# How should one interpret $\vec{f}=0$ in an ideal battery?

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1. Feb 1, 2015

### ghostfolk

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.

2. Feb 1, 2015

### Staff: Mentor

I don't think that this is a correct analysis, but f=0 would imply that the charge is not accelerating rather than that it is not moving.

3. Feb 1, 2015

### ghostfolk

I got it from Griffiths' Introduction to Electrodynamics. I know that would imply the charges are still moving, but how do I know that current is being produced?

4. Feb 1, 2015

### Staff: Mentor

Current = charges moving. They are the same thing. If you know that the charges are moving then you automatically know that there is current.

5. Feb 1, 2015

### ghostfolk

But if the net force on the charges are zero, shouldn't the charges move about randomly cancelling out any current?

6. Feb 1, 2015

### Staff: Mentor

No. If the net force is zero then they continue in a straight line at a constant speed.

7. Feb 1, 2015

### ghostfolk

So how do I know they started in a straight line?

8. Feb 1, 2015

### Svein

Yes, but the moment you close a circuit on the outside of the battery, the current reduces the charge and thereby the electric field. The net field will then be different from 0 and start pushing charges through the battery.

9. Feb 1, 2015

### ghostfolk

Ohh okay. You see in the textbook I was reading, it made no distinction as to whether or not the circuit was open.