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In a homogeneous Ohmic conductor, the charge continuity equation and Gauss's law imply exponential decay of the bulk charge density, $\rho (t)=\rho (0)e^{-\sigma t/ϵ}$. For an isolated finite body, bulk charge moves to its boundary; the accumulating surface charge density preserves total electric charge. The formula uses a constant permittivity $ϵ$ and the corresponding free-charge convention, or ${ϵ}_0$ for total charge in the vacuum form of Gauss's law.

Electrical conductivity $\sigma$ measures the local response of current density to electric field. In the rest frame of an isotropic Ohmic conductor, the local form of Ohm's law isLarger positive electrical conductivity gives larger current density at the same electric field. It is a bulk material coefficient, distinct from the geometry-dependent electrical conductance of a whole component and from a sheet's surface conductivity. Electrical resistivity is ${\rho }_{\mathrm{el}}=1/\sigma$ in this scalar model. ^[ref].

The local and whole-wire forms fit together particularly neatly. For uniform current in a straight homogeneous wire, with length $L$, wire cross-sectional area $A$, constant electrical conductivity and terminal electric potential difference $V$, the axial electric field has magnitude $E=V/L$. ThusFor the defined electrical resistance example $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$, $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$ and $V=5\mathrm{V}$, the local values are $E=5V/m$ and $j=2.5\times 10^{6}A/{\mathrm{m}}^2$; integrating gives $I=2.5\mathrm{A}$. One material law therefore supplies both a spatial description and a terminal relation.

The scalar form requires isotropy. In an anisotropic material the response can instead be a tensor, $J_i={\sum }_j{\sigma }_{ij}{E}_j$, and current density need not be parallel to electric field; uniaxial conductivity tensor is one existing example. These statements concern the material's rest frame. A moving electrical conductor instead uses the moving-conductor Ohm law.

An electrical insulator has sufficiently small transport of electric charge for its operating conditions. Insulation is a practical requirement involving electric potential difference, temperature, geometry and duration, rather than a universal claim of zero electrical conductivity. Electric polarization can be substantial in an electrical insulator; dielectric breakdown can end the insulating regime.

For an idealized uniform slab with small constant electrical conductivity $\sigma$, the Ohmic conductor model predicts leakage electrical resistance $R=d/(\sigma A)$ in parallel with capacitance $C=ϵA/d$. An isolated capacitor then hasThis is a derived model of slow discharge, not an assertion that every leakage mechanism obeys Ohm's law. Its time scale matches charge relaxation. The fixed-charge comparison in dielectric capacitor response needs an observation time short compared with $RC$, but long enough for electric polarization to settle. During a maintained voltage, leakage consumes power $V^2/R$ as Joule heating; delayed electric polarization can add dielectric loss under changing electric fields.^[ref]

Large relative permittivity alone does not rule out either mechanism.^[ref]

Electrical resistance measures the electric potential difference needed to drive a given electric current through a component. For an Ohmic conductor under fixed material conditions, $R=V/I$ is constant and Ohm's law gives $V=IR$. The SI unit is the ohm, $1\Omega =1V/A$. At a fixed positive electric potential difference, increasing electrical resistance reduces electric current. Electrical conductance $G=1/R$ expresses the same response reciprocally. ^[ref].

For a straight uniform wire at rest, with constant electrical conductivity $\sigma >0$, length $L$ and wire cross-sectional area $A$, the uniform Ohmic conductor model givesThe electrical resistivity ${\rho }_{\mathrm{el}}$ characterizes the material; electrical resistance also includes geometry. Doubling length doubles $R$, while doubling wire cross-sectional area halves $R$, with material and temperature fixed. The uniform-field assumption includes insulated sides and equipotential end faces, as explained in uniform current in a straight homogeneous wire. ^[ref].

