# Forces on charges moving through battery

In the attachment, which comes from a text "Introduction to Electrodynamics" by Griffiths. There are two statements: "Within an ideal source of emf (resistenceless battery), the net force on the charges is zero, so $\vec{E} = -\vec{f}_s$" and further down at the bottom we have the statement "inside the battery $\vec{f}_s$ drives current in the direction opposite to $\vec{E}$". Would the correct interpretation of this be that the charge has no net force when it moves through the battery (since $\vec{E} = -\vec{f}_s$) and so essentially the charge is moving at a constant velocity within the battery, it entered the battery at an initial velocity from being driven by the electric field force from the electric field in the rest of the circuit (where in the circuit itself it has a net force $\vec{E}$ per unit charge)? Is this the correct interpretation? Or close?

Thanks.

Electrons do not move through batteries. Not in reality, and neither in this model, I assume.

in 7.9, an integral is taken over the whole (closed) loop. inside the ideal battery, $E=-f_s$. This is not the case in the entire external circuit (there you can only skip the contribution of $E$ because it is an integral over a closed loop). Consider $\int_a^b\mathbf{f}\cdot d\mathbf{l} = 0$ inside the battery. If you extend the integral to the whole loop, you get 7.9. So the emf is really between the two terminals over the external circuit.

The reason an ideal battery is mentioned is because in a real battery $|f_s|>|E|_{ab}$, because of the internal resistance.

• In this model it does assume that the electrons move through the battery, see highlighted sentence at the bottom which states "notice, however that in the battery $\vec{f}_s$ drives the current in the direction opposite to $\vec{E}$".
– Alex
Dec 7, 2016 at 14:00
• > "Electrons do not move through batteries." How can you know this? The whole path through the battery is conductive. It is most natural, in classical model of circuit and battery, to assume the electron that comes in through the positive terminal will get through the whole battery and reach the negative terminal after which it can go back into the wire again. Apr 7, 2022 at 23:57

To clarify this is the paper Griffiths was citing. It depicts Cd++ and Hd+ flowing towards positive terminals which is where “current is being driven in an opposite direction to E.”

I thought the original person who asked this question had a pretty good understanding except that it’s unnecessary to describe current as a single particle traveling through the entire circuit.

• Where can I find the complete abstract?
– user248666
Oct 4, 2020 at 11:00
• The paper isn't completely bad, but it argues against a misunderstanding of the concept of non-electric electromotive force intensity $\mathbf E_i$ inside electrochemical cell (or a thermocouple). It is not standard to say (the misunderstanding) that inside the battery, there are electro-chemical forces that would count as part of $\mathbf E$ and thus make the condition $\mathbf E\cdot \mathbf j < 0$ impossible. The standard account for centuries is that a non-electric force $\mathbf E_i$ is present in the cell/thermocouple where it counteracts the electric field $\mathbf E$. Apr 8, 2022 at 0:02
• Otherwise the paper shows an interesting example of how diffusion and restrictive environment (membrane) can create such macroscopic effective force $\mathbf E_i$, even if there is no special microscopic force besides collisions. Apr 8, 2022 at 0:05

The current is a complete circle. Complete the circle from the external circuit through the inside of the battery. There is supposed to be negligible charge buildup inside the battery or the reaction stops. The faster the current the larger the buildup. This is probably a factor as to why the electric car has been delayed for half a century as —moving— accelerating charges produce electromagnetic radiation