Coulomb's law for steady currents

[Kashmir]

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

 

[PeroK]

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.

 

[alan123hk]

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.

 

[Delta2]

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.

 

[vanhees71]

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.

 

[Kashmir]

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?

 

[Delta2]

Seems correct to me.

 

[vanhees71]

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.

 

[Kashmir]

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?

 

[vanhees71]

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",
$$\vec{E}(\vec{x}) = \int_{V} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0} \, \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$

 

[Kashmir]

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 $dq$