# Electromotive force and electric field for batteries

There are really two forces involved in driving current around a circuit: the source $$\mathbf{f}_{s}$$ which is ordinarily confined to one portion of the loop (a battery, say), and an electrostatic force, which serves to smooth out the flow and communicate the influence of the source to distant parts of the circuit: $$\mathbf{f}=\mathbf{f}_{s}+\mathbf{E}$$ Griffiths introduction to electrodynamics-4.ed-page:304

What I am confused about is, what do $$\mathbf{f}_{s}$$ and $$\mathbf{E}$$ physically represent, and what are they in an idealized battery model? The word "smooth out" is used but I'm unable to deduce any meaning to it. Is there any direct relation to the physical battery and what exactly the driving force $$\mathbf{f}_{s}$$ in a battery?

• Closely related question here. Feb 2, 2021 at 17:03

In a battery mobile charged particles move against the macroscopic electric force due to electrostatic field. So there has to be other force there that push them. Such force per unit charge is usually called electromotive force, but that term is too general. It is more descriptive to call it "chemical electromotive forces", because they arise as a result of chemical reactions in the battery. There are other kinds of electromotive forces.

This electromotive force reach is limited to the internals of the battery. It can't push current in the rest of the circuit, in the wires.

Dynamic current in a wire is distributed through the cross-section evenly. Due to Ohm's law, such evenly distributed current means that electric field is the same throughout the cross-section. Surface charges on the battery and wires and other components distribute in such a way that electric field is constant on the cross-section.

Maybe this is what Griffiths means by "smoothing".

First things first, $$f_b$$ is not electrostatic in nature. Here is an elementary explanation of how a battery works, this might give you a feel of what $$f_b$$ is:

• A cell does not "produce" charge. It gives energy to the charge carriers ($$e^-$$) and they use this energy to run a load connected to the circuit.
• Inside a cell, an exothermic chemical reaction takes place and eventually develops charges on anode and cathode.
• An electric field is set up from positive diode to negative due to this reaction in electrolyte.
• This exothermic energy released forces the electron to move from POSITIVE to NEGATIVE diode and in so doing, increase its potential energy.
• It is this potential energy that the electron uses to run through the circuit.

So you see, the battery did exert a "force" on the electron to push it against it's own electric field. It is this force that is called the $$f_b$$.

What's funny is that the potential difference of the battery through which the electron moves to gain its energy is actually called the EMF - electromotive force (misnomer); it is not a force, it is potential difference. If you were to vectorise this potential difference and attribute a force to it, that would be $$f_b$$. Read the first sentence again.

Regarding the smoothening out, as Ján mentioned, it might refer to the current per cross-section area becoming constant throughout the wire while it is not so in the battery.

What I am confused about is, what do fs and E physically represent, and what are they in an idealized battery model?

The force $$\mathbf f_s$$ is more commonly called the electromotive force (EMF). It is the external energy which is provided to the circuit. For example, in a battery it is the energy from the chemical reaction that pushes charges "against" the direction that it would naturally want to move (e.g. according to Ohm's law). In a battery the EMF is highly localized, it only exists at the surfaces of the electrodes. In the rest of the circuit the force on a charge is provided by the E field.

FYI, my personal opinion is that I do not like the equation $$\mathbf f = \mathbf f_s + \mathbf E$$ because it invites confusion like yours. The symbol $$\mathbf f$$ usually refers to a force with SI units of N and the symbol $$\mathbf E$$ usually refers to an E field with SI units of V/m = N/C, and the EMF here represented by $$\mathbf f_s$$ usually has SI units of V = J/C. So I cannot see how an author could expect students to not come away confused from reading this equation.

Regarding the "smooths out" comment. That is actually talking about the effect of surface charges. Consider a simple battery connected by two wires to a resistor. Without the wires, the terminals of the battery are slightly charged and produce an electric dipole field which varies over space. This field, like any dipole field, falls off rather rapidly away from the battery. However, when we attach the wires they obtain a surface charge which makes it so that the dipole field does not fall off so rapidly. In fact, it brings the full voltage of the battery smoothly over to the resistor. In the absence of the smoothing effect of the wires the potential across the resistor is miniscule, but with it the full voltage is smoothly applied across the resistor.