3
$\begingroup$

In a circuit there are two forces that act on the charges to keep the current uniform through out,$\vec{f}=\vec{E}+\vec{f_s}$, where $\vec{E}$ is the electrostatic field and $\vec{f_s}$ is the electric field produced by chemical reactions. Inside an ideal battery, $\vec{E}$ and $\vec{f_s}$, oppose each other such that $\vec{f}$, is zero. What exactly would be the correct way of thinking of $\vec{f}=0$? I mean obviously the charges are moving, but it looks like in the battery they are not.

$\endgroup$
  • $\begingroup$ $\vec{f}_s$ isn't always an electric field produced by chemical reactions. It is just a force per unit charge produced by the battery. It could be a magnetic force, a chemical force, a mechanical force, etc. $\endgroup$ – Timaeus Feb 9 '15 at 13:52
1
$\begingroup$

In an ideal battery, there is no energy "loss" inside the battery during operation, but it is probably literally not the case that $\vec{f}=\vec{0}$ pointwise inside the battery, even in an ideal (no power loss) situation. Let's carefully review what happens inside a circuit and a battery in electrostatic equilibrium.

Force (per unit charge) is different than motion, you can have zero force, but still have motion, and we definitely have motion. But we also do actually have forces in some places. If you have a current, then there should be a force to make charges turn corners, you can get that from an electrostatic force that has some excess charge at the outside corner and opposite excess charge at the inside the corner, then the mobile charges whip around the corner. You might have one region where there are fewer mobile charges, but moving quicker, and before that you have a region with more charges moving slower, so you should have a charge imbalance in the between region to get that transition from slower to faster. All these things happen in order to even out the current so that charge stops piles up, so it piles up where it needs to be to produce to an equilibrium where it doesn't pile up any further. This equilibrium still has charges and they still have forces, but the forces just produce the motion that makes the steady current from the existing situations with the same existing steady current.

For instance you can have a rod moving orthogonal to itself and both those things (direction of the rod, and direction of the motion) mutually orthogonal to a $\vec{B}$-field. Then charge piles up at the ends (equal and opposite) until the electrostatic force cancels the magnetic force. That's equilibrium. If you connected the ends of the rod to a circuit, that equilibrium breaks as charge distributes itself so that the mobile charges get the forces they need to negotiate every corner and every speed up and every slow down. And then there is a new equilibrium, and that's what we want to analyze.

I think it's best not to think that $\vec{f}=\vec{0}$ everywhere inside the battery, after all, maybe the battery has some corners too. It's best to think about energy. So have an actual current flow into the battery for a small time $\Delta t$. We want an equal amount of charge to flow out for the same small time $\Delta t$ to avoid a pileup of charge inside the battery. So an equal amount of charge flows in and out, so an equal number of mobile charges (electrons) flow in and out. So the current is the average speed of this same number of electrons. So the work done from a to b should be zero, so while $\vec{f}\neq\vec{0}$ we do have $\int_a^b\vec{f}\cdot d\vec{\ell}=0$. Wait, where did the assumption of the ideal battery come in? A nonideal battery can have $\int_a^b\vec{f}\cdot d\vec{\ell}\neq 0$ if that net work done doesn't go to the kinetic energy of those same number of electrons entering/exiting the circuit. For instance, if the battery has some resistance, and hence some resistive losses due to joule heating.

Thus even in the ideal situation, $\vec{f}\neq\vec{0}$ can hold inside the battery (there might be corners for instance), but for the ideal situation $$\int_a^b\vec{f}\cdot d\vec{\ell}= 0.$$

And that result is exactly what we need in situations where people (sometimes wrongly) assume that $\vec{f}=\vec{0}$ inside the battery. For instance:

$$ 0=\int_a^b\vec{f}\cdot d\vec{\ell}= \int_a^b(\vec{f}_s+\vec{E})\cdot d\vec{\ell}= \int_a^b\vec{f}_s\cdot d\vec{\ell}+ \int_a^b\vec{E}\cdot d\vec{\ell}, $$

So $\int_a^b\vec{f}_s\cdot d\vec{\ell}=- \int_a^b\vec{E}\cdot d\vec{\ell}$, which is what we actually want and need to get, for instance:

$$V=-\int_a^b \vec{E} \cdot d\vec{\ell}=\int_a^b \vec{f}_s \cdot d\vec{\ell}=\oint \vec{f}_s \cdot d\vec{\ell}=\mathscr E.$$

Where the second to last equality comes from the fact that the battery only exerts a force inside the battery and the last equality is the definition of an emf.

It's the averaged result of $\vec{f}=\vec{0}$ that is used, and it holds for the average $\vec{f}$ (two things have the same spatial average if their integrals are the same). So $\vec{f}_s\neq -\vec{E}$ inside the battery, but if we integrate over the whole battery we do get $\int_a^b\vec{f}_s\cdot d\vec{\ell}=- \int_a^b\vec{E}\cdot d\vec{\ell}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.