Why is there a voltage drop in a metal with no electric field present?

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NS1
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Metals are highly effective at screening electric fields. If we place two contacts reasonably far away from each other on a piece of metal and apply a voltage bias, the charge carriers in the section that is far enough from both the contacts should be unaffected by the electric field. Why then is the electrical resistance affected by the electric field between the contacts (through the electron mobility)? Why do we measure a voltage drop (e.g. in a four probe measurement) when there is no electric field, and the potential profile in the metal bulk should be flat?
 
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Thanks for the reply. In the Drude model the electric field is assumed to be constant and uniform. My question is why is this a good assumption, given the screening by the electrons in the contact region? Is this because the circuit is closed, and so there is no accumulation of the electrons at the contacts? Is correct to say that the electric field in a metal under bias decays linearly between the contact during DC conduction?
 
The Wiki says "The simplest analysis of the Drude model assumes that electric field E is both uniform and constant" - This is a model of the behavior and used for analysis. The "Constant and uniform" are the conditions set up in order to do the analysis. But this is looking at an E field only ( not a circuit).

IMO you are looking at two different things - in the case of screening E fields we are typically looking at a conductor floating in the presence of an E field. Here the free charges can move within the conductor - to effectively cancel out the externally applied field. Any internal field would result in more charge moving.

In the case of electrodes applied to a metal sheet - here we have a circuit, and a source / sink of electrons. Here, for BASIC DC analysis we need to assume a few things ( or account for things like the temperature impact on resistivity) - but since the electrons can not "accumulate" to negate the applied E field - you can and will set up an E field in the metal - you will have a voltage gradient. ( Note : it could be said that in an IDEAL conductor this will not happen because there is no resistivity in the material - but this is not a real world case, and in analysis does not lend itself to any better understanding)

In your OP you state "electrical resistance affected by the electric field" - as for this statement, believe you may be off - I do not know of any cases of this, while I do not know everything, I base this on the lack of devices that take advantage of this phenomena... it would seem to be very useful in sensors and other cases. But if we apply 10x the potential, the resistance does not change, perhaps in very high orders of magnitude there are some issues?? IDK... Can you point to an example where there is theory or real world case of this?
 
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Windadct said:
The Wiki says "The simplest analysis of the Drude model assumes that electric field E is both uniform and constant" - This is a model of the behavior and used for analysis. The "Constant and uniform" are the conditions set up in order to do the analysis. But this is looking at an E field only ( not a circuit)...
You're right, I was mixing two different cases (open vs. closed circuit). Regarding the effect on resistance, this was an editing error, caused by rewriting the question.
Thanks!