# Moving charges in electrostatics

• I
According to a popular book on electrodynamics a special case of electrostatics is- ''source charges are stationary (though the test charge may be moving)''.
My question is- now that the test charge is moving, how is it a special case of electrostatics anymore?
Also many times we deal with questions which involve finding trajectory or kinetic energy of a charge placed freely in a uniform electric field. We conveniently use electrostatics in this case too.
How can we bring in electrostatics when the charges are moving?
Would the last example still be a case of electrostatics if the initial velocity of the moving charge is closer to speed of light?

ergospherical
The assumption implied by the notion of a test charge is that it's of sufficiently negligible charge so as to not have any effect on the charge distribution generating the background field. This means that the background field is still given by the time independent solutions to Maxwell's equations with ##\mathbf{J} = \mathbf{0}##, specifically ##\nabla \cdot \mathbf{E} = \rho / \epsilon_0## and ##\nabla \times \mathbf{E} = \mathbf{0}##.

• DaveE, hutchphd and vanhees71
The assumption implied by the notion of a test charge is that it's of sufficiently negligible charge so as to not have any effect on the charge distribution generating the background field. This means that the background field is still given by the time independent solutions to Maxwell's equations with ##\mathbf{J} = \mathbf{0}##, specifically ##\nabla \cdot \mathbf{E} = \rho / \epsilon_0## and ##\nabla \times \mathbf{E} = \mathbf{0}##.
what about the case when we have a particle of sufficient charge? why do we still use concepts of electrostatics to find say the trajectory of the moving particle?

ergospherical
what about the case when we have a particle of sufficient charge? why do we still use concepts of electrostatics to find say the trajectory of the moving particle?
Then you're no longer doing electrostatics, but instead either magnetostatics [steady currents] or electrodynamics. • rudransh verma
Then you're no longer doing electrostatics, but instead either magnetostatics [steady currents] or electrodynamics. I get your point but many times we ignore the effects of magnetostatics or electrodynamics(specially in high school physics). Is it because the velocity of moving charge is not comparable to speed of light?

ergospherical
Without a specific example it's difficult to say, but it's probably not anything to do with the speed of the charge as compared to that of light. Most likely you're just dealing with examples where the source charge distribution is significantly larger than the charge whose dynamics you're trying to analyse, so the model of a test charge moving in a background field is a good approximation.

• DaveE, Delta2, hutchphd and 1 other person
Without a specific example it's difficult to say, but it's probably not anything to do with the speed of the charge as compared to that of light. Most likely you're just dealing with examples where the source charge distribution is significantly larger than the charge whose dynamics you're trying to analyse, so the model of a test charge moving in a background field is a good approximation.
So let's say we have a test charge which is negligibly smaller than the source charge, if its moving at speed near to that of speed of light, would we consider the effect of magnetic field in this case? Because in the test charge's frame of reference its the source charge which is moving and since its sufficiently high, it would produce a magnetic field?

ergospherical
So let's say we have a test charge which is negligibly smaller than the source charge, if its moving at speed near to that of speed of light, would we consider the effect of magnetic field in this case? Because in the test charge's frame of reference its the source charge which is moving and since its sufficiently high, it would produce a magnetic field?

In this approximation you will just neglect the action of the test charge on the field, so it's just a case of considering the equations of motion in the background field generated by the source charge. If the source charge is moving in some frame of reference then it will generate a magnetic field, however note also that in the rest frame of the test charge the velocity of the test charge is zero, ##\mathbf{v} = \mathbf{0}##, so there won't be any magnetic force. In the rest frame of the source charge, the background field will be purely electric.

• In this approximation you will just neglect the action of the test charge on the field, so it's just a case of considering the equations of motion in the background field generated by the source charge. If the source charge is moving in some frame of reference then it will generate a magnetic field, however note also that in the rest frame of the test charge the velocity of the test charge is zero, ##\mathbf{v} = \mathbf{0}##, so there won't be any magnetic force. In the rest frame of the source charge, the background field will be purely electric.
thanks for help. really appreciate it :)

2022 Award
Because in the test charge's frame of reference its the source charge which is moving and since its sufficiently high, it would produce a magnetic field?
That electric fields are electromagnetic fields in other frames is true at any speed. That's how electric motors work, and they don't move anywhere close to the speed of light. But we are working in a frame where the charges (except the test ones) aren't moving, so this fact is irrelevant - we have pure electric fields.

Essentially you can ignore the fields of the moving charges when they don't have a measurable effect on any other charge. Typically this is done by simply stating that the "test" charges are very weak and light where the other charges are strong and either fixed or very heavy.

• That electric fields are electromagnetic fields in other frames is true at any speed. That's how electric motors work, and they don't move anywhere close to the speed of light. But we are working in a frame where the charges (except the test ones) aren't moving, so this fact is irrelevant - we have pure electric fields.

Essentially you can ignore the fields of the moving charges when they don't have a measurable effect on any other charge. Typically this is done by simply stating that the "test" charges are very weak and light where the other charges are strong and either fixed or very heavy.
Just a quick clarification on the last paragraph.
Even though the fields of the strong charges will dominate, the field in this region won't be purely electrostatic right? Due to the very small and almost negligible magnetic field of the moving charges?

2022 Award
Even though the fields of the strong charges will dominate, the field in this region won't be purely electrostatic right? Due to the very small and almost negligible magnetic field of the moving charges?
You are cheating here a bit, by changing my "negligible" into an "almost negligible". If it's almost negligible, it's not negligible.

You are correct that the field of a moving charge includes a magnetic component. But in this case the system would still be described as electrostatic because the magnetic component is negligible - that is, it's indistinguishable from zero in the context of this experiment. That may sound a little like I'm just playing word games, but it's a serious point. There's always experimental error and you can never say something is exactly zero - only that it's so near zero you can't measure it.

So I'd say that electrostatics is always an approximation. The "non-moving" charges are always moving a little bit and the test charge doesn't have zero charge, so there is always a little bit of magnetic field. The trick in physics is recognising correctly when it's safe to say "that effect is so weak I can ignore it".

• DaveE, ergospherical, vanhees71 and 1 other person
Gold Member
2022 Award
There's of course also the one approximation step between magnetostatics and full Maxwell theory, the socalled quasistatic approximation, where one neglects the "displacement current" in the Maxwell-Ampere Law,
$$\vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j},$$
which still decouples the magnetic from the electric field components.

This is usually justified, if you are interested in the field only "close to the sources", where "close" means at distances from the sources smaller than the typical wavelengths of the electromagnetic waves.

• ergospherical, Ibix, Delta2 and 1 other person
Homework Helper
Gold Member
@vanhees71 in the so called quasistatic approximation we also set $$\nabla\times\mathbf{E}=0$$ or we leave faraday's law unchanged that is $$\nabla\times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$$

• vanhees71
• 