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Charge conservation says that the total signed electric charge of an isolated collection stays constant. More generally, a collection closed to electric charge exchange has constant total electric charge, even if energy crosses its boundary. This is the charge bookkeeping relevant to a closed system or an isolated system; isolation from energy exchange is not required. Positive and negative electric charge count with opposite signs. Creating or removing an electron and a Positron together is compatible with charge conservation, because their combined electric charge is zero. ^[ref].

Imagine a fixed box initially containing electric charge $Q(0)=10\mu \mathrm{C}$. Stipulate incoming conventional current $I_{\mathrm{in}}=3\mu \mathrm{A}$ and outgoing conventional current $I_{\mathrm{out}}=1\mu \mathrm{A}$ throughout $0\le t\le 2\mathrm{s}$, with no other boundary transport. Since one ampere is one coulomb per second, the stored electric charge grows at $2\mu C/s$:If instead the outgoing electric current also equals $3\mu \mathrm{A}$, the stored electric charge remains $10\mu \mathrm{C}$ while electric charge passes through. These are stipulated transport examples; constant terminal electric currents alone do not guarantee a time-independent interior charge density.

Locally, charge conservation is the charge continuity equation, an instance of the general continuity equation:Here $\rho (\mathbf{r},t)$ is signed volume charge density and $\mathbf{J}(\mathbf{r},t)$ is current density. A positive divergence of current density means net outward transport from a small neighborhood, so its charge density decreases. For a fixed control volume $V$ with closed surface $\partial V$ and outward unit normal $\mathbf{n}$, the equivalent integrated balance isThe surface integral is signed: incoming positive electric charge contributes negatively to outward electric current, and outgoing positive electric charge contributes positively. Integrate the charge continuity equation over fixed $V$ and apply the divergence theorem to obtain this balance. Zero net boundary transport gives constant $Q_V$; the global and local statements describe the same conservation law at different scales. ^[ref].

In a steady transport state, charge density and current density are time-independent, so the charge continuity equation reduces to $\nabla \cdot \mathbf{J}=0$. At an electrical circuit junction, choose each terminal electric current $I_k$ positive outward and assume the terminals account for all boundary transport. ThenKirchhoff's current law, ${\sum }_k{I}_k=0$, follows when accumulation of electric charge at the junction is negligible. During storage, retain the same balance with its nonzero derivative; charge conservation continues to hold. This makes charge conservation useful for checking electrical circuit models and for tracking the charging of a capacitor. ^[ref].

The beautiful consistency with Maxwell equations is that the electric field and magnetic field already enforce this transport balance. For sufficiently smooth fields in vacuum SI, take the divergence of the Ampère-Maxwell equation, use that the divergence of a curl is zero, and substitute Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Constant vacuum permittivity ${ϵ}_0$ givesThe displacement current term supplies the part needed when electric charge accumulates. Historically, Maxwell added that term to extend the steady-current equation; see the existing charge conservation from Maxwell equations for this connection. ^[ref].

The local form of conservation of electric charge relates charge density to current density. Taking the divergence of the Ampère-Maxwell equation and substituting Gauss's law proves the equation, using that the divergence of a curl is zero.

In vacuum, displacement current is the term ${\mathbf{J}}_D={ϵ}_0{\partial }_t\mathbf{E}$ in the Ampère-Maxwell equation. It has the units of current density, but need not describe electric charge crossing the region. In Maxwell equations in matter, the corresponding term is ${\partial }_t\mathbf{D}$, with $\mathbf{D}$ the electric displacement field.

Consider a slowly charging parallel-plate capacitor with vacuum gap, plate area $A$, and charge $Q(t)$ on its positive plate. Neglect fringing and assume the charging timescale is long compared with the light-travel time across the apparatus. The nearly uniform gap electric field has magnitude $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isFor $A=0.010{\mathrm{m}}^2$ and charging electric current $I=1.0\mathrm{mA}$, this gives $dE/dt\simeq 1.13\times 10^{10}\mathrm{V}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$ and $J_D=0.10A/{\mathrm{m}}^2$. No conducting electric current crosses the gap. A spanning surface cutting the wire and another passing between the plates nevertheless give the same magnetic circulation when the full Ampère-Maxwell equation is used. ^[ref].

