# Rationale for the Electrostatic Field

This is from div grad curl and all that, h.m. schey.

By trivial the author, I think, just means dividing the electrostatic force by the test charge to get the electrostatic field. What I need are concrete examples of (a) and (b), which would make the author's intent clear.

(a) could mean that for a given distribution of charges we can calculate the field without worrying about the effect these charges have on the field created by other charges within this distribution. Isn't that just the principle of superposition?

(b) could mean that once we know the field, we don't have to worry about how those charges that produced the field are arranged in order to calculate the effect of the field on some charge placed in this field. I need an example of that.

• Aug 25, 2021 at 8:30
• I feel you are misusing the book, the book was meant to be a book about vector calculus. Trying to really understand the detials of physics example is gonna distrac you from actual vector calculus. There are other books more apt for electrostatics like purcell. Aug 25, 2021 at 8:31

Consider a electrostatic dipole - two charges $$\pm q$$ separated by a small distance $$a$$. Each charge produces an electric field. The reason the field approach works is because I can now describe/talk about the field generated by each charge independently. If, I choose to add the fields to find the net field at an arbitrary point (which is what one is often interested in), then I can do so thanks to superposition. However, my understanding of '(a)' is that it is the field that allows us to first talk about individual charges at all. If we had to deal with forces then I would need to necessarily talk about the effect of charge #1 on charge #2. The field in contrast is just the effect of charge #1. On anything and everything. It might seem like a small distinction, but as you have noticed, this coupled with superposition is quite powerful, although they're not quite the same thing.

For part (b), suppose you have a charged particle $$Q$$ in an external electric field. If you are given the external field, then the field itself is a dynamical object that can interact with the charge $$Q$$. Knowing how the external field was produced will not help me answer what the effect of the field on $$Q$$ is. However if the field description is replaced by a force description, to know the force on $$Q$$ I will necessarily need to know the other charge - if for no other reason, but of a mathematical necessity thanks to Coulomb's law. This is important because complicated charge distributions can produce relatively simple electric fields based on the geometry of the distributions. And because Couloumb's law is valid for point charges, it is easier to work with an external field affecting a charged particle than to work with a charge distribution that causes the same field, interacting with a charged particle.

As an example, one can consider an infinite plane charge with a constant surface charge density. This configuration generates a constant electric field along the normal to the plane. Knowing this electric field, it is very easy to find out it's effect on a charge. However, it is much harder to work directly with the infinite plane charge and a single point charge since now, to find the force from Couloumb's law one has to write down and perform a nasty integral.

Within the realm of electrostatics these may seem like trivial/insignificant differences but working with fields and not knowing/caring about source charges unless one is explicitly interested in them/in how the field is generated is useful, because the notion of fields generalises to electrodynamics, and to many other concepts in physics.

• Can we not replace "field" with "force": calculate the FORCE due to a given distribution of charges without worrying about the effect these charges have on other charges in the vicinity? Since forces also obey the principle of superposition. Aug 25, 2021 at 7:58
• What you are saying is not wrong, and it is possible. However, it is computationally hard. The problem rapidly increases in computational difficulty based on the kind of charges and/or distributions one works with. Within the context of electrostatics the differences may seem small, like I mentioned. However, as one generalises to electrodynamics and beyond we see that electric charges (currents) can also generate magnetic fields and magnetic fields can cause electric fields. In such circumstances one finds using fields not just easier, but also more insightful. Aug 25, 2021 at 8:39
• In principle, if one accounts for the fact that the effects of charges travel with a finite velocity, one can do all of EM with force laws (Ex: physics.stackexchange.com/a/661329/221739). When we begin to use fields, we can also use potentials. And the reason we always define quantities per unit charge/mass is to describe/understand the effect for what it is, and not because of what caused it. The idea is to understand elements of the 'effects of charges' that are universal regardless of which charge caused it or how. In doing so, we understand how interactions and dynamics work. Aug 25, 2021 at 8:43
• What "other" charges does he mean when he says "without worrying about the effect these charges have on other charges"? Aug 25, 2021 at 8:55
• I cannot read the author's mind, but from my understanding this is the idea. If I have $q_A$ and $q_B$ and I am interested in the field generated by $q_A$ I can do so without worrying about what $q_B$ is doing. If I want to know the force exerted by $q_A$, that question is meaningless unless I specify $\textit{on what}$ the force is exerted, which is presumably $q_B$ in this case. Aug 25, 2021 at 9:02