# Coulomb force and the self electric field

## Main Question or Discussion Point

The well-known connection between the electric field and Coulomb's law is that

1. The magnitude of the electic field at distance r from a point-charge Q is $E=Q/r^2$
2. The definition of E is that the force exerted the field on a point-charge q is $F=qE$.

These two equalities yield the $F=qQ/r^2$ Coulomb's law. It's important that we don't care of the electric field of q, only with of Q.

But.

In the second volume of Feynman Lectures, Feynman calculates the force acting on one of the plates of the capacitor using the whole electric field, coming from both plates together. Then he makes the following correction in the calculation:

One would immediately guess that the force acting on one plate is the charge
Q on the plate times the field acting on the charge. But we have a surprising factor
of one-half. The reason is that E0 is not the field at the charges. If we imagine that
the charge at the surface of the plate occupies a thin layer, as indicated in Fig. 8-4,
the field will vary from zero at the inner boundary of the layer to E0 in the space
outside of the plate. The average field acting on the surface charges is Eo/2. That
is why the factor one-half is in Eq. (8.18).
I wonder if this logic can be applied for deriving the Coulomb's law too.
Of course we cannot apply this logic for point-charges because the electric field at the place of the charge is infinite, but we can apply to charged metal spheres, and we get the Coulomb's law as the limit when the radius of the sphere tends to zero.

So, my question is that if I calculate the common electric field of a point-charge Q and a metal sphere with charge q, and I apply the half of this electric field on the sphere, then will I get the same force as I get if I take into account only the electric field of the point charge but don't devide it by 2?

Last edited:

Related Classical Physics News on Phys.org
Jano L.
Gold Member
I apply the half of this electric field on the sphere, then will I get the same force as I get if I take into account only the electric field of the point charge but don't devide it by 2
No, that is not correct way to calculate forces. It may give correct answer in some cases, but it is not general. In case of capacitor plates, the force is (field due to one plate) x (charge of the other plate). Due to simple geometry this gives one half (total field between plates) x (charge on one plate), but in other cases 1/2 need not appear in such way.

In your example, if the sphere and charge do not move, the force on the sphere is determined by the electric field of the point charge only. If any of them moves, the field of the sphere may have non-negligible contribution to the total force acting on the sphere.

So there is no theoretical basis of Feynman's solution? Is it theoretically wrong?

It may give correct answer in some cases, but it is not general.
Are you absolutely sure that in the case of the point-charge and charged metal sphere it gives a wrong answer? How can you see this?

Jano L.
Gold Member
Take the case when the charge of the point ##q_p## is much larger than the charge of the sphere ##q_s##. The total field ##\mathbf E## is then almost equal to the field of the point alone ##\mathbf E_p##. If the sphere is at ##\mathbf r_s## far from the point, the force on the sphere is approximately

$$q_s \mathbf E_p(\mathbf r_s).$$

If we used the 1/2 rule, we would get ##1/2q_s \mathbf E##, which for our case is almost ##1/2q_s\mathbf E_p##.The factor of 1/2 would be wrong here. It works only when ##\mathbf E = 2\mathbf E_p##, like for capacitor plates.

1 person
Thanks, this is convincing. Is it a known bug in Feynman Lectures? Are there yet more serious bugs in them?

Jano L.
Gold Member
Yes, but I do not know of any compiled list of bugs. Actually, such list seems like a good idea. Feynman's lectures are great, young people are still being redirected to them often, so a list of serious bugs with short explanation and updates (they are really old: written in 60's) would really contribute to physics education.

Jano L.
Gold Member
Thank you for the link, having the Lectures online is really great for quick searching and also service to all students who can't afford them.

1 person
Jano L.
Gold Member
Yeah, that's it, much more interesting than the errata. The best way to correct errors is to put them into revised text, and it seems that authors have done this with the known errors. But still there may be more errors, so you may write to the maintainers about your 1/2 problem if you did not already, perhaps they will incorporate a comment about it in the future publication of the 2nd volume.

I've found an errata page here: Errata for The Feynman Lectures on Physics, but this bug isn't listed here. I've reported this error them.
That's because it is not an error. Feynman's argument is sound. He's using the continuity of the field after the self interaction has been removed. That can also be applyed to the sphere example. But it must be applied point by point and then all the little forces are added as vectors, which means an integral will have to be evaluated. The case of the Capacitor is simpler because the field is uniform, all the little forces are in the same direction so the vector addition produces a force proportional to the total charge in the capacitor.

Jano L.
Gold Member
That's because it is not an error. Feynman's argument is sound.
Feynman's argument depends on the assumption that the charge density is uniform (see the figure with plot of E). He also uses expression "average field". Where does ##1/2E_0## come from in his explanation? From the linear increase of ##E## and from averaging it over distance. But those are very special assumptions. The charge density does not have to be uniform.

He's using the continuity of the field after the self interaction has been removed. That can also be applyed to the sphere example. But it must be applied point by point and then all the little forces are added as vectors, which means an integral will have to be evaluated. The case of the Capacitor is simpler because the field is uniform, all the little forces are in the same direction so the vector addition produces a force proportional to the total charge in the capacitor.
I do not know where you think self-interaction has been removed. Feynman does not mention self-interaction. I think what you say above may be correct, but I think it is nowhere to be found in the Feynman attempt at explanation.

The argument based on the extended layer of charge can be made convincing in the following way. I will show that the force on the layer is independent of the actual distribution of charge in the layer.

Let the layer have thickness ##t##. Introduce coordinate ##x## such that metal surface has ##x=0##, rightmost surface of charge has ##x=t##.

Let the charge density in the layer be described by any function ##\rho(x)##, electric field by ##E(x)## . The force is given by

$$F = A \int_0^t \rho(x) E(x) dx.$$

From the Gauss law and symmetry, ##\rho(x) = \epsilon_0 \partial_x E(x)##. Integrating per partes, we find out that the result depends only on the value of field at ##x=t## and is equal to

$$F = \frac{1}{2}\epsilon_0 E_0^2 A.$$

Since ##Q = \epsilon_0 E_0 A##, this is equal to ##\frac{1}{2}QE_0##, i.e. the force is as if the field ##1/2E_0## acted on the whole plate charge.

So the argument with distributed charge in the layer can be made, but it is different from what Feynman claims.

Even so, I have to say that the argument is unnecessarily complicated: we had to introduce metal, extended layer of arbitrary non-zero thickness ##t## and use Maxwell equations and integrals.

Compare this to the following simple argument:

The left plate does not act on itself, since in static case the internal forces cancel out. Thus the total force on it is only due to the right plate, which produces field ##\frac{1}{2}E_0## at the left plate. Thus the force is ##\frac{1}{2}QE_0##.

In this we do not need to introduce metal, layer thickness or integration; the only important things are the charges on the plates and the Coulomb law.