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Derivation

Poynting's Theorem and Field Energy

D-166 Home PU-204 Threads energy · fields · waves Depends on Assembly of Maxwell's Equations, lorentz-force-law
Statement

Starting from the microscopic Maxwell equations and the Lorentz force law, we derive the local conservation law for electromagnetic energy: the rate of change of the field energy density \( u = \tfrac{1}{2}\varepsilon_0 E^2 + \tfrac{1}{2\mu_0} B^2 \) plus the divergence of the Poynting vector \( \vec{S} = \tfrac{1}{\mu_0}\,\vec{E}\times\vec{B} \) equals minus the rate at which the fields do work on charges per unit volume, \( \partial u/\partial t + \nabla\cdot\vec{S} = -\,\vec{J}\cdot\vec{E} \). Both \(u\) and \(\vec{S}\) emerge from the derivation rather than being postulated.

Why it matters

Poynting's theorem promotes "energy is conserved" from a global bookkeeping statement to a local field equation of the continuity type: energy cannot vanish here and reappear there without a flux carrying it across every intermediate surface. It is the electromagnetic instance of the general pattern \( \partial_t(\text{density}) + \nabla\cdot(\text{current}) = -(\text{sink}) \) that recurs for charge, momentum, and probability, and it is what makes the claim "light carries energy" quantitative: the intensity of an electromagnetic wave is the time-averaged Poynting flux.

Practically, the theorem is how one traces power flow in circuits, waveguides, antennas, and radiating systems. It yields the surprising and instructive result that the energy dissipated in a resistive wire enters through the sides of the wire from the surrounding fields, not "through the wire" with the current — a corrective to the naive picture of electrons carrying energy like a conveyor belt.

