it is to note that electric charge is the source of electric flux and if you don't have a flux source in a volume , the net output flux will be zero. (think of a fishing net in a river. Don't you agree that the net outward fluid flux is equal to the input ?) $\endgroup$ – Mo's May 7 '13 at 20:51

In an attempt to be brief: The big thing to remember is that the flux is also proportional to the area (technically, the surface integral of the field over the area). Crudely speaking, the side of the enclosed surface with exiting field lines are further away from the external charge than the side with "entering" field lines, and the surface area increases by $r^2$ (remember the formula for the surface area of a sphere). Gauss's Law is just a mathematically precise way of stating this compensation.

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(Image source: http://www.ux1.eiu.edu/~cfadd/1360/24Gauss/Gauss.html)

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    $\begingroup$ Glad to hear it, and thank you for the compliment. $\endgroup$ – David H May 7 '13 at 21:02
  • $\begingroup$ @DavidH - What if the the Gaussian surface is, say a cube? All its sides have the same area. $\endgroup$ – user37433 Apr 21 '14 at 12:11
  • $\begingroup$ @user37433 I'm not quite sure what you're asking. What are you trying to do with this cubic Gaussian surface?. $\endgroup$ – David H Apr 21 '14 at 12:19
  • $\begingroup$ @DavidH - you said that "the side of the enclosed surface with exiting field lines are further away from the external charge than the side with "entering" field lines, and the surface area increases by $r^2$". But why should it increase? As far as I know, we are free to choose any Gaussian surface. So if I choose a cube, there will be no side with bigger area. $\endgroup$ – user37433 Apr 21 '14 at 12:22

Consider this as an appendix to other answers. I hope that this will you to believe that 'the number of lines' is not only a qualitative description of the flux, but also quantitative.

Actually, with the use of Gauss' law, you can prove that the flux through a surface is proportional to the number of lines crossing it (under certain conditions). Unfortunately, I dont have time to draw neat pictures but hope that imagination will be enough.

Lets start from one charge in external field (so that all field lines start from this charge). Consider a small surface that you want to get the flux through. Lets say it is some disk. Then, you can move the boundary of this disc towards the charge along the fieldlines, so you will get a kind of deformed cone with the charge being the vertex. Now, we cant talk about the field at the charge, so we subtract a smaaaal ball around this charge from the cone. We have now two places where the field lines intersect the cone -- near the charge the lines enter into our cone, and then exit it through our surface (since the lines are tangent to the other parts of boundary by construction). Now, it is clear (from Gauss' law) that the flux is also the same. And, near the charge, it is clear that if draw many enough lines, and distribute them homogeneously around it, as far as in this small area the field is determined mostly by the charge itself, the flux is proportional to the number of lines. So, as we have shown, this flux and the number of lines are just those that pass through our surface.

When we have several charges, we have to divide our surface into pieces such that each piece moves entirely to its own charge. Actually, such a piece always has a charge to go to, but it may have two options -- to go either to a negative or a positive charge. And you have to make the total number of lines going from a charge to be proportional to its value. Surely, there is a numerical error, which goes to zero as you increase the number of lines per unit charge. And, once again, this is funny that this process described above naturally cuts any surface into pieces which can be attributed to charges (or pairs of charges).

  • $\begingroup$ Thanks for the answer. I understand your concept, but I suddenly have another question - why should it be a cone? I mean, why should we have a smaller area on one side and bigger area on another? Is it has something to do with the fact that the field is radial? Why not take a cylinder for example, with the same bases at any intersection? Is it because the body gets bigger and thus we have to compare bigger area? And could you please explain why the $\Delta\phi_1/\Delta\phi_1$ in cthulhu's answer is independent of the cosines? $\endgroup$ – grjj3 May 8 '13 at 19:41
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    $\begingroup$ @grjj3 The key idea is that you move the boundary along the field lines, so the shape is dictated by the field. You get a cone because the field lines are going from a charge. If you consider a charged plane, then you have to modify the proof a bit, but you will get a cylinder. $\endgroup$ – Peter Kravchuk May 8 '13 at 19:52
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    $\begingroup$ @grjj3, as for the cosines: the solid angle $\Delta\Omega$ is fixed, and you have to divide it by cosine to get the area and then to multiply by the same cosine which comes from the projection of the electric field. $\endgroup$ – Peter Kravchuk May 8 '13 at 19:55
  • $\begingroup$ Thanks for the explanation of the cone. What about the cosines: I don't really see in that proof the use of a solid angle, but just the use of geometric similarity. It is like he assumed that cone bases are parallel to each other. How's that possible (without talking about solid angle. I just didn't learn that subject yet although I do understand it roughly.)? Sorry for my newbie questions. The proof looks very simple and nice just like your explanation, but these cosines are the only part that confuses me. $\endgroup$ – grjj3 May 8 '13 at 20:11
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    $\begingroup$ @grjj3 See my comment to that answer. The use is that $\Delta S_1=r_1^2\Delta\Omega/\cos\alpha_1$, up to a sign. $\endgroup$ – Peter Kravchuk May 8 '13 at 20:21

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