Coulomb law dedution from Maxwell equations

In summary, to deduce the electric field produced by a single particle of charge q placed in the origin, you use the following steps: 1) Use Maxwell's equations to calculate the rotational and divergence field components, 2) Use spherical coordinates to smooth the solution, 3) Use the chain rule to find the potential at a given distance from the origin, and 4) integrate to get Coulomb's Law.
  • #1
tsuwal
105
0
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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  • #2
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.
 
  • #3
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.,
[tex]\Phi(\vec{x})=\Phi(r).[/tex]
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]
 
  • #4
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)
 
  • #5


I can provide a response to your inquiry about deducing the electric field produced by a single charged particle using Coulomb's law and Maxwell's equations. Firstly, Coulomb's law states that the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This law can be derived from Maxwell's equations, which are a set of fundamental equations that describe the behavior of electric and magnetic fields.

To deduce the electric field produced by a single particle of charge q placed at the origin, we can use the Gauss's law, which is one of Maxwell's equations. This law states that the electric flux through a closed surface is equal to the charge enclosed by that surface. In this case, we can choose a spherical surface centered at the origin and calculate the electric flux through it. Since the electric field is radial, the electric flux will also be radial and have the same magnitude at every point on the surface. Therefore, we can use the Gauss's law to deduce the electric field at the origin, which is given by q/4πε0r^2, where ε0 is the permittivity of free space and r is the distance from the particle to the point of interest.

Furthermore, to deduce the electric field in polar coordinates, we can use the gradient operator, which is a mathematical tool used to calculate the rate of change of a function in different directions. In polar coordinates, the gradient operator takes the form of 1/r ∂/∂r (rE) + 1/r sin θ ∂/∂θ (E sin θ) + ∂E/∂φ, where E is the electric field and θ and φ are the polar and azimuthal angles, respectively. By using this operator, we can calculate the electric field in polar coordinates and verify that it matches with the result obtained from Coulomb's law.

In conclusion, Coulomb's law can be deduced from Maxwell's equations, and it is sufficient to deduce the electric field produced by a single charged particle at the origin. To do so, we can use Gauss's law and the gradient operator in polar coordinates. I hope this explanation helps in your study of electromagnetism.
 

1. What is Coulomb's law and how is it related to Maxwell's equations?

Coulomb's law is a fundamental law of electromagnetism that describes the force between two electrically charged particles. It states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This law was derived by Charles-Augustin de Coulomb in the 18th century and is one of the four Maxwell's equations, which describe the behavior of electric and magnetic fields.

2. How does Coulomb's law lead to the Gauss's law in electrostatics?

Gauss's law is another one of Maxwell's equations and it states that the electric flux through a closed surface is proportional to the total charge enclosed by that surface. This law can be derived from Coulomb's law by using the concept of electric flux and the divergence theorem in vector calculus. It is an important tool for calculating electric fields and understanding the behavior of charges in electrostatics.

3. Can Coulomb's law be used to describe the force between moving charges?

No, Coulomb's law is only applicable for stationary charges. When charges are in motion, the force between them is described by the Lorentz force law, which takes into account the effects of electric and magnetic fields on the charges. This law is also a part of Maxwell's equations and is crucial for understanding the behavior of charged particles in motion.

4. How does Coulomb's law relate to the concept of electric potential?

Electric potential is a scalar quantity that describes the potential energy of a charged particle in an electric field. Coulomb's law can be used to derive the expression for electric potential, which is given by the product of the charge and the electric potential at a certain point in space. This relationship is important in understanding the behavior of charged particles and in solving problems involving electric fields.

5. Is Coulomb's law still valid in all situations?

Coulomb's law is a simplified version of the more general law of electromagnetism, which takes into account the effects of relativity. In most everyday situations, Coulomb's law is a good approximation, but at very high speeds or in the presence of strong electromagnetic fields, the more general law must be used to accurately describe the behavior of charges. However, for most practical purposes, Coulomb's law is still valid and useful.

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