Assumptions
Microscopic Maxwell equations hold in the region considered.If dropped: no field dynamics to work with; in macroscopic media one must instead use \( \vec{D}, \vec{H} \), and the identification of "field energy" versus "matter energy" becomes convention-dependent (Abraham–Minkowski-type ambiguities).
The work done on matter is given by the Lorentz force on the charges.If dropped: the sink term \( \vec{J}\cdot\vec{E} \) loses its meaning as mechanical power delivered to charges; for polarizable or magnetizable matter with hidden internal dynamics, extra energy channels appear.
Fields are differentiable (smooth enough for the vector identities used).If dropped: at surfaces of discontinuity (conductor boundaries, shock fronts) the differential form fails and one must use the integral form with pillbox/loop arguments and surface terms.
Vacuum constitutive relations, i.e. all charges — bound and free — appear explicitly in \( \rho \) and \( \vec{J} \).If dropped (macroscopic formulation): the theorem becomes \( \partial_t u_{\text{mac}} + \nabla\cdot(\vec{E}\times\vec{H}) = -\vec{J}_f\cdot\vec{E} \) with \( u_{\text{mac}} \) well defined only for linear, non-dispersive media; in dispersive media the stored-energy density must be replaced by the Brillouin expression involving \( \partial(\omega\varepsilon)/\partial\omega \).
The identification of \( \vec{S} \) is made by the minimal (curl-free ambiguity) choice.If dropped: \( \vec{S}' = \vec{S} + \nabla\times\vec{W} \) satisfies the same theorem for any field \( \vec{W} \); only the total flux through closed surfaces is fixed by this derivation. Relativistic consistency (the symmetric stress–energy tensor, \( \vec{S} = c^2 \times \) momentum density) selects the standard choice.
Derivation
1
\[ \frac{dW}{dt} = \vec{F}\cdot\vec{v} = q\left(\vec{E} + \vec{v}\times\vec{B}\right)\cdot\vec{v} = q\,\vec{E}\cdot\vec{v} \]
Power delivered to a point charge by the Lorentz force; the magnetic term drops because \( (\vec{v}\times\vec{B})\cdot\vec{v} = 0 \) — magnetic forces do no work. Uses the prior Lorentz-force result. B
2
\[ \frac{dW}{dt} = \int_{V} \vec{E}\cdot\vec{J}\; d^3r \]
Sum over a continuous distribution: with charge density \( \rho \) and velocity field \( \vec{v} \), the current density is \( \vec{J} = \rho\vec{v} \), so \( q\vec{E}\cdot\vec{v} \to \vec{E}\cdot\vec{J} \) per unit volume. This is the rate at which fields transfer energy to matter inside \( V \). A
3
\[ \vec{J} = \frac{1}{\mu_0}\,\nabla\times\vec{B} - \varepsilon_0\,\frac{\partial \vec{E}}{\partial t} \]
Solve the Ampère–Maxwell law \( \nabla\times\vec{B} = \mu_0\vec{J} + \mu_0\varepsilon_0\,\partial_t\vec{E} \) for \( \vec{J} \): we want the work rate expressed purely in fields, so the sources must be eliminated using Maxwell's equations. B
4
\[ \vec{E}\cdot\vec{J} = \frac{1}{\mu_0}\,\vec{E}\cdot\left(\nabla\times\vec{B}\right) - \varepsilon_0\,\vec{E}\cdot\frac{\partial \vec{E}}{\partial t} \]
Dot Step 3 with \( \vec{E} \); distributivity of the dot product. A
5
\[ \nabla\cdot\left(\vec{E}\times\vec{B}\right) = \vec{B}\cdot\left(\nabla\times\vec{E}\right) - \vec{E}\cdot\left(\nabla\times\vec{B}\right) \]
Standard vector identity, provable componentwise with the Levi-Civita symbol: \( \partial_i(\epsilon_{ijk}E_j B_k) = \epsilon_{ijk}(\partial_i E_j)B_k + \epsilon_{ijk}E_j(\partial_i B_k) \), and reindexing gives the two curl terms with the stated signs. Valid where \( \vec{E},\vec{B} \) are differentiable. A
6
\[ \vec{E}\cdot\left(\nabla\times\vec{B}\right) = -\,\vec{B}\cdot\frac{\partial \vec{B}}{\partial t} - \nabla\cdot\left(\vec{E}\times\vec{B}\right) \]
Substitute Faraday's law \( \nabla\times\vec{E} = -\,\partial_t\vec{B} \) into the identity of Step 5 and rearrange. This is where the second Maxwell curl equation enters — both dynamical equations are needed. B
7
\[ \vec{E}\cdot\frac{\partial \vec{E}}{\partial t} = \frac{1}{2}\frac{\partial}{\partial t}\left(E^2\right), \qquad \vec{B}\cdot\frac{\partial \vec{B}}{\partial t} = \frac{1}{2}\frac{\partial}{\partial t}\left(B^2\right) \]
Chain rule on \( E^2 = \vec{E}\cdot\vec{E} \) (and likewise for \( \vec{B} \)); this recognizes the time-derivative terms as perfect time derivatives — the signature of a stored-energy density. A
8
\[ \vec{E}\cdot\vec{J} = -\,\frac{\partial}{\partial t}\left(\frac{\varepsilon_0 E^2}{2} + \frac{B^2}{2\mu_0}\right) - \nabla\cdot\left(\frac{\vec{E}\times\vec{B}}{\mu_0}\right) \]
Insert Steps 6 and 7 into Step 4 and collect terms. Defining \( u \equiv \tfrac{1}{2}\varepsilon_0 E^2 + \tfrac{1}{2\mu_0}B^2 \) and \( \vec{S} \equiv \tfrac{1}{\mu_0}\vec{E}\times\vec{B} \), this is the theorem in differential form. A
9
\[ \frac{d}{dt}\int_V u\; d^3r + \oint_{\partial V} \vec{S}\cdot d\vec{a} = -\int_V \vec{E}\cdot\vec{J}\; d^3r \]
Integrate over a fixed volume \( V \) and apply the divergence theorem to the \( \nabla\cdot\vec{S} \) term (legal for piecewise-smooth \( \partial V \) and differentiable \( \vec{S} \)). Field energy inside \( V \) changes only by flux through the boundary or by work done on charges — a strictly local ledger. A
Result
\[ \frac{\partial u}{\partial t} + \nabla\cdot\vec{S} = -\,\vec{J}\cdot\vec{E}, \qquad u = \frac{\varepsilon_0 E^2}{2} + \frac{B^2}{2\mu_0}, \qquad \vec{S} = \frac{\vec{E}\times\vec{B}}{\mu_0} \]

