# Do charged particles move along the electric field line?

I was discussing with my friend about electric field lines and he said that:

given a test charge at rest at a point in the region of the electric field the charge will continue to move along the path of the field line passing through that point.

When asked for how reliable this statement was he couldn't say anything for or against it, stating that he hasn't thought much about it and will talk about it later.

But now this statement is bugging me a lot because I can easily see it being true in a uniform electric field, but can't say so in the case of a non-uniform one.

• So can someone tell me if a stationary charge allowed to move will follow the path of electric field lines or not?

One thing that I noticed is that if that were the case then both the force vector and velocity vector will point in the same direction.

• Sep 13, 2020 at 14:42

No, charged particles do not need to move along the path of field lines. The field lines will just show the direction of acceleration, but just because acceleration is in some direction doesn't mean the particle moves in that direction. This is true for all motion, not just charged particles in electric fields.

Of course if the charge starts at rest in a uniform field then the charge will move with the field lines. However, in general even in a uniform field this will not be the case (As a simple example think about projectile motion).

• Thanks for the response. I was talking about a test charge which is initially at rest. Will it follow that path or not.?
– user249968
Sep 13, 2020 at 13:47
• @JohanLiebert No, unless the test charge has no mass. I don't remember if test charges are always assumed to be massless. Sep 13, 2020 at 14:29
• @JohanLiebert If the field lines are straight lines, then yes, the particle will turn out to follow the field lines if it starts at rest. If the field lines turn at some point, the particle's velocity will be turned along with them, but not fast enough to actually follow the field lines (for the reason described in this answer). Sep 14, 2020 at 10:48

It is not true in general. The simplest counterexample would be the “classical” Bohr-like hydrogen atom without radiation, where the electron moves in a circular orbit in the central $$\vec E$$ of the proton. Even allowing for electron to loose energy through radiation the motion could be a spiral.

Another example would be the motion of an electron in an inkjet printer where the electron is deflected by a perpendicular $$\vec E$$ field but still maintains its velocity normal to that field.

If the particle is initially at rest it would presumably be so only punctually in time and acquire some initial velocity in the direction of $$\vec E$$ at that point, but then the general case would apply.

• Thanks for the response. I was talking about a test charge which is initially at rest. Will it follow that path or not? Also, please prove it in the case of newtonian mechanics as I have very little knowledge about quantum mechanics.
– user249968
Sep 13, 2020 at 13:50
• I have updated my answer. Sep 13, 2020 at 13:52
• The "at rest" assumption (which was in the original question) invalidates the hydrogen atom example. There, although the field is not uniform, field lines are straight and so the motion would still be along a field line (radial). Sep 14, 2020 at 21:55

This is generally not true but there are circumstances under which this is true. The field lines give the acceleration of the test particle. So initially the particle's path is aligned but as it builds up momentum it will overshoot and deviate from the field lines.

If you have a lot of friction this effect is counteracted because the particles aren't able to build up momentum. In real life there is often a lot of friction so this is probably why your intuition tells you that the particles should follow the field lines. For example if you have a pond with some leaves or dust on the surface then the leaves/dust will follow the velocity field of the water because the friction is sufficiently large.

Under these circumstances the particles always have terminal velocity which means their velocity vector is aligned with the force vector.

• This explanation is exactly to the point. Sep 14, 2020 at 13:42

Simple concrete counterexample. In Cartesian coordinates on a 2D plane, let $$\mathbf{E}=\frac{1}{x^2+y^2}\langle-y,x\rangle.$$

(I believe this is the electric field around a decaying/strengthening bundle of magnetic field lines.) It looks like this:

The electric field lines (curves that are everywhere tangent to the electric field) are circles. At time $$t=0$$, put a charge $$q=1$$ at rest with mass $$m = 1$$ at position $$(1,0)$$. Our choice of units means that $$\mathbf{E}=\mathbf{F}=\mathbf{a}$$, and expanding everything gives us \begin{align}&\frac{d^2x}{dt^2}=\frac{-y}{x^2+y^2}&&\frac{d^2y}{dt^2}=\frac{x}{x^2+y^2}&\\&\frac{dx}{dt}\Big\rvert_{t=0}=0&&\frac{dy}{dt}\Big\rvert_{t=0}=0&\\&x\rvert_{t=0}=1&&y\rvert_{t=0}=0.\end{align}

We can graph the solution and observe its non-circleness.

