Maxwell's Equations from −¼F²

Derivation

Maxwell's Equations from −¼F²

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Statement

From the Lagrangian density \(\mathcal{L}=-\tfrac{1}{4}F_{\mu\nu}F^{\mu\nu}-j^{\mu}A_{\mu}\), with \(F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\), the Euler–Lagrange equations for the field \(A_\mu\) yield the inhomogeneous Maxwell pair \(\partial_\mu F^{\mu\nu}=j^{\nu}\) (Gauss's law and the Ampère–Maxwell law), while the homogeneous pair \(\partial_{[\mu}F_{\nu\rho]}=0\) (no monopoles and Faraday's law) follows as an identity from \(F=dA\) and requires no variation.

Why it matters

This is the moment electromagnetism stops being four empirical laws stapled together and becomes the unique consequence of one scalar built from a single gauge potential. The dynamics — how charges source fields — come out of extremizing an action; the constraints — why there are no magnetic monopoles — come out for free because \(F\) is a curvature. Two of Maxwell's four equations are not laws of nature at all but mathematical tautologies about a potential.

The construction also fixes the template for every gauge theory that followed. Replacing the abelian \(F_{\mu\nu}\) by a non-abelian field strength turns this exact same \(-\tfrac14\operatorname{tr}F^2\) into Yang–Mills theory, the backbone of the Standard Model. Understanding why electromagnetism has this form is understanding why the strong and electroweak interactions have theirs.

Assumptions

The dynamical variable is the potential \(A_\mu\), not the field \(F_{\mu\nu}\).If one instead varied \(E\) and \(B\) as six independent fields, the homogeneous equations would have to be imposed by hand rather than emerging as identities, and the gauge structure would be lost.

The metric is flat Minkowski \(\eta_{\mu\nu}=\operatorname{diag}(+,-,-,-)\) and indices are raised with it.On curved spacetime the plain derivatives must become covariant ones and a \(\sqrt{-g}\) appears in the action, modifying the divergence term.

The Lagrangian is at most quadratic in \(F\) and linear in \(A\), with no explicit \(x\)-dependence.Adding terms like \((F_{\mu\nu}F^{\mu\nu})^2\) (Euler–Heisenberg) makes the equations nonlinear; explicit \(A_\mu A^\mu\) mass terms break gauge invariance and produce Proca rather than Maxwell.

The current \(j^\mu\) is externally prescribed and conserved, \(\partial_\mu j^\mu=0\).If \(j^\mu\) is not conserved the field equations are inconsistent (their divergence forces \(\partial_\nu j^\nu=0\)), so an unconserved source has no solution.

Fields and variations fall off fast enough at infinity to drop boundary terms.If surface terms survive, the naive Euler–Lagrange equation is incomplete and radiation or boundary charges must be tracked separately.

Derivation

\[ S=\int d^4x\;\mathcal{L},\qquad \mathcal{L}=-\tfrac{1}{4}F_{\mu\nu}F^{\mu\nu}-j^{\mu}A_{\mu},\qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu. \]

Write the action as the spacetime integral of the given density; the fundamental field is \(A_\mu\). A

\[ \frac{\partial\mathcal{L}}{\partial A_\lambda}-\partial_\sigma\!\left(\frac{\partial\mathcal{L}}{\partial(\partial_\sigma A_\lambda)}\right)=0. \]

Apply the Euler–Lagrange field equation (prior result) to each component \(A_\lambda\); \(\mathcal{L}\) depends on \(A_\lambda\) and its first derivatives. A

\[ \frac{\partial\mathcal{L}}{\partial A_\lambda}=\frac{\partial}{\partial A_\lambda}\!\left(-j^\mu A_\mu\right)=-j^\lambda. \]

Only the source term contains \(A_\lambda\) undifferentiated; \(\partial A_\mu/\partial A_\lambda=\delta_\mu^{\,\lambda}\). A

\[ -\tfrac14 F_{\mu\nu}F^{\mu\nu}=-\tfrac14\,\eta^{\mu\alpha}\eta^{\nu\beta}F_{\mu\nu}F_{\alpha\beta},\qquad \frac{\partial}{\partial(\partial_\sigma A_\lambda)}\big(F_{\mu\nu}F^{\mu\nu}\big)=2F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial(\partial_\sigma A_\lambda)}. \]