Reading. At every point of space, the electromagnetic energy density \(u\) can change in exactly two ways: energy can flow away as the flux \( \vec{S} \) (energy per unit area per unit time, directed along \( \vec{E}\times\vec{B} \)), or it can be handed to matter at the rate \( \vec{J}\cdot\vec{E} \) per unit volume (Joule heating when positive, generator action when negative). With no charges present, \( \partial_t u + \nabla\cdot\vec{S} = 0 \) is a strict continuity equation: field energy is locally conserved and transported.

Units check. \( [\varepsilon_0 E^2] = (\mathrm{C^2\,N^{-1}\,m^{-2}})(\mathrm{V^2\,m^{-2}}) = \mathrm{J\,m^{-3}} \) and \( [B^2/\mu_0] = \mathrm{T^2}/(\mathrm{T\,m\,A^{-1}}) = \mathrm{T\,A\,m^{-1}} = \mathrm{J\,m^{-3}} \), so \(u\) is an energy density. \( [S] = [E][B]/[\mu_0] = (\mathrm{V\,m^{-1}})(\mathrm{T})/(\mathrm{T\,m\,A^{-1}}) = \mathrm{V\,A\,m^{-2}} = \mathrm{W\,m^{-2}} \), a power flux. Finally \( [\vec{J}\cdot\vec{E}] = (\mathrm{A\,m^{-2}})(\mathrm{V\,m^{-1}}) = \mathrm{W\,m^{-3}} \): every term of the theorem is \( \mathrm{W\,m^{-3}} \). Consistent.

