Proof of Coulomb's Law using Gauss' Law

[jdstokes]

I'm trying to derive the vector field [itex]\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q\vec{r}}{r^3}[/itex] surrounding a point charge, starting with [itex]\oint_S \vec{E} \cdot \mathrm{d}\vec{A}[/itex]. My uneducated guess would be to get the magnitude of the electric field from gauss' law, then integrate to get the scalar potential, before taking the gradient to get the vector field. Is there a more elegant way to achieve this?

Thanks.

James

 

[robphy]

Use the freedom to choose any closed surface that encloses your charge to make your calculation easier. In the case of a point charge, it's easiest to choose a sphere (of radius r) centered at your point charge.

Now, at each point on that sphere, observe that, by symmetry, the Electric Field must be radial and must have the same magnitude. In addition, the area elements [tex]d\vec A [/tex] are radially outward. Use these observations to sequentially simplify your integral.

Since r is arbitrary (as long as your sphere encloses the point charge that it is centered on), your expression works for arbitrary r.

(In a "proof" that involves the potential, you probably need to first prove that it exists. That is, show that the electric field is minus the gradient of a function for this situation. What condition must be imposed on the Electric Field?)

 

[thepatientmental]

Hi James,

Yes, there is a more elegant way to derive Coulomb's Law using Gauss' Law. Let me walk you through the steps:

1. Start with the integral form of Gauss' Law:

\oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \frac{Q_{enc}}{\epsilon_0}

where \vec{E} is the electric field, \mathrm{d}\vec{A} is the differential surface area element, and Q_{enc} is the enclosed charge.

2. Consider a point charge q at the origin, surrounded by a spherical surface with radius r. The enclosed charge in this case is just q.

3. Since the electric field is radially symmetric, we can choose a Gaussian surface in the form of a sphere with radius r. This means that \vec{E} and \mathrm{d}\vec{A} are parallel at every point on the surface, and the dot product becomes \vec{E} \cdot \mathrm{d}\vec{A} = E\mathrm{d}A.

4. Substituting this into the integral form of Gauss' Law, we get:

\oint_S E\mathrm{d}A = \frac{q}{\epsilon_0}

5. The surface area of a sphere is given by A = 4\pi r^2. So, the integral becomes:

E \cdot 4\pi r^2 = \frac{q}{\epsilon_0}

6. Solving for the electric field, we get:

E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

which is the magnitude of the electric field at a distance r from a point charge q.

7. Now, to get the vector form of Coulomb's Law, we simply need to multiply this magnitude by the unit vector in the direction of \vec{r}, which is given by \hat{r} = \frac{\vec{r}}{r}. So, we get:

\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q\vec{r}}{r^3}

which is the desired result.

In conclusion, we have derived Coulomb's Law using Gauss' Law in a simple and elegant way. Hope this helps! Let me know if you have any further questions.