Raise indices with the flat metric, then differentiate the quadratic form; the factor \(2\) comes from the product rule on the two identical factors. B

\[ \frac{\partial F_{\mu\nu}}{\partial(\partial_\sigma A_\lambda)}=\frac{\partial(\partial_\mu A_\nu-\partial_\nu A_\mu)}{\partial(\partial_\sigma A_\lambda)}=\delta_\mu^{\,\sigma}\delta_\nu^{\,\lambda}-\delta_\nu^{\,\sigma}\delta_\mu^{\,\lambda}. \]

Each velocity \(\partial_\sigma A_\lambda\) is an independent variable; differentiate the definition of \(F_{\mu\nu}\) term by term. B

\[ \frac{\partial}{\partial(\partial_\sigma A_\lambda)}\big(F_{\mu\nu}F^{\mu\nu}\big)=2F^{\mu\nu}\big(\delta_\mu^{\,\sigma}\delta_\nu^{\,\lambda}-\delta_\nu^{\,\sigma}\delta_\mu^{\,\lambda}\big)=2\big(F^{\sigma\lambda}-F^{\lambda\sigma}\big)=4F^{\sigma\lambda}. \]

Contract the Kronecker deltas, then use antisymmetry \(F^{\lambda\sigma}=-F^{\sigma\lambda}\) to combine the two terms. B

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\sigma A_\lambda)}=-\tfrac14\cdot 4F^{\sigma\lambda}=-F^{\sigma\lambda}. \]

Multiply by the \(-\tfrac14\) prefactor; the source term carries no derivatives of \(A\) and contributes nothing here. A

\[ -j^\lambda-\partial_\sigma\!\left(-F^{\sigma\lambda}\right)=0\quad\Longrightarrow\quad \partial_\sigma F^{\sigma\lambda}=j^\lambda. \]

Substitute Steps 3 and 7 into the Euler–Lagrange equation of Step 2 and rearrange signs. A

\[ \boxed{\;\partial_\mu F^{\mu\nu}=j^{\nu}\;}\qquad(\text{inhomogeneous pair}). \]

Relabel dummy indices \(\sigma\to\mu,\ \lambda\to\nu\). Component \(\nu=0\) is Gauss's law; \(\nu=i\) is the Ampère–Maxwell law (prior result, integral/differential dictionary). A

\[ \partial_\mu(\partial_\nu F^{\mu\nu})=\partial_\mu j^\mu. \]

Take the divergence of the field equation. The left side is \(\partial_\mu\partial_\nu F^{\mu\nu}\): a symmetric pair \(\partial_\mu\partial_\nu\) contracted with the antisymmetric \(F^{\mu\nu}\), hence identically zero — forcing \(\partial_\mu j^\mu=0\), current conservation, as a built-in consistency condition. C

\[ \partial_{[\lambda}F_{\mu\nu]}\equiv\partial_\lambda F_{\mu\nu}+\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}. \]

Now the homogeneous pair. Write the totally antisymmetrised derivative of \(F\); substitute \(F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\) into each of the three terms. C

\[ \partial_\lambda(\partial_\mu A_\nu-\partial_\nu A_\mu)+\partial_\mu(\partial_\nu A_\lambda-\partial_\lambda A_\nu)+\partial_\nu(\partial_\lambda A_\mu-\partial_\mu A_\lambda)=0. \]

Every term pairs with an equal-and-opposite partner because mixed partials commute, \(\partial_\lambda\partial_\mu A_\nu=\partial_\mu\partial_\lambda A_\nu\). This is the Bianchi identity — it holds for any \(A_\mu\), with no equation of motion used. C

\[ \boxed{\;\partial_{[\mu}F_{\nu\rho]}=0\;}\qquad(\text{homogeneous pair}). \]

This identity is \(\nabla\cdot\mathbf{B}=0\) (no monopoles) and \(\nabla\times\mathbf{E}=-\partial_t\mathbf{B}\) (Faraday) in the 3+1 dictionary. All four Maxwell equations now stand. A

Result

\[ \partial_\mu F^{\mu\nu}=j^{\nu}\qquad\text{and}\qquad \partial_{[\mu}F_{\nu\rho]}=0. \]

Reading. Varying the potential in \(-\tfrac14F^2-j\!\cdot\!A\) says the divergence of the field strength equals the current: charges and currents source \(F\). This single covariant equation contains Gauss's law (\(\nu=0\)) and the Ampère–Maxwell law (\(\nu=i\)). The second equation is not a law but the statement that \(F\) is the curl (exterior derivative) of a potential; it delivers "no magnetic charge" and Faraday induction automatically. Conservation of charge, \(\partial_\mu j^\mu=0\), is not an extra assumption but a forced consequence of the antisymmetry of \(F\).