Limiting cases
  • Statics (\( \partial_t = 0 \), steady currents): \( \nabla\cdot\vec{S} = -\vec{J}\cdot\vec{E} \) — the Poynting flux converging on a resistor exactly supplies its Ohmic dissipation; nothing accumulates.
  • Source-free vacuum (\( \vec{J} = 0 \)): \( \partial_t u + \nabla\cdot\vec{S} = 0 \), the pure continuity equation obeyed by radiation.
  • Plane wave: with \( B = E/c \) and \( \vec{E}\perp\vec{B}\perp\hat{k} \), the electric and magnetic energy densities are equal, \( u = \varepsilon_0 E^2 \), and \( |\vec{S}| = c\,u \) — energy rides along at speed \(c\), and the time-averaged \( \langle S\rangle = \tfrac{1}{2}\varepsilon_0 c E_0^2 \) is the intensity.
  • Static crossed fields (e.g. a charged capacitor sitting in a magnet): \( \vec{S}\neq 0 \) but \( \nabla\cdot\vec{S} = 0 \) — circulating energy flux with no net transport into any closed region; only the divergence is physical here (though the circulating \( \vec{S} \) does encode real field angular momentum).
  • Ideal conductor boundary: \( \vec{E}_\parallel = 0 \) forces \( \vec{S} \) tangent-normal component to vanish — no energy enters a perfect conductor, consistent with zero dissipation.
Breaks when
  • Dispersive or lossy media described macroscopically. If one uses \( \vec{D} = \varepsilon(\omega)\vec{E} \), the candidate \( u = \tfrac{1}{2}\vec{E}\cdot\vec{D} + \tfrac{1}{2}\vec{B}\cdot\vec{H} \) is not the stored energy: for a quasi-monochromatic field the correct density is the Brillouin form \( u = \tfrac{1}{4}\frac{\partial(\omega\varepsilon)}{\partial\omega}|\vec{E}|^2 + \tfrac{1}{4}\frac{\partial(\omega\mu)}{\partial\omega}|\vec{H}|^2 \), and when \( \operatorname{Im}\varepsilon \neq 0 \) no purely field-based \(u\) exists at all — energy is irreversibly shared with the medium's internal degrees of freedom.
  • Point charges and self-fields. For a strictly point charge, \( u \) diverges (the classical self-energy \( \int u\,d^3r \to \infty \)), and applying the theorem to a charge's own field requires renormalizing the self-energy; naive application to radiation reaction produces the pathologies of the Abraham–Lorentz force.
  • Quantum regime. At the level of single photons, \( \vec{S}(\vec{r},t) \) is an operator; its expectation obeys the theorem, but "the energy is located at \( \vec{r} \)" is no longer a sharp classical statement — photodetection statistics, not the classical flux, are fundamental.
  • Surfaces of discontinuity. The differential form presumes differentiable fields; across an interface one must use the integral form, and idealized boundary conditions (perfect conductors) hide the thin layer where dissipation actually occurs.
Failure modes
  • Assigning work to the magnetic field. Writing the sink term as \( \vec{J}\cdot(\vec{E} + \vec{v}\times\vec{B}) \) — the magnetic contribution is identically zero (Step 1); all energy transfer to charges is electric.
  • Factor-of-two slips in the densities. Writing \( u = \varepsilon_0 E^2 + B^2/\mu_0 \) (dropping the halves), or conversely halving \( \vec{S} \). The halves come from the perfect-derivative step \( \vec{E}\cdot\partial_t\vec{E} = \tfrac12\partial_t E^2 \); \( \vec{S} \) has no half.
  • Confusing \( \langle S\rangle \) with \( S_{\text{peak}} \) for waves. The intensity is \( \tfrac{1}{2}\varepsilon_0 c E_0^2 \), not \( \varepsilon_0 c E_0^2 \); the time average of \( \cos^2 \) contributes the \( \tfrac12 \).
  • Reading a nonzero \( \vec{S} \) as necessarily "energy going somewhere." In static crossed fields \( \vec{S} \) circulates with zero divergence; only \( \oint \vec{S}\cdot d\vec{a} \) over a closed surface is fixed by the theorem.
  • Sign errors from Faraday's law. Dropping the minus sign in \( \nabla\times\vec{E} = -\partial_t\vec{B} \) flips the magnetic energy term's sign and destroys the perfect-derivative structure — a good self-check: if \( u \) doesn't come out positive-definite, revisit Step 6.
  • Treating \( \vec{J}\cdot\vec{E} \) as always dissipative. In a battery or generator region \( \vec{J}\cdot\vec{E} < 0 \): matter does work on the field, sourcing the outward Poynting flux that feeds the rest of the circuit.
Discussion

The theorem's deepest lesson is where the energy is. Circuit theory books talk of energy "in" capacitors and inductors; Poynting's theorem says precisely that the energy resides in the fields occupying the space between plates and inside windings, at density \( \tfrac{1}{2}\varepsilon_0 E^2 + \tfrac{1}{2\mu_0}B^2 \). In a DC circuit, energy leaves the battery through the surrounding fields — guided by the surface charges that shape \( \vec{E} \) outside the wires — and converges on the resistor from the outside. The wires act as guides for the flux, not pipes for the energy.

Structurally, Poynting's theorem is one row of a larger conservation package. Momentum conservation gives the companion statement \( \partial_t \vec{g} + \nabla\cdot(-\overset{\leftrightarrow}{T}) = -(\rho\vec{E} + \vec{J}\times\vec{B}) \) with momentum density \( \vec{g} = \varepsilon_0\,\vec{E}\times\vec{B} = \vec{S}/c^2 \) and the Maxwell stress tensor \( \overset{\leftrightarrow}{T} \). That \( \vec{g} = \vec{S}/c^2 \) is no accident: it is the electromagnetic instance of the relativistic relation between energy flux and momentum density, and it is what makes radiation pressure (\( P = \langle S\rangle/c \) for absorption) an immediate corollary.

The derivation also illustrates a recurring theme: conservation laws are not extra postulates but consequences of the dynamical equations. Both curl equations were needed — Ampère–Maxwell to eliminate the source, Faraday to close the perfect derivative. Drop Maxwell's displacement-current term and the derivation collapses: no \( \tfrac{1}{2}\varepsilon_0 E^2 \) appears and the bookkeeping fails for time-dependent fields. Historically this consistency of the energy ledger was strong internal evidence for the displacement current itself.

