= Local Joule heating rate
{title2=$p_{\rm vol}=\mathbf J\cdot\mathbf E$}
The <electric field> does <work> on matter at local <power> per <volume>
$$
p_{\rm vol}=\mathbf J\cdot\mathbf E,\qquad
[p_{\rm vol}]=\frac{\mathrm A}{\mathrm{m^2}}\frac{\mathrm V}{\mathrm m}
=\mathrm{W\,m^{-3}}.
$$
Here $\mathbf J$ is <current density> and the <dot product> includes the relative direction. Positive $\mathbf J\cdot\mathbf E$ means <energy> passes from the field to matter; negative means matter supplies <energy> to the field. In general this is a <work> rate, not automatically an irreversible heating rate.<The Feynman Lectures on Physics, Volume II by Richard Feynman and others>[§27–2, equations 27.3–27.5 and the intervening charge-force argument: work on matter per unit volume and its sign in the field-energy balance]{ref}
For a stationary <isotropic> <Ohmic conductor> with real scalar <electrical conductivity> $\sigma>0$, local <Ohm's law> gives $\mathbf J=\sigma\mathbf E$. Thus the <local Joule heating rate> is
$$
p_{\rm vol}=\sigma|\mathbf E|^2
=\frac{|\mathbf J|^2}{\sigma}
=\rho_{\rm el}|\mathbf J|^2\geq0,\qquad
\rho_{\rm el}=\frac1\sigma.
$$
The <electrical resistivity> $\rho_{\rm el}$ has units $\Omega\,\mathrm m$, and <electrical conductivity> has units $(\Omega\,\mathrm m)^{-1}$. These formulas use the material's rest frame and dissipative conduction <current density>. Moving media, <complex conductivity>, and additional storage or transport processes require their appropriate <energy> balance rather than this scalar substitution.
For <uniform current in a straight homogeneous wire> of <length> $L$, <wire cross-sectional area> $A$ and constant <electrical conductivity>, $E=V/L$, $J=I/A$ and $R=L/(\sigma A)$. The <volume integral> recovers terminal <power>:
$$
P=\int_{\text{wire}}\mathbf J\cdot\mathbf E\,d^3x
=\frac IA\frac VL(AL)=IV=I^2R.
$$
As a numerical check, the wire already used in <electrical resistance> has $L=1\,\mathrm m$, $A=10^{-6}\,\mathrm{m^2}$, $\sigma=5\times10^5\,(\Omega\,\mathrm m)^{-1}$ and $V=5\,\mathrm V$. It gives $J=2.5\times10^6\,\mathrm{A/m^2}$, $p_{\rm vol}=1.25\times10^7\,\mathrm{W/m^3}$ and $P=12.5\,\mathrm W$, agreeing with $I=2.5\,\mathrm A$ and $R=2\,\Omega$. Concentrating the same <electric current> in a smaller <wire cross-sectional area> therefore raises the local heating rate if <electrical resistivity> stays fixed.
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