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The electric field does work on matter at local power per volumeHere $\mathbf{J}$ 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.^[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 isThe electrical resistivity ${\rho }_{\mathrm{el}}$ 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:As a numerical check, the wire already used in electrical resistance has $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$, $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$ and $V=5\mathrm{V}$. It gives $J=2.5\times 10^{6}A/{\mathrm{m}}^2$, $p_{\mathrm{vol}}=1.25\times 10^{7}W/{\mathrm{m}}^3$ 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.

The Poynting vector is the electromagnetic energy flux density,In vacuum it equals $\mathbf{E}\times \mathbf{B}/{\mu }_0$. For a linear nondispersive medium with time-independent constitutive coefficients, Maxwell's equations and the identity $\nabla \cdot (\mathbf{E}\times \mathbf{H})=\mathbf{H}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{H}$ give Poynting's theorem,The source term is work done by the fields on free charges; for an Ohmic conductor it is the positive local Joule heating rate $\sigma E^2$. Integrating over a volume states that loss of stored field energy plus outward energy flow equals energy transferred to matter. In dispersive media, such as a plasma, material motion contributes to stored energy and the simple nondispersive expression cannot be used indiscriminately.

For a vacuum plane wave, $B=E/c$ and $\mathbf{E}$, $\mathbf{H}$ and the propagation direction form a right-handed triad. With real peak field $E_0$, the time-averaged wave intensity isMore generally, complex peak amplitudes give $⟨\mathbf{S}⟩=\frac{1}{2}\mathrm{Re}({\mathbf{E}}_0\times {\mathbf{H}}_0^{\ast })$. Along a transmission line, the cross-section integral of the longitudinal Poynting vector is the transmitted electrical power $VI$. Energy flows through the electromagnetic fields in the surrounding dielectric, even though the conductors guide the currents.

Distinguish electrical current density ${\mathbf{J}}_e$ from heat flux ${\mathbf{J}}_Q$. Uniform current across the circular section givesby Ohm's law. The local Joule heating rate is electrical work per unit volume and time:For a segment of length $\ell$, integration gives $H\pi a^2\ell =I^2\ell /(\sigma \pi a^2)=I^{2}R$, confirming the usual resistor power. The model takes electrical conductivity constant and ignores temperature-induced current redistribution.