At the relativistic level, \( u \) and \( \vec{S} \) are components of the symmetric stress–energy tensor: \( T^{00} = u \), \( T^{0i} = S^i/c \), and Poynting's theorem is the \( \nu = 0 \) component of \( \partial_\mu T^{\mu\nu} = -F^{\nu\lambda}J_\lambda/c \) (in Gaussian-like index conventions). This is the proper resolution of the ambiguity noted in the assumptions: while \( \vec{S} \to \vec{S} + \nabla\times\vec{W} \) leaves the theorem intact, only the standard choice makes \( T^{\mu\nu} \) symmetric and gauge invariant, which in turn is required for local angular-momentum conservation and for the fields to gravitate correctly in general relativity. Via Noether's theorem, the whole package expresses the invariance of the electromagnetic action under time and space translations.

Common misconceptions. (i) "The Poynting vector is only meaningful for waves" — it is defined for any field configuration, including statics. (ii) "Energy flows along wires with the current" — the drift velocity is \( \sim 10^{-4}\,\mathrm{m/s} \); the energy travels in the fields at nearly \(c\), entering devices sideways. (iii) "\( \vec{S} \) at a point is measurable" — only surface integrals of \( \vec{S} \) enter the physics at this level; pointwise values carry a curl ambiguity fixed only by relativistic consistency.

Worked examples

Example 1 — Where the Joule heat enters a resistive wire. A straight wire of radius \( a = 1.0\,\mathrm{mm} \) and length \( L = 1.0\,\mathrm{m} \) has resistance \( R = 0.050\,\Omega \) and carries a steady current \( I = 10\,\mathrm{A} \). Verify Poynting's theorem by computing the field-energy flux through the wire's surface.

1
\[ E = \frac{V}{L} = \frac{IR}{L} \]
Inside and just outside the wire surface, the axial electric field driving the current is the voltage drop per unit length (uniform for a homogeneous ohmic wire). Symbols first. A
2
\[ B(a) = \frac{\mu_0 I}{2\pi a} \]
Ampère's law on a circle of radius \(a\): the azimuthal magnetic field at the wire's surface. B
3
\[ \vec{S} = \frac{\vec{E}\times\vec{B}}{\mu_0} \;\Rightarrow\; S = \frac{E\,B(a)}{\mu_0} = \frac{IR}{L}\cdot\frac{I}{2\pi a} \quad \text{directed radially inward} \]
\( \vec{E} \) is axial, \( \vec{B} \) azimuthal, so \( \vec{E}\times\vec{B} \) points radially into the wire everywhere on its surface. A
4
\[ P_{\text{in}} = S \cdot 2\pi a L = \frac{I R}{L}\cdot\frac{I}{2\pi a}\cdot 2\pi a L = I^2 R \]
Multiply the (uniform) inward flux by the lateral surface area; the geometric factors cancel symbolically before any numbers enter — the answer is exactly the Ohmic dissipation. A
5
\[ E = \frac{(10)(0.050)}{1.0} = 0.50\,\mathrm{V\,m^{-1}}, \qquad B(a) = \frac{(4\pi\times10^{-7})(10)}{2\pi(1.0\times10^{-3})} = 2.0\times10^{-3}\,\mathrm{T} \]
Insert numbers with SI units. A
6
\[ S = \frac{(0.50)(2.0\times10^{-3})}{4\pi\times10^{-7}} \approx 7.96\times10^{2}\,\mathrm{W\,m^{-2}}, \qquad P_{\text{in}} = (7.96\times10^{2})(2\pi\times10^{-3}) \approx 5.0\,\mathrm{W} \]
Evaluate the flux and total inflow; compare with \( I^2R = (10)^2(0.050) = 5.0\,\mathrm{W} \). A
\[ P_{\text{in}} = I^2 R = 5.0\,\mathrm{W} \]

Reading. Every watt dissipated in the wire flows in radially through its cylindrical surface from the surrounding fields. The current sets up the geometry; the field delivers the energy.

Units check. \( \mathrm{(W\,m^{-2})\times m^2 = W} \); and \( \mathrm{A^2\,\Omega = A^2\,V\,A^{-1} = W} \). Consistent.