We can sum everything up like this: electric field lines are curves everywhere tangent to the electric field—i.e. the path a particle would take if the field determined its instantaneous velocity ($$\mathbf{v}\propto\mathbf{E}$$). But for real particles, the field determines (some part of) the acceleration ($$\mathbf{a}\propto\mathbf{E}$$), so real particles generally don't travel along field lines.

Note: this example electric field is most easily realized by using a time-varying magnetic field. However, any electric field with "curves" in it should do. E.g. an electrostatics example would be two fixed point charges forming a dipole. The following configuration is two charges -1 and 1 respectively held fixed at $$(-1,0),(1,0)$$ with the same test particle from before placed at $$(0,1)$$, with its path in green. (This time I graphed the "field lines" as actual curves rather than graphing the vector field itself. Note that the density of lines here does not correspond to the field strength like it should in an actual field line drawing.)

Again, the inertia of the moving charge means it definitely does not follow the electric field lines. The "circular" example from before was chosen simply because it is mathematically simple, not because we absolutely need a magnetic field.

given a test charge at rest at a point in the region of the electric field the charge will continue to move along the path of the field line passing through that point.

It's true while the test charged is at rest. While the charge is at rest there is no magnetic field acting on it. And it has no momentum in another direction.

It doesn't continue to be true because as soon as it has moved a little bit then it is no longer at rest.

If you could have a test charge with no momentum that would help, but it would still be susceptible to magnetic fields.

If the particle's velocity has a component normal to the field line, then it will not follow the field line. If a particle starts from rest and the electric field lines are straight, then the particle will follow the electric lines. If the field line is curved, then the particle must be experiencing a force perpendicular to the curve (that is, there must be a perpendicular component; the total force need not be normal). This is the definition: if something is not moving in a straight line, its acceleration has a component tangential to its velocity. But the electric field line is, by definition, a curve such that the electric field line is tangent to that curve. So a test particle experiencing only the electric force can't follow a curved electric field line.

Say that the electric field is described by the vector field $$\bf{E}$$. A field line $${\bf{x}}(\lambda)$$ passing at the point $${\bf{x}}_0$$ is a curve in the 3D space, which can be found by solving the equations

$$\frac{d }{d \lambda}{\bf{x}}(\lambda) = \alpha {\bf{E}}({\bf{x}}(\lambda)) \\ {\bf{x}}(\lambda = 0 ) ={\bf{x}}_0$$

where $$\alpha>0$$ is a constant of proportionality which value is fixed but unimportant.

If your particle (starting at rest at $${\bf{x}}_0$$) has to follow the same path given by $${\bf{x}}(\lambda)$$, then you must have that its velocity is

$$\frac{d }{d t}{\bf{x}}(t) = \beta \, {\bf{E}}({\bf{x}}(t)) \\ {\bf{x}}(t = 0 ) = {\bf{x}}_0$$

for some constant $$\beta$$. As you can see, this is in contrast with the Newton equation for the particle, which reads ($$q$$ and $$m$$ mare the particle's charge and mass)

$$\frac{d^2 }{d t^2}{\bf{x}}(t) = (q/m) \, {\bf{E}}({\bf{x}}(t)) \\ \frac{d }{d t}{\bf{x}}(t = 0 ) = 0 \\ {\bf{x}}(t = 0 ) = {\bf{x}}_0 \, ,$$

without considering the back reaction of EM emission from the particle itself and relativistic effects. The simple fact that the particle has "inertia" makes it drift from the path along the field lines. So, for your particle to be advected exactly along the field lines, you have to demand that its equation of motion is not the Newton one, namely go in the limit of zero inertia (and no radiation).

A charged particle following an electric field line means that the velocity (tangent to the path) and acceleration (tangent to the electric field lines) are proportional to each other at all times.Therefore, there is no normal acceleration during the motion and it moves in a straight line.

The converse is also true if we require that the velocity and acceleration are proportional to each other in the initial conditions.

So we get the final result:

A charged object follows a field line if and only if the field lines are straight lines and the initial velocity is tangent to them.

The result applies to any vector field in which the force and acceleration are proportional to the field vector.

You can understand like this.

If an electric field line be like this (a curve), then you can imagine that at every single point of the curve, the force acting on the charge shall be tangential to the curve. Since the force is always tangential, there is no centripetal component of the force acting on the charge.

Thus, it shall not be possible for the charge to move in the trajectory same as electric field line (since it cannot make curve movements due to lack of centripetal force).