Electric Field of a moving charge

  • #1
Zahid Iftikhar
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When a charge is at rest, it has an electric field only. When the charge starts moving , it is said to have accompanied a magnetic field. My question relates to its electric field while in motion. Does it still exist or not? I know in electron guns electrons are deflected while passing thru the electric fields, so it seems electric field is still there. Please guide me on this.
High regards.
 

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  • #2
Vanadium 50
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The electric field is still there.
 
  • #3
Zahid Iftikhar
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Sir, thanks for the reply.
Does it have the same intensity?
 
  • #4
Ibix
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Not in general.

The easiest thing to do if you want an algebraic answer is to write down the electromagnetic field tensor for the field of a stationary charge and then boost it.
 
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  • #5
Nugatory
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Many I-level treatments of E&M don’t use the electromagnetic tensor (which is a sadness because it’s not that hard and is by far the most elegant approach). If @Zahid Iftikhar isn’t comfortable working with tensors there are also the standard formulas for transforming electrical and magnetic fields between frames - Google will find these, and they’ll be in any first-year E&M textbook.
 
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  • #6
Zahid Iftikhar
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Thanks for the time. I ll try the way you suggested.
Regards
 
  • #7
vanhees71
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Rather than doing the Lorentz boost, it's much simpler to write down the four-potential first. I'll work in Heaviside-Lorentz units.

For the particle at rest you have (in Lorenz gauge)
$$\bar{A}^{\mu}=\begin{pmatrix}\Phi_C \\0 \\0 \\0 \end{pmatrix},$$
where
$$\Phi_C=\frac{q}{4 \pi \bar{r}}.$$
Now let's write this in manifestly covariant form. The only four-vector we have in this problem is the four-velocity of the particle ##u^{\mu}=\gamma (1,\vec{\beta})## with ##\beta=\vec{v}/c## and ##\gamma=(1-\beta^2)^{-1/2}##. In the rest frame ##u^{\mu}=(1,0,0,0)##. Thus ##A^{\mu}=\Phi_C u^{\mu}##. Now we only have to write ##\Phi_C## in a manifestly covariant way (which now is identified as a Lorentz scalar field!). Obviously we can write
$$\bar{r}^2=\vec{\bar{x}}^2=(\bar{u}_{\mu} \bar{x}^{\mu})^2 - \bar{x}_{\mu} \bar{x}^{\mu}$$. Thus we have in a general frame
$$A^{\mu}(x)=\frac{q}{4 \pi \sqrt{(u \cdot x)^2-x \cdot x}} u^{\mu}.$$
Now the electric and magnetic field components are given by (written in (1+3)-notation now)
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
The result is
$$\vec{E}= -\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} A^0 =
\frac{q}{4 \pi} \frac{\gamma(\vec{x}-\vec{v} t)}{[(u \cdot
x)^2-x^2]^{3/2}}, \quad \vec{B}=\vec{\nabla} \times \vec{A} = \frac{q}{4 \pi}
\frac{\vec{u} \times \vec{x}}{[(u \cdot x)^2-x^2]^{3/2}}=\vec{\beta} \times \vec{E}.$$
 
  • #8
Zahid Iftikhar
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Thanks Sir for your effort and time. Mathematics is my downside. I find difficulty in drawing something from these equations. I am apologetic for this weakness of mine.
 
  • #9
tom_
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Let me answer your question as an electrical engineer.

First you should realize that ##\vec{B}## and ##\vec{E}## cannot be measured directly. One can only measure their effects and these consist in the fact that the fields exert forces onto charges. By definition, E fields act on resting and moving charges, B fields act only on moving charges.

So your true question is, which force would a moving point charge (velocity ##\vec{v}_s##) exert on another moving point charge (velocity ##\vec{v}_d##)? The answer, derived from the Maxwell equations, is:

