# How can I tell which end of a wire will have a higher potential?

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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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.
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 $\vec {BA}$