As a defined wire example, take $L=1\mathrm{m}$, $A=1\mathrm{m}{\mathrm{m}}^2$ and $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$. Then $R=2\Omega$; applying electric potential difference $5\mathrm{V}$ drives electric current $2.5\mathrm{A}$. With the same length and material, a second wire of wire cross-sectional area $2\mathrm{m}{\mathrm{m}}^2$ has $R=1\Omega$ and carries $5\mathrm{A}$ at that same electric potential difference. Fixing electric potential difference here differs from fixing electric current in the current density comparison: stating what is held fixed is essential.

Joule heating is the dissipative conversion of electrical work into internal energy in an electrical conductor. In a metal, the electric field supplies energy to charge carriers, which transfer it to the material. This makes an immersed resistor a useful calibrated heater and makes electrical resistance a source of losses in wiring. Wind-driven electric currents are a proposed source of heating of an inflated giant planet.

For a stationary, purely resistive Ohmic conductor, use the passive sign convention: $V$ is the terminal electric potential difference from the entry to the exit of positive electric current $I$. The absorbed electrical power and Ohm's law giveThe last two expressions assume $V=IR$ at the relevant material temperature. For constant electrical resistance and electric current over a duration $\Delta t$, the dissipated energy is $E_{\mathrm{diss}}=I^{2}R\Delta t$; otherwise use $E_{\mathrm{diss}}=\int V(t)I(t)dt$. In SI, $\mathrm{V}\mathrm{A}=\mathrm{W}=J/s$.^[ref]

Absorbed electrical power $VI$ is more general than Joule heating: a capacitor can store energy, a battery can change its chemical energy, and an electrical circuit can deliver mechanical work or emit light. Even a heated resistor need not show an indefinitely rising temperature: in a thermal steady state, its dissipated power leaves by heat transfer. With the heater inside a calorimeter boundary, electrical input crosses as work; the subsequent heat transfer to the room is a different term in the first law of thermodynamics.

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.

Ohm's law states that an Ohmic conductor has electric current proportional to its applied electric potential difference, with constant electrical resistance under fixed material conditions:Here $V=\varphi (A)-\varphi (B)$ is the terminal electric potential difference, and positive $I$ enters at $A$ and exits at $B$. This sign convention makes $VI$ the electrical power absorbed. The straight-line relation through the origin lets one constant predict the response at many applied electric potential differences. Merely defining $R=V/I$ at one nonzero operating point does not establish Ohm's law: a component may have a ratio that varies with applied electric potential difference. Material and temperature must be specified and held fixed; many components have a nonlinear response. ^[ref].

For the ideal resistor model $R=2\Omega$ at electric potential difference $V=5\mathrm{V}$,This power is converted to internal energy by Joule heating; a watt is a joule per second. If the model holds for time $10\mathrm{s}$, the transferred energy is $125\mathrm{J}$. Doubling electric potential difference doubles electric current and quadruples power for the same fixed $R$. The model keeps material temperature fixed even while calculating Joule heating; appreciable changes in temperature may require a changing electrical resistance. ^[ref].

In 1827, Georg Ohm published his mathematical account of electrical transport in Die galvanische Kette, mathematisch bearbeitet. On printed page 36, his relation $S=A/L$ expresses electric current in terms of total driving tension and a reduced length incorporating electrical conductivity and geometry. This is historical notation: his $A$ is not the wire cross-sectional area used here, and his $L$ is not simply the wire's physical length. The striking idea is that material response and geometry can be combined into one quantity governing an electrical circuit. ^[ref].

In SI units, the two curl Maxwell equations areUse the vector identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ to obtainIntegration over the fixed volume and the divergence theorem give Poynting's theorem:The first term is the rate of change of stored electromagnetic energy. The surface integral is the outward energy flux carried by the Poynting vector. The right side is minus the power delivered by the field to matter; for an Ohmic conductor it is minus the Ohmic heating rate. Thus energy lost from the field either flows out or does work on charges.

