This has been a really great confusion for me now ....

Many places I have read in books that when a potential difference is applied across the ends of a wire a constant electric field is generated inside it which drives the current through it...

My question is how is this electric field generated? Why is it constant in magnitude throughout the wire? What is the mechanism of flow of current inside the wire?


What all things happen inside the conductor just after closing the switch and before the low of constant current through it?

  • $\begingroup$ physics.stackexchange.com/q/31482 $\endgroup$ – stochastic13 Oct 22 '13 at 14:03
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    $\begingroup$ If you want "detailed answers" you're going to have do better than "plz hlp me". $\endgroup$ – dfg Oct 22 '13 at 14:30
  • $\begingroup$ Consider a spring.If you drag and hold it,it will store the work done by you in dragging it, as potential energy at that point.If you leave the spring from the dragged position,the spring moves back to the initial position delivering the potential energy for its motion(In other words it converts potential energy into kinetic energy).Now you can imagine that,there is a energy difference between dragged position and equilibrium position.This energy difference is called potential difference..... $\endgroup$ – Immortal Player Oct 22 '13 at 16:05
  • $\begingroup$ [continuation of the previous comment]Similarly,when you make one end of the conductor to be at higher potential, as like spring in the dragged position,and the other end of the conductor to be at lower potential as like spring in undragged position(equilibrium position).In the way dragged spring tending to move to equilibrium position,the electrons in the conductor tend to move to lower potential energy state,this makes electron to move in the conductor to constitute the current(in simple words). $\endgroup$ – Immortal Player Oct 22 '13 at 16:24
  • $\begingroup$ its due to surface charge gradient . physics.stackexchange.com/questions/5277/… $\endgroup$ – user31782 Oct 27 '13 at 15:51

Actually the electric field inside the wire isn't the same, at least initially. This is what the skin effect is all about.

When you first apply a potential to a wire, the electric field is a thin shell only on the outside of the wire. This causes current to flow in a thin layer along the outer surface of the wire. Since the conducting material of the wire has some non-zero resistance, this current in the outer shell causes a voltage drop, which means a layer a little further into the wire will "see" the electric field, which causes current deeper into the wire, which causes a electric field deeper into the wire, etc. In steady state, the electric field inside the wire is uniform along the crossection of the wire.

The fact that it takes time before most of the material in a wire is conducting current means that the effective resistance of the wire starts out high and exponentially converges to its DC (steady state) resistance over time. Taking this further, it also means that the wire looks like a different resistance to different frequencies.

For any particular resistivity of the wire material and a particular frequency, a skin depth can be calculated. For external purposes, the wire can be thought of as being a hollow pipe with the thickness of the pipe wall being this skin depth. This can then be used to calculate the apparent resistance at a particular frequency.

This is a real issue, even for frequencies as low as 60 Hz. Power companies are well aware of this, which is one reason electric power cables are not made arbitrarily large. After the radius gets to around the skin depth, making the wire thicker yields less return for the amount of material used. This is why high current 60 Hz transmission lines have multiple cables for each phase (another reason is to decrease corona losses, but that's another topic).

This is one of the advantages of DC power lines. There are disadvantages to DC power transmission too, so it's usually done for long distances where the savings in cable material outweigh the other costs.

  • $\begingroup$ Thanks for the picture of electric field starting out at the surface and then being "seen" inside progressively. Can you elaborate on how this "shielding" takes place? $\endgroup$ – Milind R Sep 25 '15 at 14:31
  • $\begingroup$ @mili: I'm not really sure what to add. A inner chunk of wire can only see the voltage immediately across it. For there to be a voltage across its outer edge, the wire material on the other side of that edge must be conducting, else there would be no voltage by Ohm's law. One layer has to conduct before the next layer has a voltage drop across it, which then causes it to conduct. In reality there aren't discrete layers, so it is actually a exponential decay of the full current propagating into the interior of the wire. $\endgroup$ – Olin Lathrop Sep 25 '15 at 15:04

With no potential difference applied to the ends of the wire the electrons in the wire feel no net force because the environment around them is symmetric i.e. the density of other charges around them is the same in all directions.

When you apply a potential difference you inject electrons into the wire at the end connected to the anode and you withdraw electrons from the wire at the end connected to the cathode. This means the electrons near the ends are no longer in a symmetric environment and they feel a net force. Electrons near the anode end will flow away from the higher electron density near the anode, and electrons at the cathode end will flow towards the reduced electron density near the cathode. This wave of motion propagates through the wire at between a tenth and a third of the speed of light depending on the wire, so it quickly passes through the whole wire and all the electrons in the wire start moving.

It's very common to model electric circuits by water flowing through pipes. This is known as the hydraulic analogy. The hydraulic equivalent to your question is asking how water starts moving through a pipe when you attach a pump to it, and the answer is basically the same. The pump pushes water into the pipe, that water pushes along the adjacent bit of water and so on until all the water is flowing. The electrons in a metal wire behave the same way.

  • $\begingroup$ Can development of potential difference across a resistor be explained according to this ? $\endgroup$ – Shashaank May 20 '17 at 11:01

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