Example 2 — Sunlight as a Poynting flux. The solar irradiance above Earth's atmosphere is \( \langle S\rangle = 1361\,\mathrm{W\,m^{-2}} \). Treating sunlight as a plane wave, find the peak field amplitudes \( E_0, B_0 \) and the mean energy density.

1
\[ \langle S \rangle = \frac{\langle E B\rangle}{\mu_0} = \frac{E_0 B_0}{2\mu_0} = \frac{1}{2}\,\varepsilon_0 c\, E_0^2 \]
For a plane wave \( B_0 = E_0/c \) and the time average of \( \cos^2 \) is \( \tfrac12 \); use \( 1/(\mu_0 c) = \varepsilon_0 c \). B
2
\[ E_0 = \sqrt{\frac{2\langle S\rangle}{\varepsilon_0 c}}, \qquad B_0 = \frac{E_0}{c}, \qquad \langle u \rangle = \frac{\langle S\rangle}{c} \]
Rearrange for the amplitudes before inserting numbers; the last relation is the plane-wave limiting case \( |\vec{S}| = c\,u \). A
3
\[ E_0 = \sqrt{\frac{2(1361)}{(8.854\times10^{-12})(2.998\times10^{8})}} = \sqrt{1.025\times10^{6}} \approx 1.01\times10^{3}\,\mathrm{V\,m^{-1}} \]
Numerical evaluation; \( \varepsilon_0 c = 2.654\times10^{-3}\,\mathrm{\Omega^{-1}} \) (the vacuum admittance). A
4
\[ B_0 = \frac{1.01\times10^{3}}{2.998\times10^{8}} \approx 3.4\times10^{-6}\,\mathrm{T}, \qquad \langle u\rangle = \frac{1361}{2.998\times10^{8}} \approx 4.5\times10^{-6}\,\mathrm{J\,m^{-3}} \]
The magnetic amplitude is tiny in tesla, yet carries exactly half of \( \langle u\rangle \) — equality of electric and magnetic energy in a plane wave. A
\[ E_0 \approx 1.0\times10^{3}\,\mathrm{V\,m^{-1}}, \quad B_0 \approx 3.4\,\mu\mathrm{T}, \quad \langle u\rangle \approx 4.5\times10^{-6}\,\mathrm{J\,m^{-3}} \]

Reading. Full sunlight corresponds to a kilovolt-per-metre-scale oscillating electric field and a magnetic field an order of magnitude smaller than Earth's; a cubic metre of sunlit space holds only microjoules, but it is replenished at speed \(c\), giving \( 1.36\,\mathrm{kW} \) through every square metre.

Units check. \( \sqrt{\mathrm{(W\,m^{-2})/(\Omega^{-1})}} = \sqrt{\mathrm{W\,\Omega\,m^{-2}}} = \sqrt{\mathrm{V^2\,m^{-2}}} = \mathrm{V\,m^{-1}} \); \( \mathrm{(W\,m^{-2})/(m\,s^{-1}) = J\,m^{-3}} \). Consistent.