For the specified transverse plane wave, both divergence Maxwell equations hold because $E_y$ depends only on $x,t$. Faraday's equation requiresHere we take the magnetic wave associated with the electric wave, without an additional static background field. Ampère's vacuum equation then gives $k^2={\mu }_0{ϵ}_0{\omega }^2$, equivalentlyThus all four vacuum Maxwell equations are satisfied. The Poynting vector is $\mathbf{P}=(k{E}_0^2/({\mu }_0\omega )){\cos }^2(kx-\omega t){\mathbf{e}}_x$, and averaging over a period givesFor propagation in the positive $x$ direction, $\omega =ck$ and this is $E_0^2{\mathbf{e}}_x/(2{\mu }_{0}c)$. It represents the mean energy transmitted per unit area per unit time, the wave's time-averaged energy flux per unit area, directed along its propagation.

Take the divergence of the Ampère-Maxwell equation, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{ϵ}_0{\partial }_t\mathbf{E}$. Since the divergence of a curl is zero vanishes, and Gauss's law gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the result isThis is the charge continuity equation.

At an internal point of a homogeneous Ohmic conductor, Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Since $\sigma$ is constant, the charge continuity equation becomes ${\partial }_t\rho =-(\sigma /{ϵ}_0)\rho$. ThereforeThis is charge relaxation. Here $\rho$ is the total charge appearing in the vacuum form of Gauss's law; in a homogeneous dielectric description of free charge the corresponding constant permittivity replaces ${ϵ}_0$.

For a finite isolated body, charge travels to the surface. The bulk formula applies at internal points, not across the interface where electrical conductivity jumps and surface charge density accumulates. Integrating the charge continuity equation over the body, including its surface charge, gives constant total electric charge because no electric current crosses the external boundary. Bulk decay and conservation of electric charge are therefore compatible.

Insert $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$ into the given identity. For a linear, time-independent, nondispersive medium, $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$, givingThe divergence theorem then yieldsThe left side is outward electromagnetic power flow. The current term is power delivered to matter, becoming Joule heating for an Ohmic conductor, and ${\partial }_{t}u$ is the change in stored electric plus magnetic energy. The minus sign says outward flux reduces stored energy unless supplied locally.

Follow the drawing's coordinates: lower strip at $z=0$, upper strip at $z=d$, current $I$ along $+x$ in the upper strip and back along the lower. The wording “both strips lie in the $xy$-plane” must mean parallel to that plane, since the drawing separates them in $z$. Neglect the side and end fringes using $l\gg a\gg d$. Then $I=V_0/(R+R_L)$ and the top-strip potential is $V(x)=V_0-IRx/l$, while the lower strip is equipotential at zero.

The harmonic function matching those two surfaces isIndeed ${\nabla }^2\varphi =0$, and it has the required potentials at $z=0,d$. ThusThe two current sheets reinforce their magnetic field between the strips; its direction follows the right-hand rule. The Poynting flow in resistive parallel strips follows from their cross product:Power flows along the gap towards the load and upwards into the resistive strip. Across a transverse area $ad$, $P_x(x)=IV(x)$ falls from $I{V}_0$ to $I^2{R}_L$. The upper surface receives ${\int }_0^{l}a{S}_z(z=d)dx=I^{2}R$, exactly the strip's heating; the ideal lower strip receives none. In the charge-free gap, ${\partial }_x{S}_x+{\partial }_z{S}_z=0$, consistent with a static field and no local dissipation. The small longitudinal electric field cannot be discarded if this upward energy transfer is to be represented.

In a homogeneous Ohmic conductor, a steady electric current density obeys $\nabla \cdot \mathbf{j}=0$ and $\nabla \times \mathbf{j}=0$. Uniform axial current in a straight cylinder follows if its side is insulated and its flat end faces are equipotentials. The electric potential then solves Laplace's equation with those boundary conditions, and uniqueness makes it linear in the axial coordinate. Steadiness alone does not impose these contact conditions: nonuniform end potentials can produce nonuniform current even with an insulating side. Bending also generally redistributes current; in an annular conductor the azimuthal current density is proportional to the inverse radius.

The wire cross-sectional area is the area of a slice perpendicular to the wire's axis. For a round wire of radius $a$, it is $A=\pi a^2$. It enters both the uniform-flow formula $I=jA$ for current density and the formula $R=L/(\sigma A)$ for the electrical resistance of a uniform Ohmic conductor.