$$
\vec{F}_M(\vec{r},\vec{v}_s,\vec{v}_d) = \frac{c\,\left(c^2-v_s^2\right)\,\left(\vec{r} + \frac{1}{c^2}\,\vec{r}\times\vec{v}_s\times\vec{v}_d\right)}{\sqrt{\left(c^2 - v_s^2\right) + \left(\frac{\vec{r}}{r}\,\vec{v}_s\right)^2}^{3}}\,\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0\,r^3}.
$$
You can get to this formula by inserting E and B field (see answer above) into the formula of the Lorentz force. For low velocities ##\vec{v}_s## and ##\vec{v}_d## this formula turns into
$$
\vec{F}_{M}(\vec{r},\vec{v}_s,\vec{v}_d) \,\approx\, \vec{F}_W(\vec{r},\vec{v}_s) + \frac{1}{c^2}\,\left(\vec{r}\times\vec{v}_s\times\vec{v}_d - \frac{v_s^2\,\vec{r}}{2}\right) \frac{q_s\,q_d}{4\,\pi\,\varepsilon_0\,r^3}.
$$
with
$$
\vec{F}_{W}(\vec{r},\vec{v}) = \left(1 + \frac{v^2}{c^2} - \frac{3}{2} \left(\frac{\vec{r}}{r}\,\frac{\vec{v}}{c}\right)^2\right)\,\frac{q_s\,q_d}{4\,\pi\,\varepsilon_0}\,\frac{\vec{r}}{r^3}
$$
and the differential speed ##\vec{v} = \vec{v}_d - \vec{v}_s##. This formula is called Weber force (Weber electrodynamics).

The Weber force formula works very well for its alone in practice. If the forces of all charge carriers are integrated in a current-carrying wire, one can see how the magnetic forces result from the speed differences between electrons/atoms in relation to a resting/moving test charge. The electric fields are canceled out and only the portion of the field that results from the distortions remains.
 
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  • #10
Zahid Iftikhar
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Thank you very much for the reply. I am proud of people at PF for being so helpful in clarifying concepts.
What I get out of your explanation is that once charges set into motion, their electric field is mainly canceled out and but just a portion remains due to distortions. But the magnetic field is predominant in this situation. I am trying to improve upon use of Maxwell Equations, Lorentz Transformations etc.
High Regards.
Zahid
 
  • #11
ndriana
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No, as long as it has an electric charge, it will always have an electric field. Remember, motion is relative, so although your electron is moving, it is always at rest with respect to itself.
The only reason they don't talk about it when introducing magnetism is the equation gets complicated and instead of learning magnetism, the study jumps into electromagnetism.
 
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  • #12
tom_
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What I get out of your explanation is that once charges set into motion, their electric field is mainly canceled out and but just a portion remains due to distortions.

No. Take formula ##\vec{F}_W## and set the relative velocity ##\vec{v}=\vec{0}##. You see that the Weber force becomes the Coulomb force, i.e. the net electromagnetic force is only caused by the E-field for ##\vec{v}=\vec{0}##.

Set now ##\vec{v}=(0.01,0,0)c##. From this follows ##v^2=0.0001c^2## and ##v^2/c^2 = 0.0001##. The second term with ##\vec{v}## also contains a ##c## below the fraction line. You see, for small relative velocities, a tiny correction of the electric force occurs. This is magnetism.

In a current-carrying metallic conductor, the electrical components compensate each other (The metal ions have the same speed-independent field as the moving electrons). What remains is the distortion field of the electrons (terms with v and c), so only a force can be observed here when test charges move in relation to the wire.

Just think about it and play with the formulas. Is a simple effect.
 
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  • #14
vanhees71
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The complete field is given in #7...
 
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  • #15
davidpeng949
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Summary: Stationary charge has only electric field, a moving charge has a magnetic field, why not electric field?

When a charge is at rest, it has an electric field only. When the charge starts moving , it is said to have accompanied a magnetic field. My question relates to its electric field while in motion. Does it still exist or not? I know in electron guns electrons are deflected while passing thru the electric fields, so it seems electric field is still there. Please guide me on this.
High regards.
i think it is a great question, since it leads to an deeper question that, therefore, the magnetic field law should be able to be derived mathematically from the Coulomb law and velocity, nothing more.
 
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  • #16
vanhees71
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There's an electromagnetic field. The split in electric and magnetic components is frame dependent. The purely "electric" Coulomb field of a static charge distribution in some frame of reference has also magnetic components in another frame, where the charge distribution moves.
 
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  • #17
Zahid Iftikhar
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Really serious consideration. So matter is not that simple. It involves relativity.
Thanks for the time sir.
High Regards
 
  • #18
vanhees71
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Yes, indeed. As got unnoticed for about 50 years after its discovery Maxwell theory is the paradigmatic example for a relativistic classical field theory. Only when treated as such it becomes completely consistent with the rest of physics, particularly the space-time model describing all observations best (that's special relativity and Minkowski space as long as you can neglect gravity and general relativity and a pseudo-Riemannian manifold if gravity has to be included).
 
