
#1
Aug2806, 05:33 PM

P: 460

I was wondering, how would you prove Gauss's law (either mathematically or intuitively)? I mean, I know that sometimes people take it as the fundamental law (one without proof) but how would you derive such a law from Coulomb's law?
Any help would be extremely appreciated. I have been fussing over this topic for a month now. 



#2
Aug2806, 05:44 PM

P: 1,194




#3
Aug2806, 07:49 PM

P: 460





#4
Aug2806, 08:15 PM

P: 1,194

Proof of Gauss's LawStokes theorem is used to transform the integral forms of maxwell's third equation (Faraday's law) and fourth equation (ampere's law) into their differential forms. 



#5
Aug2906, 08:46 AM

P: 460

So in the end there is no rigorous proof of Gauss' law (the fact that it works for ANY closed surface)? 



#6
Aug2906, 09:00 AM

Sci Advisor
HW Helper
P: 1,930

d\Omega=r.dA/r^3. Then \int qdA.r/r^3=4\pi q. For q outside the closed surface, the sign of the solid angle has to be used. 



#7
Aug2906, 09:32 PM

P: 2,265

[tex] \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} [/tex] is saying the same thing as [tex] \oint_S \mathbf{E} \cdot d\mathbf{A} = \int_V \frac{\rho}{\epsilon_0} dV = \frac{Q}{\epsilon_0} [/tex] one is the differential form, the other is the integral form. 



#8
Aug3006, 04:30 PM

P: 460





#9
Aug3006, 04:38 PM

Sci Advisor
HW Helper
P: 2,004

Here's the idea of the proof in short, you can work out the details yourself:
(I): Show that [itex]\nabla \cdot \vec F =0[/itex] if F has the form: [itex]\hat F=K\hat r/r^2[/itex] ([itex]r\neq 0[/itex]). (II)The divergence theorem holds for simple solid regions, but can be extended to hold for regions that are finite unions of simple solid regions. Any volume region with a 'hole' in it can be viewed as a union of two simple solid regions. (For example: A solid sphere with a cavity in it, the outer surface is the surface of the sphere and the inner one is the surace of the cavity.) (III) Show that the flux through such a surface is zero (if the inner surface surrounds the origin). This means the flux through both surfaces is equal. Since the surfaces were arbitrary, it means the flux through any connected closed surface that surrounds the origin has the same value. It thus suffices to calculate the flux through a sphere. 



#10
Aug3006, 04:41 PM

Sci Advisor
HW Helper
PF Gold
P: 4,768

[tex] \nabla \cdot \frac{\vec{r}\vec{r}'}{\vec{r}\vec{r}'^3} = 4\pi \delta^3(\vec{r}\vec{r}') [/tex] rigorous? Just curious. 



#11
Aug3006, 08:32 PM

P: 2,265

As an illustrative example, let's think about the concept of power and intensity. At the beach, when you feel the sun's radiation on your skin, what you are measuring is how much energy (in a particular band of frequencies) is impinging upon your skin in a particular amount of time. If you're out in the sun longer, more energy falls on your skin (this is the concept of power: how much energy per unit time) and, in the same intensity of sunlight, the bigger you are, the more of that sunlight you will scoop up: more area exposed to the sun, more power that falls on that area. So, whether it is light or sound waves (or something else), intensity is amount of power that falls on a unit area that is positioned perpendicular to the oncoming wave. it's watts/m^{2}. double the area, you get twice as much power. Double the intensity of radiation, you also get twice as much power. Okay, consider a [itex]P_0[/itex] watt light bulb and assume we're measuring all of the radiation (infared, too) at a distance of [itex]r[/itex] meters. Imagine a sphere of radius of [itex]r[/itex] surrounding and centered at the light bulb. Conservation of energy says that the entire [itex]P_0[/itex] watts is distributed over the entire surface area of that sphere , which is [itex]4 \pi r^2[/itex] and the argument of symmetry says that every one of those m^{2} gets the same amount of power. And every little snippet of area is directly facing the light bulb and is perpendicular to the wave front. So, at that distance the power, [itex]P_A[/itex], falling on an area of [itex]A[/itex] is [tex] P_A = \frac{P_0}{4 \pi r^2} A [/tex] the factor [tex] I = \frac{P_0}{4 \pi r^2} [/tex] is the intensity of radiation and is proportional to the power of the point source, as one would expect. Note also, that with a constant source power, the intensity is also proportional to the reciprocal of the square of the distance from the source. A natural inversesquare behavior. Note also that if one were to integrate the intensity (which is [itex] \frac{P_0}{4 \pi r^2} [/itex] over the entire spherical surface area (which is [itex] 4 \pi r^2 [/itex]) i would be adding up all the little snippets of power falling on the little snippets of area and that would add to the total power [itex] P_0 [/itex] emitted by the point source. Doesn't matter how big the sphere is. If the sphere is twice as big, it has 4 times the surface area, but the intensity is ^{1}/_{4} as large and it comes out to be the same total power. Adding up all of the little snippets of power impinging upon the surface of the sphere equals the power emitted by the source inside and we can thank the Conservation of Energy law for that simple fact. Now here's a little example: Someone, somewhere measured the intensity of radiation from the sun (the power falling on a square meter facing the sun out in a vacuum in Earth orbit) to be: 1360.77 W/m^{2}. We also know that we are 149.6 x 10^{9} meters (about 93 million miles) from the sun. From that we can calculate how big of a light bulb that the sun is: [tex] I = \frac{P_0}{4 \pi r^2} [/tex] [tex] 1360.77 \ \mbox{W/m}^2 = \frac{P_0}{4 \pi (149.6 \times 10^9 \ \mbox{m})^2} [/tex] results in: [tex] P_0 = 1360.77 \ (\mbox{W/m}^2) 4 \pi (149.6 \times 10^9 \ \mbox{m})^2 = 3.827 \times 10^{26} \ \mbox{W} [/tex] Now this is intensity, not electric field, but both share the same mathematics. Both are proportional to the source agent (power of point source vs. electric charge of point source) and both are a inversesquare relationship with distance. Now, instead of calling this quantity "intensity" which is power per unit area, for electric charges this natural inversesquare quantity is called "flux density" or "electrostatic flux density", and if we were talking about this other inversesquare phenomenon called gravity, it would be "gravitational flux density". These are those field lines that they like to draw and you don't like. But it doesn't matter, if experimental evidence shows that the electrostatic force follows the inversesquare law precisely (and it does, experiment confirms the exponent of [itex]r^{n}[/itex] to be be [itex]1.99999999 < n < 2.00000001[/itex], then whether or not you like it, that density of lines of field model is valid, even if it comes from our imagination. If the E field is inversesquare, we can imagine a point source radially emitting a bunch of field lines (a.k.a. "total flux"), the quantity of which is proportional to the amount of charge, and the density of those field lines crossing a perpendicular surfaces (a.k.a. "flux density") is proportional to the number emitted and inversely proportional to the square of the distance. Now we're moving toward proving: [tex] \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\epsilon_0} [/tex] but I want you to get used to these first concepts before moving on. If you don't like any of the above, I ain't gonna waste time continuing. Another installment to follow. 