Problems
  1. A \( 5.0\,\mathrm{mW} \) laser pointer produces a uniform beam of diameter \( 1.0\,\mathrm{mm} \). Find the intensity \( S \), the peak electric field \( E_0 \), the peak magnetic field \( B_0 \), and the time-averaged energy density in the beam.
    Solution Beam area: \( A = \pi r^2 = \pi (0.50\times10^{-3})^2 = 7.85\times10^{-7}\,\mathrm{m^2} \). Intensity: \( S = P/A = (5.0\times10^{-3})/(7.85\times10^{-7}) = 6.37\times10^{3}\,\mathrm{W\,m^{-2}} \). Peak field: \( E_0 = \sqrt{2S/(\varepsilon_0 c)} = \sqrt{2(6.37\times10^{3})/(2.654\times10^{-3})} = \sqrt{4.80\times10^{6}} = 2.19\times10^{3}\,\mathrm{V\,m^{-1}} \). Magnetic amplitude: \( B_0 = E_0/c = (2.19\times10^{3})/(3.00\times10^{8}) = 7.3\times10^{-6}\,\mathrm{T} \). Mean energy density: \( \langle u\rangle = S/c = (6.37\times10^{3})/(3.00\times10^{8}) = 2.1\times10^{-5}\,\mathrm{J\,m^{-3}} \). Note the pocket laser's intensity exceeds sunlight's by a factor of about 4.7 — hence the eye hazard despite the tiny total power.
  2. A parallel-plate capacitor with circular plates of radius \( R = 5.0\,\mathrm{cm} \) and separation \( d = 2.0\,\mathrm{mm} \) is charging. At one instant the (uniform) field between the plates is \( E = 5.0\times10^{5}\,\mathrm{V\,m^{-1}} \) and is rising at \( dE/dt = 1.0\times10^{12}\,\mathrm{V\,m^{-1}\,s^{-1}} \). Compute the Poynting flux at the rim (\( r = R \)) and the total power flowing in through the cylindrical side surface, and verify it equals \( dU/dt \) of the stored energy.
    Solution Induced magnetic field at the rim (Ampère–Maxwell with no conduction current between plates): \( B(R) = \tfrac{1}{2}\mu_0\varepsilon_0 R\, dE/dt = \dfrac{R}{2c^2}\dfrac{dE}{dt} = \dfrac{(0.050)(1.0\times10^{12})}{2(8.99\times10^{16})} = 2.78\times10^{-7}\,\mathrm{T} \), azimuthal. Poynting flux at the rim: \( S = EB/\mu_0 = (5.0\times10^{5})(2.78\times10^{-7})/(1.257\times10^{-6}) = 1.11\times10^{5}\,\mathrm{W\,m^{-2}} \), directed radially inward (axial \( \vec{E} \), azimuthal \( \vec{B} \)). Power in: \( P = S\cdot 2\pi R d = (1.11\times10^{5})(2\pi)(0.050)(2.0\times10^{-3}) = 69.5\,\mathrm{W} \). Stored-energy rate: \( U = \tfrac12\varepsilon_0 E^2 \pi R^2 d \Rightarrow dU/dt = \varepsilon_0 E \dfrac{dE}{dt}\pi R^2 d = (8.854\times10^{-12})(5.0\times10^{5})(1.0\times10^{12})\pi(0.050)^2(2.0\times10^{-3}) = 69.5\,\mathrm{W} \). The two agree: the capacitor's energy enters sideways through the fringing region, not down the wires.
  3. A coaxial cable (inner radius \( a = 1.0\,\mathrm{mm} \), outer radius \( b = 3.0\,\mathrm{mm} \), perfect conductors) carries DC power from a \( V = 12\,\mathrm{V} \) source at current \( I = 2.0\,\mathrm{A} \). In the gap, \( E = \dfrac{V}{r\ln(b/a)}\,\hat{r} \) and \( B = \dfrac{\mu_0 I}{2\pi r}\,\hat{\phi} \). Show by explicit integration that the total Poynting flux down the cable equals \( VI \), and evaluate \( S \) at the inner conductor's surface.
    Solution Poynting vector: \( \vec{S} = \dfrac{\vec{E}\times\vec{B}}{\mu_0} = \dfrac{VI}{2\pi r^2 \ln(b/a)}\,\hat{z} \) — axial, in the direction of power flow, nonzero only in the gap \( a < r < b \). Total power: \( P = \displaystyle\int_a^b \frac{VI}{2\pi r^2\ln(b/a)}\,2\pi r\,dr = \frac{VI}{\ln(b/a)}\int_a^b \frac{dr}{r} = \frac{VI}{\ln(b/a)}\ln\frac{b}{a} = VI = (12)(2.0) = 24\,\mathrm{W} \). At \( r = a \): \( S(a) = \dfrac{(12)(2.0)}{2\pi (1.0\times10^{-3})^2 \ln 3} = \dfrac{24}{2\pi(1.0\times10^{-6})(1.099)} = 3.5\times10^{6}\,\mathrm{W\,m^{-2}} \). All 24 W travels through the dielectric gap; the perfect conductors carry none of it (\( \vec{E} = 0 \) inside them, so \( \vec{S} = 0 \) there).
