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

• ghostfolk

#### 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.

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.

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.
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?

I know that would imply the charges are still moving, but how do I know that current is being produced?
Current = charges moving. They are the same thing. If you know that the charges are moving then you automatically know that there is current.

Current = charges moving. They are the same thing. If you know that the charges are moving then you automatically know that there is current.
But if the net force on the charges are zero, shouldn't the charges move about randomly cancelling out any current?

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

No. If the net force is zero then they continue in a straight line at a constant speed.
So how do I know they started in a straight line?

Inside an ideal battery, E⃗ \vec{E} and fs→\vec{f_s}, oppose each other such that f⃗ \vec{f}, is zero.
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.

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