# How does a copper wire encase electric field lines inside it while conducting current?

When we connect the positive and negative terminals of a cell with a copper wire, the electric field lines travel through the wire. What is it about the copper wire that it is able to encase almost all of the field lines originating from the positive terminal of the battery within itself?

P.S. After going through a few resources recommended by fellow users in answers to other similar questions on this site, I've compiled an answer to this question in hopes that it can be improved upon by your comments and edits.

the electric field lines travel through the wire and are almost null outside its surface.

This is incorrect. Copper is a very good conductor. Therefore the electric field inside the copper wire is very low in amplitude, and the electric field outside the wire is much stronger.

It guides the electric field by imposing a boundary condition (the tangential components of the field must go to zero) on the electric field at its surface. Physically this happens because charge can flow freely in the copper to terminate any electric field lines that reach the surface of the copper from the surrounding space.

• I don't think this is convincing. A wire with current flowing in it is not electrostatic, or even in equilibrium. It's not obvious that the field inside is very low in amplitude compared to the field outside. Simulations suggest the fields are the same order of magnitude. Also, I can't quite follow what you are trying to say in sentence starting "Physically this ..." Commented Oct 21, 2021 at 18:43

The field lines inside the wire run from the positive charges in the positive electrode to the negative charges in the negative electrode. The battery serves to maintain this charge density inside the electrodes constant.

In order to guide the field lines inside the wire, there must reside charges on the surface of the wire, distributed in a gradient.

At curved surfaces, the charge distribution is more dense:

There will be a dipole like field outside of the wire due to these surface charges as shown below.

Inside a resistor, the field must be greater than that in the conducting wire. This field is contributed by the surface charges in the wire-resistor interface.