Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Gauss's law says that the flux through a closed surface which contains neither a sink nor a source will be zero.

It's quite clear that all field lines will have to exit somehow, but the strength of the E-field is also proportional to the inverse of the distance squared.

So if, for example, we have a cube, and the E field is perpendicular to one of the sides, the electric flux through that one side will be $A * E$ = $A$ * $kq \over r^2$. But on the opposite side, the distance from the source of the E field will be larger, so the magnitude of the E field should be smaller.

Where is my misconception? Thank you.

EDIT: Okay, the point charge was just an example.

I'll ask it differently:

All the proofs I've seen of this concept state that "all field lines that enter the closed surface must also leave the closed surface, hence the total flux will be zero".
But how does this account for the differences in the distances of the sides of the closed surface from the source of the charge?

Can someone refer me to a proof or give an explanation of why the differences in distances always balance out with the differences in area in order to give you a zero result?

share|improve this question
1  
If the electric field is perpendicular to one of the sides of the cube, the charge configuration can't actually be a point charge, so $E\neq kq/r^2$. –  David Z Feb 2 '12 at 19:18
    
Okay, I've edited the question –  fiftyeight Feb 2 '12 at 22:07
add comment

4 Answers 4

up vote 4 down vote accepted

The OP wants an intuitive answer to an intuitive obstacle to seeing its truth. Well, the intensity of the flux is like how many lines we draw per unit area. No one line « loses strength » so to speak. (There is no dissipation, no friction.) If it is a point source, the lines are not parallel, they diverge, and the greater distance between the lines leads to their lesser density, and so less field strength. But each line keeps its strength...

So all the lines enter one face, and most but not all leave at the parallel, far wall....which shows that the field strength there is a little smaller. But the diverging lines leave the other walls after all...and there, too, their density is less, but there is more area, so it all adds up.

share|improve this answer
add comment

You should take into account the flux through all six faces of the cube. If the field is perpendicular to one face (for example, if it is homogeneous), then the field is not "relative to the inverse of the distance squared", as it is for a field of a point charge.

share|improve this answer
    
Okay, I've edited the question to be more general. –  fiftyeight Feb 2 '12 at 22:08
add comment

This is a simple result of the differential form of Gauss' law $\nabla \cdot \mathbf{E} = \rho$ and the divergence theorem $\iiint_V \nabla\cdot \mathbf{F} dV = \iint_S \mathbf{F}\cdot\mathbf{n} dS$

If there is no charge in the region, then the LHS is zero so the total flux must be zero as well. If you want a proof of the divergence theorem, there is a fairly straightforward one here: http://www.proofwiki.org/wiki/Divergence_Theorem

share|improve this answer
    
This post is true, but doesn't figure out where the misconception was. –  joseph f. johnson Feb 4 '12 at 17:33
add comment

For a point charge :The example that you have given is not real, it is only an approximation. In reality you can never have parallel flux lines. As the electric field lines would always diverge. If you consider a charge and two concentric spheres, then as the distance from the centre increases the electric field(E) reduces but the area(A) increases. Therefore EA remains constant. Only on a very small area on the surface of the sphere you can approximate the area to be flat. And if you want another flat area so that you form a rectangle then you need to draw another concentric sphere so close to the first one that, the deviation that you find in the law is by the amount that you are approximating.Therefore the error is your won choice and not the flaw in the gauss's law.

For a distribution of charges the electric field need not go down with small distance for example if you consider a positively charged infinite plate, where the electric field will not reduce if you go away from the surface.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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