I have the following setup:
A <---- wire ----> B
$V_b - V_a = \Delta v = \text{a positive value}$
I have two questions:
- Which end of the wire has a higher potential?
- What is the direction of the electric field inside the wire?
Tell me more
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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. 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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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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