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Coulomb law dedution from Maxwell equations

by tsuwal
Tags: coulomb, dedution, equations, maxwell
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Feb17-13, 10:35 AM
tsuwal's Avatar
P: 103
I'm starting my study in eletromagnetism and I would like to know how do you deduce the eletric field produced by a single particle of charge q placed in the origin.
The magnetic field is constant so by Maxwell equations, the rotacional is 0 and the divergence is constant.

Is this enough to deduce the field? I think I could deduce it but I would need to do it in polar coordinates, however, I don't know how to use rotacional and divergence in polar coordinates..
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Feb17-13, 10:46 AM
P: 11,631
I don't know how to use rotacional and divergence in polar coordinates..
There are formulas for those operations in spherical coordinates.
To avoid singularities, you could distribute the charge in a sphere of radius r. div(E)=ρ (with some prefactor depending on the unit system) and rot(E)=0 will give the correct result. It is not sufficient to fix the field everywhere (you can always add constant fields and electromagnetic waves, for example), but you can require that the field is time-independent and vanishes for large distances, for example.
Feb17-13, 11:44 AM
Sci Advisor
P: 2,354
Your idea is correct. In electrostatic cases, i.e., all charges at rest forever, there are only electric field components and everything is time independent. Then you have [itex]\vec{B}=0[/itex] and the electric Maxwell equations simplify to (using Heaviside-Lorentz units):
[tex]\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{E}=0.[/tex]
Here, the charge density is given by a [itex]\delta[/itex] distribution:
[tex]\rho(\vec{x})=q \delta(\vec{x}).[/tex]

We can expect that the solution is smooth everywhere except at the origin, where the singularity caused by the point charge sits. This is a simply connected region of space, and thus the vanishing curl of the electric field tells you that it has a scalar potential:
[tex]\vec{E}=-\vec{\nabla} \Phi.[/tex]
Plugging this into the first equation leads to
[tex]\Delta \Phi(\vec{x})=-q \delta(\vec{x}).[/tex]
Due to rotational symmetry around the origin, the potential is a function of the distance from the origin only, i.e.,
Now you don't need to know the formulae for the differential operators in shperical coordinates (although it's good to look them up in some textbook), but you can derive, what you need easily in this case.

From the chain rule, you first find, using Cartesian coordinates
[tex]\vec{\nabla} \Phi(r)=\Phi'(r) \vec{\nabla} r=\Phi'(r) \frac{\vec{x}}{r}[/tex]
and then
[tex]\Delta \Phi(r)=\frac{\mathrm{d}}{\mathrm{d} r} \left (\frac{\Phi'(r)}{r} \right ) \vec{\nabla} r \cdot \vec{x}+ \frac{\Phi'(r)}{r} \vec{\nabla} \cdot \vec{x}=r \frac{\mathrm{d}}{\mathrm{d} r} \left (\frac{\Phi'(r)}{r} \right ) +3 \frac{\Phi'(r)}{r}= \frac{1}{r^2} \left (r^2 \Phi'(r) \right)'.[/tex]
For [itex]r>0[/itex] you have
[tex]\frac{1}{r^2} \left (r^2 \Phi' \right)'=0.[/tex]
This you can integrate successively:
[tex]r^2 \Phi'(r)=C_1 \; \Rightarrow \; \Phi(r)=-\frac{C_1}{r}+C_2.[/tex]
The integration constant [itex]C_2[/itex] is physically irrelevant and is usually chosen to be [itex]C_2=0[/itex], so that the potential vanishes at infinity.

The constant [itex]C_1[/itex] must be chosen such that you get the right charge from Gauß's Law. To that end we take
[tex]\vec{E}=-\vec{\nabla} \Phi=-\frac{C_1}{r^2} \frac{\vec{x}}{r}.[/tex]
Now integrating Gauß's Law over a sphere of radius [itex]R[/itex] gives, due to Gauß's integral theorem
[tex]\int_{K_R} \mathrm{d}^3 \vec{x} \rho(\vec{x})=q=\int_{\partial K_R} \mathrm{d}^2 \vec{A} \cdot \vec{E}=-4 \pi C_1 \; \Rightarrow \; C_1=-\frac{q}{4 \pi}.[/tex]
This gives finally Coulomb's Law:
[tex]\Phi(r)=\frac{q}{4 \pi r}, \quad \vec{E}=-\vec{\nabla} \Phi=\frac{q}{4 \pi r^2} \frac{\vec{x}}{r}.[/tex]

Feb22-13, 02:48 PM
tsuwal's Avatar
P: 103
Coulomb law dedution from Maxwell equations

Thanks but i didn't get this step:

∇⃗ Φ(r)=Φ′(r)∇⃗ r=Φ′(r)x⃗ r
and also this one:

ΔΦ(r)=ddr(Φ′(r)r)∇⃗ r⋅x⃗ +Φ′(r)r∇⃗ ⋅x⃗ =rddr(Φ′(r)r)+3Φ′(r)r=1r2(r2Φ′(r))′.
(sorry, when I quote what what I copy appears like this)

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