How can electric field be defined as force per charge, if the charge makes its own, singular electric field?

Question

The electric field $\bf{E}$ represents how much force would act on a particle at a certain position per unit charge. However, if we actually place a particle in that position, the electric field will have a singularity there (because of the $\frac{1}{r^2}$ in Coulomb's law). Isn't this kind of a paradox? In my eyes, this makes the concept of electric field useless, because it cannot be used to calculate the force on a particle.

Answer 1

It's true that a point particle with finite charge is problematic in electromagnetism because of the infinite field and associated energy near such a particle. However, we don't need that concept in order to make a defining statement about the electric field. Rather, we can use $$ {\bf E} = \lim_{r \rightarrow 0} \frac{\bf f}{q} $$ where $\bf f$ is the force on a charged sphere of radius $r$ with a finite charge density $\rho$ independent of $r$, and $q = (4/3) \pi r^3 \rho$ is the charge on the sphere. This charge $q$ will tend to zero as the radius does, and it does so sufficiently quickly that no infinities arise and everything is ok.

Answer 2

You're forgetting one thing: a particle cannot feel its own electric field, so a point charge that generates a $1/r^2$ field doesn't do anything unless acted upon by an external field. You also can't place a particle at $r=0$ of another particle's $1/r^2$ electric field, because, well, there's already a particle there. (Also, how are you going to get it there, even if you could? It takes so much energy to even get close that you're leaving the realm of classical electromagnetism when you try.)

Answer 3

Isn't this kind of a paradox?

Consider two point charges, $q_1$ and $q_2$, in the vacuum with separation vector $\mathbf{r}_{12}$. Coulomb's Law for the force on charge $q_2$:

$$\mathbf{F}_2=q_2\frac{q_1}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}_{12}}}{|\mathbf{r}_{12}|^2}=q_2\mathbf{E}_1$$

Thus, the force on charge $q_2$ is due to the electric field of charge $q_1$ only. Similarly,

$$\mathbf{F}_1=q_1\frac{q_2}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}_{21}}}{|\mathbf{r}_{21}|^2}=q_1\mathbf{E}_2$$

the force on charge $q_1$ is due to the electric field of charge $q_2$ only. This easily generalizes to $N$ point charges; the force on charge $q_n$ is the vector sum of the forces due to electric field of each of the other $N-1$ charges.

You may (or may not) be familiar with the notion of a test charge which 'feels' the electric field due to other charges but has no significant electric field. Armed with this abstraction, one can say that the (total) electric field at a point is the force per unit charge at that point. Indeed, from the Wikipedia article Electric field

The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

(emphasis mine)

Answer 4

You should distinguish the interaction if a charge with another charge from its self-interaction. For the first case there is no issue. For the second case there are issues. For a classical point particle the self-interaction energy diverges, so you will have to assume a finite radius. If you assume a homogeneus spherical distribution and equate the self energy to the rest energy you find about 2.8 femtometer for an electron. See https://en.m.wikipedia.org/wiki/Classical_electron_radius. However there is no experimental evidence for a finite value of the electron radius. As far as high energy physicists know it is a point particle.

Answer 5

The basic reason here is that "electric field", like all theories in physics and science, is a model for reality, not "reality" itself, to which we have no direct access, and thus what one should really be asking here is what it is intended to model and how best to conceptualize that model. A poster here named @Cort Ammon likes to emphasize points along this line. Science doesn't tell us what things "really" are - insofar as its "pictures" are concerned, it actually really is a social construct, sorry, but those "useless" degrees aren't useless. What isn't a construct, is what it tells us we can and can't do, or what will happen if we do something, when those actually happen. Photons may be a social construct and if you want you can imagine little green gnomes instead, but it is not a construct that if you put your hand under a magnifier focusing the Sun (DON'T), you'll get a nasty lil' mark. The former is a framework we have in our heads to think about the latter, why and what happens during it, perh. also why that such burn may have a slightly greater chance of cancer in the future (and that works a bit better than green gnomes do :g:).

An electric field is such a construct whose initial purpose is to model one kind of phenomenon: how an electrically-charged object moves when placed in a region of space that didn't before have one, by (simplifiedly) putting little arrows at each place in space that point in the direction in which it seems compelled to move, and making the contours of such mathematically precise with the tools originally developed by Rene Descartes and Pierre de Fermat known as analytic geometry.

Because of that, it is an answer to a kind of "counterfactual" question, and hence what the arrows of the electric field you are describing in your post, then, where you've added the second object as producing field, do is to describe how a hypothetical third object not already present there would move. If we want to talk of the motion of the second object, we thus need to consider the field from the first alone, without adding its "own" field in as you just did.

You might also compare Cort Ammon's answer at the top of here:

Why doesn't a particle exert force on itself?

as this touches on the philosophical blurb at the beginning.