Units check. In Heaviside–Lorentz natural units (\(c=\varepsilon_0=1\)), \([A_\mu]=[\text{energy}]\), so \([F_{\mu\nu}]=[\partial A]=[\text{energy}]^2\) and \([\mathcal{L}]=[F^2]=[\text{energy}]^4\), the correct dimension for a 4D Lagrangian density since \(\int d^4x\,\mathcal{L}\) is dimensionless. The field equation reads \([\partial F]=[\text{energy}]^3=[j^\nu]\), matching a 4-current density in these units. Restoring SI, \(\partial_\mu F^{\mu\nu}=\mu_0 j^\nu\) with \([F]\) split into \(E\ (\mathrm{V\,m^{-1}})\) and \(B\ (\mathrm{T})\) via \(F^{0i}=E^i/c\), and both sides carry \(\mathrm{T\,m^{-1}}=\mathrm{kg\,A^{-1}s^{-2}m^{-1}}\).

Limiting cases

Vacuum, \(j^\mu=0\): \(\partial_\mu F^{\mu\nu}=0\) together with the Bianchi identity gives \(\Box A^\nu=0\) in Lorenz gauge — free electromagnetic waves travelling at \(c\).
Statics, \(\partial_t\to 0\): the \(\nu=0\) equation collapses to \(\nabla\cdot\mathbf{E}=\rho/\varepsilon_0\) and the spatial part to \(\nabla\times\mathbf{B}=\mu_0\mathbf{J}\); the displacement current vanishes.
Massive photon (Proca): adding \(+\tfrac12 m^2 A_\mu A^\mu\) modifies Step 3 by \(+m^2A^\lambda\), giving \(\partial_\mu F^{\mu\nu}+m^2A^\nu=j^\nu\); as \(m\to0\) Maxwell is recovered.
Non-abelian limit: promoting \(A_\mu\) to a Lie-algebra field and \(F\to\partial A-\partial A+g[A,A]\) turns the derivation into Yang–Mills, \(D_\mu F^{\mu\nu}=j^\nu\); the abelian case here is \(g\to0\).

Breaks when

Strong fields (\(E\gtrsim E_{\text{crit}}=m_e^2c^3/e\hbar\approx1.3\times10^{18}\,\mathrm{V\,m^{-1}}\)): vacuum polarization adds Euler–Heisenberg \((F^2)^2\) terms, the equations become nonlinear, and light-by-light scattering appears. The pure \(-\tfrac14F^2\) form is only the leading term of an effective action.
Magnetic monopoles present: the homogeneous pair is an identity only because \(F=dA\) globally. If \(F\) is not globally exact (a Dirac monopole), \(\partial_{[\mu}F_{\nu\rho]}=g_m\,k^{\nu}\neq0\) and the "automatic" equation acquires a magnetic source.
Non-conserved or off-shell sources: if \(\partial_\mu j^\mu\neq0\), Step 10 shows the field equations are self-contradictory — no \(A_\mu\) solves them — so the framework simply has no solution.
Curved or medium-filled spacetime: in a gravitational field or a polarizable medium the flat-metric contraction of Step 4 is wrong; one needs \(\sqrt{-g}\,g^{\mu\alpha}g^{\nu\beta}\) or the macroscopic \(H^{\mu\nu}=F^{\mu\nu}\) split into \(D\) and \(H\), changing the divergence term.

Failure modes

Losing the factor of 4 (or 2): forgetting that \(F_{\mu\nu}F^{\mu\nu}\) has two \(F\)'s (product rule → 2) and that \(\partial F/\partial(\partial A)\) has two antisymmetric terms (→ another 2). Dropping either leaves a stray \(\tfrac12\) in the field equation.
Treating \(\partial_\mu A_\nu\) and \(\partial_\nu A_\mu\) as one variable: when differentiating \(F_{\mu\nu}\) with respect to \(\partial_\sigma A_\lambda\), both index orderings must be varied independently; skipping the second gives \(2F^{\sigma\lambda}\) instead of \(4F^{\sigma\lambda}\).
Varying \(E\) and \(B\) instead of \(A_\mu\): then the Bianchi identity is not automatic and students "derive" only two equations, wrongly concluding the Lagrangian is incomplete.
Sign confusion from the metric: using \((-,+,+,+)\) without adjusting the Lagrangian sign flips \(F^{0i}\) and can turn attractive into repulsive; the overall \(-\tfrac14\) is tied to the signature choice.
Claiming charge conservation is an independent postulate: it is forced by the antisymmetry of \(F^{\mu\nu}\) (Step 10), not added by hand.
Forgetting the source term contributes to \(\partial\mathcal{L}/\partial A_\lambda\) but not to \(\partial\mathcal{L}/\partial(\partial_\sigma A_\lambda)\): misplacing \(j^\lambda\) into the derivative slot corrupts the whole equation.

