Coulomb's law for steady currents

In summary, the conversation discusses whether Coulomb's law holds in a case of steady currents. While it may not hold in the general case, if the charge density is constant over time, it suggests that the charges are not moving and Coulomb's law may be used. The conversation also explores the conditions for when Coulomb's law applies, such as in electrostatics and magnetostatics. In the case of magnetostatics, the electric and magnetic fields decouple and the electric field can be described by a superposition of Coulomb fields. However, if the charge distribution is not spherically symmetric, higher multipole moments may also contribute to the electric field.
  • #1
Kashmir
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If I've steady currents i.e ##\frac{\partial}{\partial t} J=0## , does coulombs law hold in this case to find the electric field?
Since this isn't the case of electrostatics so it might not hold, but if we look at the charge density it is the same for all time, this suggests that the charges are in a way sitting still so we might use coulmbs law. What is correct in this case. Thank you
 
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  • #2
A charge creates an electric field whether it's moving or not. You can transform the electric field in the rest frame of the charge to the electromagnetic field in the frame where it's moving.
 
  • #3
Kashmir said:
If I've steady currents i.e ∂∂tJ=0 , does coulombs law hold in this case to find the electric field?
Since this isn't the case of electrostatics so it might not hold, but if we look at the charge density it is the same for all time, this suggests that the charges are in a way sitting still so we might use coulmbs law.
Although I'm not sure, I think you may be right under certain conditions. :smile:
 
  • #4
In the most general case it is $$\vec{E}=-\nabla V-\frac{\partial \vec{A}}{\partial t}$$.

In this case since we have ##\frac{\partial J}{\partial t}=0## it follows pretty easily (remember that in the lorentz gauge we have ##\nabla^2\vec{A}-\frac{1}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2}=\mu_0\vec{J}##)that also ##\frac{\partial \vec{A}}{\partial t}=0## hence we have $$\vec{E}=-\nabla V$$ that is a condition that holds in electrostatics. So yes you can treat the problem like it is electrostatics, though there are charges moving, the contribution to electric field is only from the scalar potential ##V##, like it happens in electrostatics.
 
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  • #5
If everything is time-independent you get magnetostatics, i.e.,
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E} = \rho, \quad \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0.$$
Thus the electric and the magnetic field equations decouple completely, and the sources for the electric field is the charge distribution. So as far as the electric field is concerned it's as in electrostatics.
 
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  • #6
vanhees71 said:
If everything is time-independent you get magnetostatics, i.e.,
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E} = \rho, \quad \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0.$$
Thus the electric and the magnetic field equations decouple completely, and the sources for the electric field is the charge distribution. So as far as the electric field is concerned it's as in electrostatics.
The main thing that decouples the equations is that ##B## is independent of time in magnetostatics. This gives us ##\nabla \times E=0## and as always we also have ##\nabla \cdot E=\frac{\rho}{\epsilon_{0}}##
The solution to both equations is the coulomb field.

So coulombs law holds in magnetostatics.

Is this reasoning correct?
 
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  • #7
Seems correct to me.
 
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  • #8
Though it's only a Coulomb field (outside of the charge distribution) when the charge distribution is spherically symmetric around some center, i.e., when with the corresponding choice of the reference frame, ##\rho=\rho(r)##. If this is not the case, you get also higher "multipole moments" than the Coulomb field (which is the monopole term) in the multipole expansion for the fields outside of the charge distribution.
 
  • #9
vanhees71 said:
Though it's only a Coulomb field (outside of the charge distribution) when the charge distribution is spherically symmetric around some center, i.e., when with the corresponding choice of the reference frame, ##\rho=\rho(r)##. If this is not the case, you get also higher "multipole moments" than the Coulomb field (which is the monopole term) in the multipole expansion for the fields outside of the charge distribution.
I can't understand this. We agreed that in magnetostatics ##E## is given by Coulombs law, now why do you say that there exist other terms?
 
  • #10
No, why should it be. If you have a dipole you have a dipole but not a Coulomb field. The general solution is a "superposition of Coulomb fields",
$$\vec{E}(\vec{x}) = \int_{V} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0} \, \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$
 
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  • #11
vanhees71 said:
No, why should it be. If you have a dipole you have a dipole but not a Coulomb field. The general solution is a "superposition of Coulomb fields",
$$\vec{E}(\vec{x}) = \int_{V} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0} \, \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$
Oh yes, what I meant by coulomb field is the general superposition of each charge element ##dq##
 
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Related to Coulomb's law for steady currents

What is Coulomb's law for steady currents?

Coulomb's law for steady currents, also known as Ampere's law, states that the magnetic field created by a steady electric current is directly proportional to the magnitude of the current and inversely proportional to the distance from the current.

How is Coulomb's law for steady currents different from Coulomb's law for static charges?

Coulomb's law for steady currents applies to moving charges, while Coulomb's law for static charges applies to stationary charges. Additionally, the direction of the magnetic field in Coulomb's law for steady currents is perpendicular to the direction of the current, while in Coulomb's law for static charges, the electric field is parallel to the direction of the force between the charges.

What is the mathematical formula for Coulomb's law for steady currents?

The mathematical formula for Coulomb's law for steady currents is B = (μ0/4π) * (I/r), where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the distance from the current.

How does Coulomb's law for steady currents relate to electromagnetic waves?

Coulomb's law for steady currents is one of the fundamental laws of electromagnetism and is used to understand the behavior of electromagnetic waves. Electromagnetic waves are created by the acceleration of electric charges, which is described by Coulomb's law for steady currents.

What are some real-life applications of Coulomb's law for steady currents?

Coulomb's law for steady currents has many practical applications, such as in the design of electric motors, generators, and transformers. It is also used in the field of medical imaging, specifically in magnetic resonance imaging (MRI) machines. Additionally, Coulomb's law for steady currents is essential in understanding the behavior of electrical circuits and power transmission systems.

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