The local consistency check is short. Without displacement current, taking the divergence of $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$ would force $\nabla \cdot \mathbf{J}=0$, allowing no local accumulation of electric charge. With displacement current, the divergence of a curl is zero and Gauss's law giveThis is precisely the charge continuity equation. Electric charge conservation fixes the required divergence of the added term; it does not by itself exclude other divergence-free additions. The specific Maxwell equations also yield the electromagnetic wave equation, linking this consistency correction to propagation at the speed of light.

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.

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

For a twice continuously differentiable vector potential $A$, write the curl with the Levi-Civita symbol:The second derivatives are symmetric in $i,j$, whereas ${ϵ}_{ijk}$ is antisymmetric, so the contraction is zero. Equivalently, writing out the three components shows that each mixed derivative cancels its reversed-order partner. ThusThis is divergence of a curl is zero. The regularity hypothesis is what licenses commutation of the mixed partial derivatives; it must not be omitted for a singular potential without specifying a weaker derivative interpretation.

In a magnetostatic equilibrium, inertia is absent and all forces balance. A force-free magnetic field is the limiting case in which the Lorentz force density vanishes:Thus the current density is parallel to the magnetic field, so for some scalar force-free parameter $\alpha (\mathbf{r})$,Using Ampère's law in magnetostatics givesTaking the divergence and using both the divergence of a curl is zero and $\nabla \cdot \mathbf{B}=0$ yieldssoThe force-free parameter is therefore constant along every magnetic field line.

Compute the curl component by component from the given vector field:For example, $B_x={\partial }_y{A}_z-{\partial }_z{A}_y=0-(-2xz)$ and $B_z={\partial }_x{A}_y-{\partial }_y{A}_x=(1-z^2)-(z^2-1)$. Its divergence isas expected because the divergence of a curl is zero.

Take $C$ counterclockwise when viewed from positive $z$, the disc $D$ upward oriented, and the hemisphere $H$ outward oriented. These choices give both surfaces the same positive boundary orientation. Parametrize $C$ by $r(t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$. On it, $A=(-\sin t,\cos t,0)$ and $dr=(-\sin t,\cos t,0)dt$, so the line integral isOn $D$, use polar coordinates $(x,y)=(r\cos \varphi ,r\sin \varphi )$ and $d\mathbf{S}=(0,0,1)rdrd\varphi$. Since $B_z=2$ there, the surface flux isOn the unit hemisphere, the outward unit normal vector is $n=(x,y,z)$. Since $x^2+y^2=1-z^2$, we have $B\cdot n=4z(1-z^2)$. In spherical polar coordinates, $z=\cos \theta$ and $dS=\sin \theta d\theta d\varphi$, giving the explicit surface integralIndeed the inner integral is $[{\sin }^4\theta {]}_0^{\pi /2}=1$.

These equalities illustrate Stokes theorem twice: $B=\nabla \times A$ and either upward $D$ or outward $H$ spans the oriented circle $C$, so each surface flux equals the same line integral. For the divergence theorem, close the upper half-ball by adjoining the downward oriented base, namely $-D$. Its outward surface flux isThe minus sign on the base is required because its outward normal vector points down. Reversing a specified orientation reverses that integral's sign.

Where $\mathbf{A}=|\mathbf{v}{|}^2\mathbf{w}$ is twice continuously differentiable, the identity divergence of a curl is zero gives $\nabla \cdot (\nabla \times \mathbf{A})=0$. All regular portions of the parameter interval lie in this region, henceas an improper line integral. Besides the $\mathrm{arctanh}(x^2)$ endpoint problem from (i), the curve touches $y=1$ at $\theta =\pi /2$, where the printed real $\arcsin y$ expression has no smooth open neighbourhood. The identity applies on the intervening open segments, and its zero value has a unique continuous extension along the curve. A literal classical integrand at every parameter is not supplied by the printed field.