Discussion

The deepest lesson is the asymmetry between the two Maxwell pairs. The inhomogeneous equations are dynamical: they are Euler–Lagrange equations, they depend on the source, and they could in principle be different if we chose a different Lagrangian. The homogeneous equations are kinematical: they are true of any antisymmetric tensor that is an exterior derivative, independent of any action or source. Writing \(F=dA\) makes \(dF=d^2A=0\) automatic, because the exterior derivative is nilpotent. Two of Maxwell's four equations are geometry, not physics.

Gauge invariance is the hidden organizing principle. The Lagrangian is unchanged under \(A_\mu\to A_\mu+\partial_\mu\chi\) because \(F_{\mu\nu}\) is, and the source term shifts by \(-j^\mu\partial_\mu\chi\), which integrates to a boundary term precisely when \(\partial_\mu j^\mu=0\). So gauge invariance and charge conservation are two faces of the same coin — a concrete instance of Noether's theorem linking a continuous symmetry to a conserved current. This is why a photon mass term \(m^2A_\mu A^\mu\), which breaks gauge invariance, is also the thing that spoils the neat conservation argument.

Physically, \(\partial_\mu F^{\mu\nu}=j^\nu\) unifies electricity and magnetism as a single object seen from different frames: what one observer calls a static \(\mathbf{E}\), another moving observer resolves partly into \(\mathbf{B}\), because \(F^{\mu\nu}\) is a Lorentz tensor and its components mix under boosts. The Lagrangian \(-\tfrac14F_{\mu\nu}F^{\mu\nu}=\tfrac12(E^2-B^2)\) is a Lorentz scalar; the other invariant \(F_{\mu\nu}\tilde F^{\mu\nu}\propto\mathbf{E}\cdot\mathbf{B}\) is a total derivative and does not affect the equations of motion, though it matters topologically.

At the advanced level, the whole structure is the abelian corner of a universal construction. Demanding that a matter field's local phase \(\psi\to e^{iq\chi(x)}\psi\) be a symmetry forces the introduction of a connection \(A_\mu\) with exactly the transformation \(A_\mu\to A_\mu+\partial_\mu\chi\), and the only gauge-invariant, Lorentz-invariant, renormalizable kinetic term one can build from its curvature is \(-\tfrac14F^2\). Electromagnetism is therefore not one option among many but the essentially unique theory of a massless spin-1 field coupled to conserved charge — a rigidity that generalizes directly to the \(SU(3)\times SU(2)\times U(1)\) of the Standard Model, where the same \(-\tfrac14\operatorname{tr}F^2\) governs gluons and weak bosons.

Common misconceptions. (i) That all four Maxwell equations "come from the Lagrangian" — only the inhomogeneous pair does; the other two are identities from \(F=dA\). (ii) That \(A_\mu\) is a mere calculational convenience with no physical content — the Aharonov–Bohm effect shows \(A_\mu\) has observable consequences beyond \(F_{\mu\nu}\). (iii) That charge conservation must be assumed separately — it is a theorem here, forced by the antisymmetry of \(F^{\mu\nu}\).