  4. A long solenoid (\( n = 1000 \) turns per metre, radius \( a = 2.0\,\mathrm{cm} \)) carries a current being ramped at \( dI/dt = 100\,\mathrm{A\,s^{-1}} \); at the instant considered \( I = 5.0\,\mathrm{A} \). Compute the Poynting flux at \( r = a \) (just inside the windings) and show the total inflow per unit length equals the rate of change of stored magnetic energy per unit length.
    Solution Interior field: \( B = \mu_0 n I = (4\pi\times10^{-7})(1000)(5.0) = 6.28\times10^{-3}\,\mathrm{T} \); \( dB/dt = \mu_0 n\,dI/dt = 1.257\times10^{-4}\,\mathrm{T\,s^{-1}} \). Induced azimuthal electric field at \( r = a \) (Faraday, circular loop): \( E(a) = \tfrac{a}{2}\,\dfrac{dB}{dt} = (0.010)(1.257\times10^{-4}) = 1.26\times10^{-6}\,\mathrm{V\,m^{-1}} \). Poynting flux (azimuthal \( \vec{E} \), axial \( \vec{B} \) — inward): \( S = \dfrac{E(a)B}{\mu_0} = \dfrac{(1.26\times10^{-6})(6.28\times10^{-3})}{1.257\times10^{-6}} = 6.28\times10^{-3}\,\mathrm{W\,m^{-2}} \). Inflow per unit length: \( P' = S\cdot 2\pi a = (6.28\times10^{-3})(2\pi)(0.020) = 7.9\times10^{-4}\,\mathrm{W\,m^{-1}} \). Energy-rate check: \( U' = \dfrac{B^2}{2\mu_0}\pi a^2 \Rightarrow \dfrac{dU'}{dt} = \dfrac{B}{\mu_0}\dfrac{dB}{dt}\,\pi a^2 = \dfrac{(6.28\times10^{-3})(1.257\times10^{-4})}{1.257\times10^{-6}}\pi(0.020)^2 = (0.628)(1.257\times10^{-3}) = 7.9\times10^{-4}\,\mathrm{W\,m^{-1}} \). Agreement confirms the theorem: the inductor's energy flows in radially through the bore as the current builds.
  5. A region contains a uniform static electric field \( E = 1.0\times10^{5}\,\mathrm{V\,m^{-1}} \) (from a charged capacitor) perpendicular to a uniform static magnetic field \( B = 0.50\,\mathrm{T} \) (from a permanent magnet). (a) Compute \( |\vec{S}| \). (b) Nothing is heating up and no energy is being delivered anywhere — reconcile this with your nonzero answer. (c) What quantity, fixed by Poynting's theorem, vanishes here?
    Solution (a) \( |\vec{S}| = \dfrac{EB}{\mu_0} = \dfrac{(1.0\times10^{5})(0.50)}{1.257\times10^{-6}} = 4.0\times10^{10}\,\mathrm{W\,m^{-2}} \) — an enormous formal flux, perpendicular to both fields. (b) For static uniform fields, \( \vec{S} \) is (locally) constant, so \( \nabla\cdot\vec{S} = 0 \), and with \( \vec{J} = 0 \) and \( \partial_t u = 0 \) Poynting's theorem reads \( 0 + 0 = 0 \): satisfied trivially. In the real finite geometry the \( \vec{S} \) lines close on themselves — energy circulates in closed loops, and through any closed surface \( \oint\vec{S}\cdot d\vec{a} = 0 \). No region gains or loses energy. (c) The divergence \( \nabla\cdot\vec{S} \) (equivalently, the net flux through every closed surface) vanishes; only this divergence, not the pointwise value of \( \vec{S} \), is constrained by the theorem. The circulating flux is not entirely fictitious, though: it corresponds to a genuine electromagnetic angular momentum \( \int \varepsilon_0(\vec{r}\times(\vec{E}\times\vec{B}))\,d^3r \) stored in the static fields — famously recoverable in the Feynman disk paradox.