#12
Aug3106, 04:13 AM

Sci Advisor
HW Helper
P: 1,930





#13
Sep106, 11:42 AM

P: 460





#15
Sep106, 01:06 PM

P: 1,194





#16
Sep206, 11:06 PM

P: 460

I did the calculation for [tex] \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}[/tex] and I get [tex]\frac{1}{x^2+y^2+z^2}[/tex].
Also, why should the divergence be zero, don't we have a radial field? 



#17
Sep306, 03:28 AM

Sci Advisor
HW Helper
P: 2,004

Well, that's one of the special properties of inverse square field. It may look like it has a divergence, but it doesn't and it's the underlying reason for Gauss' law.
You could do it in cartesian coordinates: [tex]\vec F(x,y,z)=K \frac{\hat r}{r^2}=K \frac{\vec r}{r^3}=K \frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2+y^2+z^2)^{3/2}}[/tex] but that's really a bit of a hassle. Try it in spherical coordinates, whhich makes it trivial. 



#18
Sep306, 09:24 AM

P: 1,194

The divergence of a point in space is the NET outward flux through an arbitrarily small volume, per unit volume, enclosing this point, where the small volume approaches zero. BY DEFINITION, a radial field HAS NO DIVERGENCE except at r = 0. All flux going into all arbitrarily small volumes must go out of the small volume, causing a net outward flux of zero, or, a divergence of zero at all points. However, if you determine the divergence at the point where there is charge, you are at the point where flux lines EMANATE, so through an arbitrarily small volume enclosing this point there IS a net outward flux. Gauss's law in differential form states that the divergence of the electric flux density is equal to the volume charge density at that point. We can apply the integral form of Gauss's law and find the net outward flux through a surface enclosing a charge distribution, and then take the limit of the outward flux as the volume approaches zero. However, you might notice that this limit is zero. In fact, this limit is zero for all charge distributions and all fields, which is quite obvious. This is where divergence of a point comes in. It is, as I stated above, the net outward flux of an arbitrarily small volume, PER UNIT VOLUME, enclosing a point in a vector field. IF we find the limit of the integral form of gauss's law to a volume where the size of that volume goes to zero about a point, the result is zero. However, if we find the divergence, we get the limit of the charge per unit volume, at that point, which is a finite number. The derivation for Gauss's law, in its most simple form, is a calculation of the net outward flux through a closed surface, where that closed surface encloses a point charge, and this calculation is performed using a closed surface with radial symmetry, and the center of this surface is centered at the point charge. Due to symmetry, E is a constant and da is a constant, and both vectors E and da are at an angle of 0 degrees, so the flux simply becomes Q/ep_0. So, we have proved that for a point charge with a spherical closed surface enclosing it, the net outward flux of the Efield is Q/ep_0. We can rewrite this by finding the net outward flux of the Dfield for this configuration, and we simply get Q. The issue, as you stated, is applying this to all charge distributions and all volumes, ans gauss's law says we can do this. Well, I suppose you could just say that no matter what surface you have enclosing the charge, your net ourward flux will be the same, and therefore enclosed charge will be the same. But this is where I bring in intuition.... Gauss's law is just something that is second nature to me I suppose. 


Register to reply 
Related Discussions  
Gauss's law in 2D  Calculus & Beyond Homework  6  
Gauss's Law  Advanced Physics Homework  5  
Gauss's Law  Classical Physics  9  
Gauss's Law  Introductory Physics Homework  1  
Gauss's law proof  Classical Physics  54 