0
$\begingroup$

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.

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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 ..." $\endgroup$
    – garyp
    Commented Oct 21, 2021 at 18:43
2
$\begingroup$

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.

enter image description here

At curved surfaces, the charge distribution is more dense:

enter image description here

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

enter image description here

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.

enter image description here

Links to resources

The Electric Field Outside a Stationary Resistive Wire Carrying a Constant Current.

Voltage and Surface Charges. What Wilhelm Weber already knew 150 years ago.

Surface charges on circuit wires and resistors play three roles.

Veritasium video -1

Veritasium video -2

Improve this answer
$\endgroup$
0
$\begingroup$

A battery pulls electrons from its positive terminal and put them on the negative terminal. From there, they distribute themselves along the wire. A uniform wire must have a uniform flow of current. That requires a uniform electric field and a uniform charge gradient. So the wire does not “concentrate” a field from charges on the terminals, it is subject to the field from charges which are distributed along the wire. One end of the wire has an excess of electrons and the other end has a deficiency. I have read on this site (but have not verified) that in order to produce a uniform field inside the wire, these “extra” charges must reside at the surface of the wire. (Electrons do repel each other.) There will be a dipole like field outside of the wire, going from the positively charged part to the negative. (This is necessary to satisfy Gauss' law for short segments with a charge gradient.)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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