# Is it really true that "no charge builds up" in a circuit?

1. Jul 23, 2014

### confundido

This is a question we were asked quite some time ago:

Consider a circuit shaped as a rectangle, with two very long straight wires (parallel to one another), connected on one end by a battery and on the other end by a resistor.

We were asked to show--and this is not difficult--that if the resistance is large enough, the wires will attract rather than repel. (Due to the electrostatic force being stronger, in this case, than the magnetic force). In order for that to work out the wires have to actually be charged, not neutral. Is that the case? What is going on here? It has lead me to re-think my understanding of potential within a circuit altogether. I had previously thought of it as there being the same amount of charge on each side of the wire, but with each charge having V/q less energy of whatever type than the electrons on the other. I now think that is wrong. (Is it?)

Last edited: Jul 23, 2014
2. Jul 23, 2014

### Simon Bridge

A very large resister could be a break in the wire - then the two (real) wires are basically a capacitor. One wire would end up negative and the other positive.
Wouldn't the two wires attract then?

If the two wires are parallel, but the current is running in opposite directions, then wouldn't the magnetic force between them be repulsive?

I'm a little slow lately...

3. Jul 23, 2014

### Staff: Mentor

"No charge builds up in a circuit" can be worded more precisely as "the charge density at all points in the circuit is independent of time". It's true only if the system is static and has been allowed to settle down into a steady state, and it does not preclude different charge densities in different parts of the system (consider, for example, any system with non-zero capacitance somewhere).