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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

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that's too advanced for me ... i'm studying electrostatics at the moment. could you reedt by removing that <- images -> below in my question?? –  hasExams Jul 6 '12 at 19:27
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edit with the images is on its way. Hold on while a moderator approves it. –  Emilio Pisanty Jul 6 '12 at 19:28
    
Thanks the second link was helpful and answers my question!! –  hasExams Jul 6 '12 at 19:50
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Please don't accept wrong answers to a question, it makes the site less usable for others. If you aren't sure about which answer is correct, ask for clarification. –  Ron Maimon Jul 8 '12 at 20:29
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up vote 6 down vote accepted

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.

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can you finally look at the edit?? –  hasExams Jul 6 '12 at 20:04
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@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. –  Alfred Centauri Jul 6 '12 at 22:59
    
@testuser I'm not exactly sure what you mean by the external current. –  Emilio Pisanty Jul 6 '12 at 23:27
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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). –  Emilio Pisanty Jul 7 '12 at 11:23
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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. –  Ron Maimon Jul 8 '12 at 6:37
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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.

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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) –  Satwik Pasani Jun 22 '13 at 15:35
    
Moreover, the distribution of these charges wouldnt change the potential difference accross the battery or whatever the original source of field was? –  Satwik Pasani Jun 22 '13 at 15:43
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@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. –  Ron Maimon Jun 22 '13 at 16:27
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Classical
You can find Electric Field by this formula: $$E=\frac {-dV}{dl}$$ with direction along length $l$. E is Electric Field and V is potential.

Modern
As Electric Field is vector and potential is scalar, you'd need Gradient operator (a vector operator which acts on scalar). Its perfect relationship is: $$\vec E= - \vec \nabla {V}$$


Lines of this Electric Field can be drawn along the length of conductor. And, it wouldn't be affected by external electrostatic field (answer of last question).

Explanation
Electrostatic Field can't exist in conductor due to electrostatic equilibrium. So, there wouldn't be potential difference between any point inside conductor from the formula $$dV= \vec E \cdot d \vec s$$ (because $\vec E$ is zero).
If you pick two points on surface, there's no chance of potential difference because surface of a conductor is equipotential surface. Mathematically, as $\vec E$ is perpendicular to surface, its projection (what's scalar multiplication all about) in the direction of $d \vec s$ would be Zero.

As there's no potential change due to this part, Electric Field along length of conductor would be intact.

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