Electric field in a wire Reading through my lecture notes I have written that the electric field $E$ drives a current $I$ around a wire such that $E = \frac{V}{L}$ where $L$ is the length of the wire and $V$ is the potential difference across the wire. Where does this come from?

 A: *

*Step 1 is to find the relation between the resistance $R$, the conductivity $\sigma$ of the material, and the cross-section of your wire.  

*Step 2 is to find the relation between the electric field and the current density $J$.  This involves the conductivity $\sigma$.

*Step 3 is to relate the current density $J$ to the net current $I$ in your wire.


If your text discusses Poynting vectors it should have all the above information as well.
A: Notice for constant E the equation is magnitude of potential difference V = EL. You get the form E = V/L. Also J=E/resistivity. Multiply this by are of wire on both sides, and the numerator and denominator of right side by L. Use areaxJ=I the current and area/(Lxresistivity)=1/R. You get the V=IR 
A: Imagine a wire with resistance $R$ and length $L$ that has a potential difference $V_{0}$ applied to it.  The potential at the positive end is $V_{0}$, and the potential at the negative end is zero.  Thus, one can formulate $V$ as a function of $x$, with $x$ being the distance from the "positive end" of the wire (set to zero) and ending at $L$, which is the length of the wire.  Because $V$ decreases linearly over $L$, $V(x) = V_{0} - \frac{V_{0}}{L}x$.  For example, if $V_{0} = 100$ V and $L = 10$ V, then $V(0) = 100$ V, $V(L) = 0$ V, and $V(L/2) = 50$ V (the last because at the half way point of the wire, the potential is \frac{1}{2} of what we started with).
Now, we know that $E = -dV/dx$.  Since $V$ is a function of $x$, we can differentiate $V$ with respect to $x$ and find that $E = \frac{V_{0}}{L}$.  Et viola.
