Proving Gauss' Law for Any Topologically Closed Surface

In summary, the Gauss' law states that the total flux through a closed surface is only dependent on the charge inside the surface, regardless of the shape of the surface. This is due to the 1/r2 property of Coulomb's law, where the increase in area with r2 balances out the weakening force. The proof of this law can be shown using the divergence theorem and the fact that the divergence is zero in regions without charges. This means that even for oddly shaped surfaces, the flux through it will still be zero if it encloses no charges.
  • #1
zezima1
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The essence of Gauss' law is that the total flux through a closed surface only depends on the charge inside the surface. So two spheres with different radii will have the same flux. This is of course due to the 1/r2 property of coulombs law. Because, since the area increases with r2 this precisely makes up for the force getting weaker proportional with 1/r2.

But in my book, I can read that Gauss' law holds for all kinds of surfaces. How do you show that, i.e. that any topologically closed surface has the same property as the surface of the sphere? Or is it actually something very mathematical, which physicists tend to be less rigourous about?
 
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  • #2
hi zezima1! :smile:

hint: what is div of [itex]\boldsymbol{\hat{r}}[/itex]/r2 ? :wink:

(ie (x,y,z)/r3)
 
  • #3
Zero everywhere expect at the origin, where it is best described by the dirac delta. But I have never really understood the physical meaning behind that, and I don't understand how the divergence comes into the picture (just started electrodynamics :) )
so.. explain please :)
 
  • #4
we can use the divergence theorem

for any closed region that does not include charges (where div is a delta),

div = 0, so the total flux through the boundary is zero​

in particular, we can replace a sphere (as in your original question) by any surface without changing the total flux :wink:

(and physically, this means that the field is like a fluid, it can change shape, but the amount entering a region always equals the amount leaving it)
 
  • #5
zezima1 said:
The essence of Gauss' law is that the total flux through a closed surface only depends on the charge inside the surface. So two spheres with different radii will have the same flux. This is of course due to the 1/r2 property of coulombs law. Because, since the area increases with r2 this precisely makes up for the force getting weaker proportional with 1/r2.

But in my book, I can read that Gauss' law holds for all kinds of surfaces. How do you show that, i.e. that any topologically closed surface has the same property as the surface of the sphere? Or is it actually something very mathematical, which physicists tend to be less rigourous about?

The same question was in my mind when i was in high school and first read 'guass law for electrostatics'. And i did not have any knowledge of vector calculus. In high school we only study calculus of one variable.

Though now I have a better mathematical understanding of guass law. But i did understood guass law at that time only. The concept is quite simple. That is a charge always creates electric line of forces. Whether you keep the charge in a rectangular box or sphere. The line of forrces coming out of the charge would be same. So the the number of line of forces coming out of any irregular/regular surface taken anywhere around that charge is constant. That is what gauss law is integral_E.dA=constant
 
  • #6
Okay Tiny Tim, I think you are right that the divergence theorem can be used to make it more clear but you have to explain more. In your example you consider a closed surface with no charge. But I'm asking for the proof of Gauss' law with Q charge inside the closed surface.
 
  • #7
zezima1 said:
Okay Tiny Tim, I think you are right that the divergence theorem can be used to make it more clear but you have to explain more. In your example you consider a closed surface with no charge. But I'm asking for the proof of Gauss' law with Q charge inside the closed surface.

eugh! :yuck: … that would involve that pesky dirac delta :frown:

isn't it easier to enclose all the charges in little spheres,

and then use coulomb's law to find the flux through the spheres (as in your first post), and the divergence theorem to extend that to general surfaces? :smile:
 
  • #8
okay I think I understand what you meant better now :)

However I'm still not completely sure if I understand it. You say: suppose we have some number of charges within a given, oddly shaped surface. We can always enclose the charges with smaller spheres through which we know, that the flux will obey Gauss' law. But I don't see how you from that and the divergence can proove that the divergence is the same through our oddly shaped surface.

The flux is defined as a dot product, and since a sphere is the only surface which has only radially outpointing area normals it is very easy to show Gauss' law this. But for a given surface you will have area normals pointing in all kinds of direction, such that the flux through some of the tiny areas will be less than the case where they are not tilted with respect to the field lines. So you have to somehow convince me that even though the flux through these infinitesimal areas will be less, there will be more of them such that the two effects exactly cancel (after all a sphere encloses a volume with the least possible surface area used). I am pretty sure, that there must be something that proves this with mathematics rigorously, probably in topology.
 
  • #9
zezima1 said:
… I don't see how you from that and the divergence can proove that the divergence is the same through our oddly shaped surface.

from the divergence theorem:

the flux through a closed surface equals the integral of the divergence over the enclosed volume

since we know the divergence is zero everywhere that doesn't include a charge (from coulomb's law for one charge, and lots of zero is still zero :wink:),

and "our oddly shaped surface" includes no charge, the flux through it must be zero :smile:
 

What is Gauss' law for any surface?

Gauss' law for any surface, also known as Gauss' law in integral form, is a fundamental law of electromagnetism that relates the electric field passing through a closed surface to the total charge enclosed by that surface. It states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space.

What is the significance of Gauss' law for any surface?

Gauss' law for any surface is significant because it allows us to calculate the electric field at any point in space by knowing the distribution of charge in that space. It is also a fundamental principle in the study of electromagnetism and has many practical applications, such as in the design of electrical circuits and devices.

How is Gauss' law for any surface used to calculate the electric field?

Gauss' law for any surface can be used to calculate the electric field at a point by integrating the electric flux over a closed surface surrounding that point. This is done by choosing a suitable Gaussian surface, which is a hypothetical surface that encloses the point of interest and simplifies the calculation of the electric flux.

What are the conditions for Gauss' law for any surface to hold true?

Gauss' law for any surface holds true under two conditions: first, the electric field must be constant and perpendicular to the surface at every point on the surface, and second, the charge enclosed by the surface must be static (not changing with time). If these conditions are met, then the electric flux through any closed surface will be equal to the total charge enclosed by that surface divided by the permittivity of free space.

Does Gauss' law for any surface apply to all types of surfaces?

Yes, Gauss' law for any surface applies to all types of surfaces, including curved surfaces, as long as the two conditions mentioned above are met. It is a general law that can be applied to any closed surface, regardless of its shape or orientation.

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