Worked examples

Example 1 — Recover Gauss's law from the covariant field equation.

\[ \partial_\mu F^{\mu\nu}=\mu_0 j^\nu,\qquad \text{take }\nu=0. \]

Select the time component; in SI the field equation carries \(\mu_0\) on the source. A

\[ \partial_\mu F^{\mu 0}=\partial_i F^{i0},\qquad F^{i0}=\frac{E^i}{c},\qquad j^0=c\rho. \]

The \(\mu=0\) term vanishes by antisymmetry (\(F^{00}=0\)); insert the standard identifications of the field-strength components. B

\[ \partial_i\frac{E^i}{c}=\mu_0(c\rho)\;\Longrightarrow\;\nabla\cdot\mathbf{E}=\mu_0 c^2\rho=\frac{\rho}{\varepsilon_0}. \]

Use \(c^2=1/(\mu_0\varepsilon_0)\) to convert the prefactor; symbolic result before numbers. B

\[ \rho=8.85\times10^{-6}\,\mathrm{C\,m^{-3}},\quad \varepsilon_0=8.854\times10^{-12}\,\mathrm{F\,m^{-1}}\;\Rightarrow\;\nabla\cdot\mathbf{E}=\frac{8.85\times10^{-6}}{8.854\times10^{-12}}. \]

Plug in a sample uniform charge density to evaluate the divergence numerically. A

\[ \nabla\cdot\mathbf{E}\approx1.00\times10^{6}\ \mathrm{V\,m^{-2}}. \]

Reading. The \(\nu=0\) component of the single covariant law is exactly Gauss's law; with \(\rho\approx8.85\,\mu\mathrm{C\,m^{-3}}\) the field divergence is \(10^{6}\,\mathrm{V\,m^{-2}}\), confirming the units of \([\nabla\!\cdot\!\mathbf{E}]=\mathrm{(V\,m^{-1})/m}\).

Example 2 — Displacement current from the \(\nu=i\) component.

\[ \partial_\mu F^{\mu i}=\mu_0 j^i,\qquad \partial_\mu F^{\mu i}=\partial_0 F^{0i}+\partial_k F^{ki}. \]

Take a spatial component and split the sum into time and space derivatives. B

\[ F^{0i}=-\frac{E^i}{c},\quad \partial_0=\frac1c\partial_t,\quad F^{ki}=-\varepsilon^{kij}B_j\;\Rightarrow\;-\frac{1}{c^2}\partial_t E^i+(\nabla\times\mathbf{B})^i=\mu_0 j^i. \]

Insert component identities; the spatial curl reassembles from \(\partial_k F^{ki}\). C

\[ \nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\,\partial_t\mathbf{E}. \]

Rearrange and use \(1/c^2=\mu_0\varepsilon_0\); the last term is Maxwell's displacement current. A

\[ \partial_t E=1.0\times10^{12}\,\mathrm{V\,m^{-1}s^{-1}}\;\Rightarrow\;\mu_0\varepsilon_0\,\partial_t E=\frac{1.0\times10^{12}}{(3.0\times10^{8})^2}. \]

Evaluate the displacement-current density for a rapidly changing field (e.g. inside a charging capacitor). A

\[ \mu_0\varepsilon_0\,\partial_t\mathbf{E}\approx1.1\times10^{-5}\ \mathrm{A\,m^{-2}}. \]

Reading. The spatial components of \(\partial_\mu F^{\mu\nu}=\mu_0 j^\nu\) are the Ampère–Maxwell law, displacement current included. A field changing at \(10^{12}\,\mathrm{V\,m^{-1}s^{-1}}\) contributes an effective current density of \(\sim10^{-5}\,\mathrm{A\,m^{-2}}\), which is what keeps \(\nabla\!\times\!\mathbf{B}\) consistent between the plates of a capacitor where \(\mathbf{J}=0\).

