Deriving Coulomb's Law from Gauss' Law

In summary, the author tries to solve an electric field equation using Maxwell's equations, but winds up getting incorrect results.
  • #1
Ralru
2
0

Homework Statement



When I try working out the example below from PlanetPhysics, I wind up with 2PI rather than 4PI in my answer. Should I be considering my result valid for only a hemisphere and double it for a sphere--or am I just making a mistake in my math?

"As an example of the statement that Maxwell's equations completely define electromagnetic phenomena, it will be shown that Coulomb's law may be derived from Gauss' Law for electrostatics. Consider a point charge. We can obtain an expression for the Electric Field surrounding the charge. We surround the charge with a "virtual" sphere of radius R, then use Gauss' law in integral form."

"We rewrite this as a volume integral in spherical polar coordinates over the "virtual" sphere mentioned above, which has the point charge at its centre. Since the electric field is spherically symmetric (by assumption) the electric field is constant over this volume" (http://planetphysics.org/encyclopedia/DerivationOfCoulombsLawFromGaussLaw.html).




Homework Equations


[URL]http://www.fortunesofwar.com/C1.bmp[/URL]
[URL]http://www.fortunesofwar.com/C2.bmp[/URL]
[URL]http://www.fortunesofwar.com/C3.bmp[/URL]



The Attempt at a Solution


I first integrate Er sin theta dtheta between the limits of 0 and PI which gives Er(-cos theta) =2Er
Then I integrate 2Er dphi between the limits of 0 and 2PI giving 4'PI'Er
And then I integrate 4'PI'Er dr between the limits of 0 and R, giving 1/2 4'PI'Er^2 evaluated between 0 and R, giving 2'PI'ER^2.

I must be missing something here (it's been a while), so I'd appreciate knowing where I am making my mistake.

Thanks
 
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  • #2
You're not making a math mistake. The example is wrong. The electric field isn't constant over the volume, and the integral shouldn't be over the volume of the sphere. The integral should be over the surface area, where the electric field is of constant magnitude.
 
  • #3
Thank you very much. It's nice to know that I have not gone completely insane yet. It's been quite a while since I played with this stuff--and after a long hiatus, one begins to doubt his/her own ability. Thanks again.
 

1. How are Coulomb's Law and Gauss' Law related?

Coulomb's Law and Gauss' Law are both fundamental laws of electromagnetism. While Coulomb's Law describes the force between two charged particles, Gauss' Law relates the electric field to the distribution of charges. This means that Coulomb's Law can be derived from Gauss' Law by mathematically manipulating the equations.

2. What is the mathematical expression for Gauss' Law?

The mathematical expression for Gauss' Law is ∮E · dA = Qenc0, where ∮E is the electric field flux, dA is the area element, Qenc is the enclosed charge, and ε0 is the permittivity of free space.

3. Can Coulomb's Law be derived from Gauss' Law for any charge distribution?

Yes, Coulomb's Law can be derived from Gauss' Law for any charge distribution. This is because Gauss' Law is a general law that applies to all charge distributions, while Coulomb's Law is a specific case for point charges.

4. What are the steps for deriving Coulomb's Law from Gauss' Law?

The steps for deriving Coulomb's Law from Gauss' Law are:
1. Choose a Gaussian surface that encloses the point charge.
2. Calculate the electric field flux through the surface using Gauss' Law.
3. Express the enclosed charge in terms of the point charge.
4. Equate the two expressions for the electric field flux and solve for the electric field.
5. The resulting equation will be Coulomb's Law.

5. Why is it useful to derive Coulomb's Law from Gauss' Law?

Deriving Coulomb's Law from Gauss' Law allows for a deeper understanding of the relationship between electric fields and charges. It also allows for the application of Gauss' Law in more complex situations where Coulomb's Law may not be directly applicable. Additionally, it provides a more general form of Coulomb's Law that can be used for any charge distribution.

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