10
$\begingroup$

Electrons move from higher potential to lower potential. When a conductor is connected to battery, electron move from negative terminal to positive terminal.

But the battery itself forms a Electric field like below enter image description here If a free electron were there on negative terminal it would follow electric field to the positive terminal.

My Question is how is Electric Field set up in the wire so that electrons pass from negative terminal to positive terminal in a conductor? Does it follow external electric field like below or set's up it's own electric field? enter image description here how can i visualize Electric Field inside conductor when Potential difference is applied across it. How can current be explained in terms of Electrostatics ( I hope it does not sound funny)

I think my problems is this http://www.physicsforums.com/showthread.php?t=159205

EDIT:: suppose two highly charges plates is connected by the conductor (that allows low current), Can i presume that the internal current inside conductor C is independent of external current? If the electric field is inside conductor, then electric field is in same shape as conductor. enter image description here

Improve this question
$\endgroup$
5

2 Answers 2

Reset to default
8
$\begingroup$

The wire sets up charges on the surface to channel the electric field exactly along the path of the wire. This isn't surprising, if the electric field doesn't exactly follow the path of the wire, charges are shunted to the surface, and these charges will then move the electric field so that it is parallel to the wire. The amount of charge required to do this shunting is tiny, it's a negligible capacitance of the wire that depends in a crazy nonlocal way on the shape of the wire, the type of battery, and the other conductors around.

But the answer is just yes: the metal conducts charges to the surface just until the electric field is going parallel to the wire along the entire length of the wire, no matter how many times it doubles back.

Improve this answer
$\endgroup$
3
  • $\begingroup$ This is understandable. But then whys is the field E constant along the conductor? Or is it not? (i know i am a bit late in asking the question considering the time of your comment but this has me bothered from quite a while) $\endgroup$
    – stochastic13
    Commented Jun 22, 2013 at 15:35
  • $\begingroup$ Moreover, the distribution of these charges wouldnt change the potential difference accross the battery or whatever the original source of field was? $\endgroup$
    – stochastic13
    Commented Jun 22, 2013 at 15:43
  • 2
    $\begingroup$ @SatwikPasani: Because the current is constant, and the voltage adjusts itself quickly so that the heat production from the current matches the potential drop--- it's the only self-consistent solution, and it's set up when you close the circuit at the speed of light. The distribution of charges doesn't change the voltage difference across the battery, because this is determined by the chemical reactions in the battery, this is the initial driver to set up the voltage differences in the wire. $\endgroup$
    – Ron Maimon
    Commented Jun 22, 2013 at 16:27
7
$\begingroup$

Electrons will flow against the electric field lines because their charge is negative, and the electric field thus exerts a force $\mathbf{F}=q\mathbf{E}$ on them which is in the opposite direction. Thus electric field lines inside the wire go from the positive to the negative terminal and the electron flow goes from the negative to the positive terminal. Electric current goes, consistently with both of the above (because the electron charge is negative), from the positive to the negative terminal.

The electric field lines will twist with the conductor if you bend it into some weird shape. (This is due to slight charge buildups on the wire bends and is beautifully explained by Purcell.) For the situation you describe, the electric field lines and the wire pretty much match already so just draw some more lines. You've already explained current flow in terms of electrostatics in a circuit like this! the only snag is what the state of affairs is inside the battery, but that's another story.

Improve this answer
$\endgroup$
10
  • $\begingroup$ can you finally look at the edit?? $\endgroup$
    – hasExams
    Commented Jul 6, 2012 at 20:04
  • 2
    $\begingroup$ @SS, you're only considering an ideal conductor. If there is a current in a non-ideal conductor, there must be an E field in the conductor. $\endgroup$
    – Alfred Centauri
    Commented Jul 6, 2012 at 22:59
  • $\begingroup$ @testuser I'm not exactly sure what you mean by the external current. $\endgroup$
    – Emilio Pisanty
    Commented Jul 6, 2012 at 23:27
  • 3
    $\begingroup$ Electrostatic fields do exist in this situation since all charge movements are slow and magnetic effects are negligible. Plus, the situation is static in the sense that although charges move, the electric field and the current do not change with time. Since the situation is independent of $c$, it has to be an electrostatic situation (which you can get from any electrodynamic situation by taking $c\rightarrow\infty$ and neglecting magnetic effects). $\endgroup$
    – Emilio Pisanty
    Commented Jul 7, 2012 at 11:23
  • 2
    $\begingroup$ This is only incorrect in claiming that the electric field around the wire will be negligible. On any path in the air from the positive to the negative terminus, the integrated electric field is the voltage. $\endgroup$
    – Ron Maimon
    Commented Jul 8, 2012 at 6:37

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