Problems

Starting from \(\mathcal{L}=-\tfrac14 F_{\mu\nu}F^{\mu\nu}\) with \(j^\mu=0\), show the field equation in Lorenz gauge \(\partial_\mu A^\mu=0\) reduces to a wave equation, and state the propagation speed.
Solution
The field equation is \(\partial_\mu F^{\mu\nu}=\partial_\mu(\partial^\mu A^\nu-\partial^\nu A^\mu)=\Box A^\nu-\partial^\nu(\partial_\mu A^\mu)\). Imposing \(\partial_\mu A^\mu=0\) kills the second term, leaving \(\Box A^\nu=0\), i.e. \(\left(\tfrac1{c^2}\partial_t^2-\nabla^2\right)A^\nu=0\). This is the wave equation; solutions propagate at speed \(c\), the speed of light. Each of the four components \(A^\nu\) satisfies an independent massless wave equation.
Show that the extra piece \(F_{\mu\nu}\tilde F^{\mu\nu}\) (with \(\tilde F^{\mu\nu}=\tfrac12\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}\)) added to \(\mathcal{L}\) does not change the equations of motion.
Solution
Write \(F_{\mu\nu}\tilde F^{\mu\nu}=\tfrac12\varepsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}\). Using \(F=dA\), \(\varepsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}=\varepsilon^{\mu\nu\rho\sigma}(2\partial_\mu A_\nu)(2\partial_\rho A_\sigma)=4\,\partial_\mu(\varepsilon^{\mu\nu\rho\sigma}A_\nu\partial_\rho A_\sigma)\), since the leftover term \(\varepsilon^{\mu\nu\rho\sigma}A_\nu\partial_\mu\partial_\rho A_\sigma=0\) by symmetry of \(\partial_\mu\partial_\rho\) against the antisymmetric \(\varepsilon\). Being a total divergence, it integrates to a boundary term and its Euler–Lagrange variation vanishes identically. (It is topological — the instanton density — and matters only globally.)
A Proca field has \(\mathcal{L}=-\tfrac14 F_{\mu\nu}F^{\mu\nu}+\tfrac12 m^2 A_\mu A^\mu-j^\mu A_\mu\). Derive its field equation and show it forces \(\partial_\mu A^\mu=0\) when \(j^\mu\) is conserved.
Solution
The mass term adds \(\partial\mathcal{L}/\partial A_\lambda\supset m^2A^\lambda\) to Step 3, giving \(\partial_\mu F^{\mu\nu}+m^2A^\nu=j^\nu\). Take the divergence: \(\partial_\nu\partial_\mu F^{\mu\nu}=0\) (antisymmetry) so \(m^2\partial_\nu A^\nu=\partial_\nu j^\nu\). With \(\partial_\nu j^\nu=0\) and \(m\neq0\), this yields \(\partial_\nu A^\nu=0\). Note the Lorenz condition is here a consequence, not a gauge choice — the mass term breaks gauge freedom, so \(A_\mu\) has three physical polarizations.
Verify the units of the coupling term \(j^\mu A_\mu\) match \(F_{\mu\nu}F^{\mu\nu}\) in SI, given \([A_0]=\mathrm{V}\) and \([\mathbf{A}]=\mathrm{T\,m}\). (Take the Lagrangian density in \(\mathrm{J\,m^{-3}}\).)
Solution
The energy-density form is \(\mathcal{L}=-\tfrac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu}-j^\mu A_\mu\). Field term: \([\,F^2/\mu_0\,]=\big[(B)^2/\mu_0\big]=\mathrm{T^2/(H\,m^{-1})}=\mathrm{J\,m^{-3}}\) (since \(B^2/2\mu_0\) is energy density). Source term: charge density \(\rho\) times potential, \([\rho A_0]=\mathrm{(C\,m^{-3})(V)}=\mathrm{(C\,m^{-3})(J\,C^{-1})}=\mathrm{J\,m^{-3}}\); likewise \([\mathbf J\cdot\mathbf A]=\mathrm{(A\,m^{-2})(T\,m)}=\mathrm{(A\,m^{-2})(kg\,A^{-1}s^{-2}\,m)}=\mathrm{kg\,s^{-2}m^{-1}}=\mathrm{J\,m^{-3}}\). Both terms are \(\mathrm{J\,m^{-3}}\), as required.
For a point charge \(q=1.6\times10^{-19}\,\mathrm{C}\) at rest, use the \(\nu=0\) field equation in integral form to find \(\mathbf{E}\) at \(r=1.0\times10^{-10}\,\mathrm{m}\). Comment on which Maxwell pair you used.
Solution
Integrate \(\nabla\cdot\mathbf E=\rho/\varepsilon_0\) over a sphere (Gauss's theorem): \(\oint\mathbf E\cdot d\mathbf A=q/\varepsilon_0\Rightarrow E(4\pi r^2)=q/\varepsilon_0\Rightarrow E=\dfrac{q}{4\pi\varepsilon_0 r^2}\). Numerically \(E=\dfrac{(8.99\times10^9)(1.6\times10^{-19})}{(1.0\times10^{-10})^2}=\dfrac{1.44\times10^{-9}}{1.0\times10^{-20}}=1.4\times10^{11}\,\mathrm{V\,m^{-1}}\). This uses only the inhomogeneous (dynamical) pair \(\partial_\mu F^{\mu\nu}=j^\nu\); the homogeneous pair (\(\nabla\cdot\mathbf B=0\), Faraday) plays no role for a static point charge, illustrating the dynamical/kinematical split at the heart of the derivation.