Register to reply 
Gauss' law for any surface 
Share this thread: 
#1
Apr2212, 01:26 PM

P: 123

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/r^{2} property of coulombs law. Because, since the area increases with r^{2} this precisely makes up for the force getting weaker proportional with 1/r^{2}.
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? 


#2
Apr2212, 03:11 PM

Sci Advisor
HW Helper
Thanks
P: 26,160

hi zezima1!
hint: what is div of [itex]\boldsymbol{\hat{r}}[/itex]/r^{2} ? (ie (x,y,z)/r^{3}) 


#3
Apr2212, 04:15 PM

P: 123

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
Apr2212, 04:40 PM

Sci Advisor
HW Helper
Thanks
P: 26,160

Gauss' law for any surface
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 zeroin particular, we can replace a sphere (as in your original question) by any surface without changing the total flux (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
Apr2212, 04:42 PM

P: 383

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
Apr2312, 03:02 AM

P: 123

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
Apr2312, 03:34 AM

Sci Advisor
HW Helper
Thanks
P: 26,160

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


#8
Apr2312, 06:32 AM

P: 123

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
Apr2312, 08:16 AM

Sci Advisor
HW Helper
Thanks
P: 26,160

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


Register to reply 
Related Discussions  
Help with gauss, Sheet vs conducting surface  General Physics  18  
Surface integral without using Gauss' theorem  Calculus & Beyond Homework  2  
Gauss's Law Flux through surface  Introductory Physics Homework  4  
Gauss Sphere Surface / U = W / f=ma=qE  Introductory Physics Homework  7  
Gauss' Law (imaginary surface)  Advanced Physics Homework  3 