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  • #20
davidpeng949
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The fundamental differences between static electric field and magnetic field are the following:
(1) The sources, “e-particle Q” vs. “e-current Qv”;
(2) The way the fields induced by source, “divE” vs. “curlB”;
(3) The nature of the fields, “vector field E”vs. “axial vector field B”;
(4) Effects of the fields on a test e-particles, “qE”vs. “qvxB”.
A physics student may ask questions: Why the motion of e-particles induces magnetic fields? Does the generation of magnetic fields relate with the Coulomb’s law? A teacher’s answer is that magnetism is the combination of electric field with Special Relativity and does not relate with the Coulomb’s law. The argument is that the teacher’s answer is not sufficient to explain the above four fundamental differences.
On the other hand, if one can derive magnetic field B from coulomb law and velocity, then all of above 4 fundamental differences explained.
 
  • #21
Ibix
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A teacher’s answer is that magnetism is the combination of electric field with Special Relativity and does not relate with the Coulomb’s law. The argument is that the teacher’s answer is not sufficient to explain the above four fundamental differences.
On the other hand, if one can derive magnetic field B from coulomb law and velocity, then all of above 4 fundamental differences explained.
I'm not at all sure what you are saying here, but it's easy enough to show that not all magnetic fields are just electric fields seen in another frame. The quantity ##E^2-B^2## is invariant (the same in all frames), so if there is a frame where ##E=0## there cannot be a frame where ##B=0##, and vice versa.
 
  • #22
vanhees71
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The fundamental differences between static electric field and magnetic field are the following:
(1) The sources, “e-particle Q” vs. “e-current Qv”;
(2) The way the fields induced by source, “divE” vs. “curlB”;
(3) The nature of the fields, “vector field E”vs. “axial vector field B”;
(4) Effects of the fields on a test e-particles, “qE”vs. “qvxB”.
A physics student may ask questions: Why the motion of e-particles induces magnetic fields? Does the generation of magnetic fields relate with the Coulomb’s law? A teacher’s answer is that magnetism is the combination of electric field with Special Relativity and does not relate with the Coulomb’s law. The argument is that the teacher’s answer is not sufficient to explain the above four fundamental differences.
On the other hand, if one can derive magnetic field B from coulomb law and velocity, then all of above 4 fundamental differences explained.
Well, as long as there is an inertial reference frame (Minkowski frame), where all charges are at rest you have a static field. In this special frame you have only electric components since in this frame the four-current simply is
##(j^{\mu})=(c \rho(\vec{x}),0,0,0)##. In this frame thus the solution for the four-potential simply is the Coulomb field, using Heaviside-Lorentz units
$$A^{\mu} = \phi(\vec{x}) (1,0,0,0), \quad \phi(\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\rho(\vec{x}')}{4 \pi|\vec{x}-\vec{x}'|}.$$
It's also very easy to give the potential in a general frame. Just write everything in manifestly covariant form:
$$\tilde{A}^{\mu}=\tilde{u}^{\mu} \tilde{\phi}(\tilde{x}).$$
Here ##\tilde{u}^{\mu}## is the four-velocity of the new frame. Of course you have to transform also the scalar Coulomb potential as a scalar potential,
$$\tilde{\phi}(\tilde{x})=\phi(x).$$
Now it's clear that in the general frame you have also magnetic components,
$$\tilde{\vec{B}}=\tilde{\vec{\nabla}} \times \tilde{\vec{A}} = -\vec{u} \times \tilde{\vec{\nabla}} \tilde{\phi}(\tilde{x}).$$
The electric field in the general frame is
$$\tilde{\vec{E}}=-\frac{1}{c} \partial_{\tilde{t}} \tilde{\vec{A}}-\tilde{\vec{\nabla}} \tilde{A}^0,$$
Since ##\tilde{A}^0=\tilde{u}^0 \tilde{\phi}## and ##\tilde{\vec{A}}=\vec{u} \tilde{\phi}## we have
$$\tilde{\vec{B}}=\frac{\tilde{\vec{u}}}{\tilde{u}^0} \times \tilde{\vec{E}}.$$
As I said before: There's one electromagnetic field, represented by the antisymmetric Faraday tensor, and whether or not in a given situation you have electric and/or magnetic components of the field is frame dependent.
 
  • #23
Zahid Iftikhar
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Yes, indeed. As got unnoticed for about 50 years after its discovery Maxwell theory is the paradigmatic example for a relativistic classical field theory. Only when treated as such it becomes completely consistent with the rest of physics, particularly the space-time model describing all observations best (that's special relativity and Minkowski space as long as you can neglect gravity and general relativity and a pseudo-Riemannian manifold if gravity has to be included).
Thanks for your time and help.
 

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