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I have the following setup:

A <---- wire ----> B

$V_b - V_a = \Delta v = \text{a positive value}$

I have two questions:

  1. Which end of the wire has a higher potential?
  2. What is the direction of the electric field inside the wire?
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up vote 3 down vote accepted

From the given information that $V_b-V_a$ is positive, we know that $V_b>V_a$. The word "potential" means simply $V$, so point B is at a higher potential than point A.

Once you know that, you can go and remind yourself of the relationship between electric field and potential (no doubt it's in your textbook). That'll give you the answer to your second question.

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You dont understand beginners problems. And High or low is simply wron. -1 – Georg Feb 22 '11 at 15:26
+1 You are totally right. – user1355 Feb 22 '11 at 15:44
I have no idea what @Georg means. This is totally right. – spencer nelson Feb 22 '11 at 18:03

Obviously $V_b$ is at a higher potential as Ted already answered. The relation between field and potential is simple.

$\vec E = - \vec \nabla V$ In this one dimensional case it is $\vec E = - \frac {dV}{dx} \hat x $ where $\hat x$ is the unit vector along the positive x direction. So you see the field is in the $- x$ direction i.e. along $\vec {BA}$

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