Separate the closed cylinder into its curved side and its two end disks, all with outward normal vectors. On the side, $\rho =R$ and $\mathbf{n}=\hat{\mathbf{\rho }}$, soAt the top, $z=h/2$ and $\mathbf{n}=\mathbf{k}$, giving $(\nabla \times \mathbf{v})\cdot \mathbf{n}=\alpha h$. At the bottom, both $z$ and the axial normal vector change sign, so the same positive value $\alpha h$ results. Each disk therefore contributes $\pi \alpha R^{2}h$. Adding all three surface integrals,This explicit cancellation is consistent with the fact that the divergence of a curl is zero. The two disk contributions add rather than cancel.

Take the curl component by component:Its divergence is zero, as expected because the divergence of a curl is zero.

The curl found in part (a) is constant, so its divergence isThis also follows from the identity divergence of a curl is zero, applicable here because the field is smooth everywhere.

On the same domain, differentiate the Cartesian components:ThusThis is also an instance of divergence of a curl is zero, since the potential is smooth off the axis. The classical field is undefined on the axis, so that identity must not be applied there without specifying an extension.

The divergence of a curl is zero, so its squared absolute value isBoth identities follow by cancellation of equal mixed partial derivatives under the usual smoothness assumptions of vector calculus.

In Cartesian coordinates, use the Levi-Civita symbol and the Einstein summation convention to expandThe contraction of the two symbols uses the Kronecker delta, and the final step is the product rule. Thus the convective acceleration identity isPut $\mathbf{\omega }=\nabla \times \mathbf{u}$. The curl of a gradient vanishes, and the curl of a cross product identity givesSince the divergence of a curl is zero, taking the curl of the first identity yieldsTwice continuous differentiability suffices for the mixed-derivative identities used here.

For a twice continuously differentiable vector field in Cartesian coordinates, the curl and divergence arePut $\mathbf{\omega }=\nabla \times \mathbf{v}$. Using the Levi-Civita symbol, the Kronecker delta contraction and the product rule,The last term vanishes because the divergence of a curl is zero: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the curl of a cross product identity becomes

For orthogonal coordinates $q_i$ with position vector $\mathbf{r}(q_1,q_2,q_3)$, the scale factors of orthogonal coordinates and local unit vectors areOrthogonality gives $d{s}^2={\sum }_i{h}_i^{2}d{q}_i^2$, and right-handedness fixes ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$. In cylindrical coordinates, $\mathbf{r}=(\rho \cos \varphi ,\rho \sin \varphi ,z)$, soThe radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\varphi$; ${\mathbf{e}}_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.

For the axial velocity profile, the cylindrical coordinates formula givesThe associated cross product isThis field has only a radial component depending on $\rho$, so its curl is zero. Also $\nabla \cdot \mathbf{u}={\partial }_z{u}_z=0$. The identity above therefore yieldsIn fact both directional derivatives vanish: the first differentiates $u_z{\mathbf{e}}_z$ only in the azimuthal direction, and the second differentiates $\mathbf{\omega }$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho =0$, despite the coordinate-basis singularity.

Take the divergence of the Ampère-Maxwell equation and use that the divergence of a curl is zero:The last step uses Gauss's law. Thus the charge continuity equation is exactlyFor a direct justification that this four-current is a four-vector, take one locally comoving population of electric charges, with rest charge density ${\rho }_0$ and four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$. Invariance of electric charge and length contraction give $\rho =\gamma {\rho }_0$ and $\mathbf{J}=\rho \mathbf{v}$, hence $J^{\mu }={\rho }_0{U}^{\mu }$. Summing populations preserves the four-vector rule. In particular, $c{\rho }^{\prime }=\gamma (c\rho -\beta J_x)$ and $J_x^{\prime }=\gamma (J_x-\beta c\rho )$. The Lorentz invariance of its four-divergence then explains why charge conservation holds in every inertial frame. A single zero-divergence condition alone would not determine a field's transformation law; the physical charge density and current density provide the additional content.