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In matter, Gauss's law for free electric charge isThe area normal points out of $\mathcal{V}$. In vacuum $\mathbf{D}={ϵ}_0\mathbf{E}$, so these become $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ and $\oint \mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enclosed}}/{ϵ}_0$.

Put charges $+Q$ and $-Q$ on the concentric conductors. Gauss's law gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$ between them, $a<r<2a$. The potential difference isThe capacitance of this vacuum spherical capacitor is thereforeThe conductors are equipotentials; a uniform dielectric filling would replace ${ϵ}_0$ by its permittivity.

In matter, Gauss's law for free electric charge isThe area normal points out of $\mathcal{V}$. In vacuum $\mathbf{D}={ϵ}_0\mathbf{E}$, so these become $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ and $\oint \mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enclosed}}/{ϵ}_0$.

Put charges $+Q$ and $-Q$ on the concentric conductors. Gauss's law gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$ between them, $a<r<2a$. The potential difference isThe capacitance of this vacuum spherical capacitor is thereforeThe conductors are equipotentials; a uniform dielectric filling would replace ${ϵ}_0$ by its permittivity.

Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:In electrostatics, the electric field is the negative gradient of the electrostatic potential, $\mathbf{E}=-\nabla V$. Substitution into the differential equation form of Gauss's law gives Poisson's equation ${\nabla }^{2}V=-\rho /{ϵ}_0$. Consequently, in a region containing no charge density,which is Laplace's equation. Charges on the bounding conductors enter through the boundary conditions; they need not vanish for Laplace's equation to hold inside the empty gap.

Take the divergence of the proposed Ampère's law. Since the divergence of a curl vanishes identically, it implies $\nabla \cdot \mathbf{J}=0$. However, local charge conservation requires the continuity equationFor example, electric current entering a charging capacitor accumulates electric charge on its plates, so the charge density is time dependent. The proposed equation cannot describe that process.

From Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Thus $\nabla \cdot ({ϵ}_0{\partial }_t\mathbf{E})={\partial }_t\rho$, which supplies precisely the missing divergence. Adding the displacement current density gives the Ampère-Maxwell equation:Its divergence now yields the correct continuity equation. Through a fixed surface, the associated displacement current is $I_D={ϵ}_{0}d{\Phi }_E/dt$, where ${\Phi }_E$ is the electric flux. In a vacuum capacitor gap, this equals the conduction electric current charging the plates, even though no charges cross the gap. Charge conservation fixes the required divergence of the correction; by itself it does not exclude adding a further divergence-free vector field.

The differential vacuum Gauss's law isIntegrate over a volume $\Omega$ with closed boundary $S$. The divergence theorem converts the left side into outward electric flux:This is the integral form. Surface charge can be treated by a thin-layer limit or by the corresponding jump in the normal field; the enclosed charge then includes the charged sheet. The surface orientation must be outward, and the field entering the integral is the total field, not merely the field of the enclosed charges alone.

Inside the uniformly charged slab, its own electric field is $\rho (z+d/2)\hat{\mathbf{z}}/{ϵ}_0$: it vanishes at the slab's midplane and has derivative $\rho /{ϵ}_0$. Everywhere below the negative sheet, that sheet adds $+\sigma \hat{\mathbf{z}}/(2{ϵ}_0)$. ThereforeIts magnitude is the absolute value of this scalar coefficient. The field changes direction atFor $z<z_0$ it points downward; for $z>z_0$ it points upward, towards the negative sheet. Thus it is neither uniform nor directed upwards throughout the slab. The derivative and the field at the lower slab boundary also agree with Gauss's law and the previously found outside field.

Let $R_1<R_2$ and put charge per unit length $+{\lambda }_q$ on the inner shell and $-{\lambda }_q$ on the facing outer conductor. For a long coaxial Gaussian surface of radius $r$ in the gap, Gauss's law givesThe electric potential difference between the shells isWith total inner charge $Q={\lambda }_{q}L$, the capacitance isThis assumes a vacuum or approximately air-filled gap and negligible fringing at the ends. A uniform linear dielectric replaces ${ϵ}_0$ by its permittivity. The expression depends on the shell-to-shell potential difference, not on the arbitrary common potential reference.

Translational and reflection symmetry parallel to the slabs imply $\mathbf{E}=E_x(x)\hat{\mathbf{x}}$. The differential Gauss's law is $d{E}_x/dx=\rho /{ϵ}_0$. Starting at the left boundary with $E_x(-a)=0$ givesThere is no singular surface charge at $x=0$, so the normal electric field is continuous there: $E_x(0)=-C/{ϵ}_0$. Integrating through the positive slab givesThe second expression indeed returns to zero at $x=b$, because the two total charges per unit area are $-C$ and $+C$. For $C>0$, the electric field throughout the interior points from the positive slab towards the negative slab, along $-x$; reversing $C$ reverses the directions. The piecewise linear profile is a useful consequence of the uniform volume charge density.

The magnetostatic Ampère's law, in vacuum, isThe path direction and surface normal are related by the right-hand rule. Taking the divergence of the differential law gives $\nabla \cdot \mathbf{J}=0$, since the divergence of a curl vanishes. However, charge conservation requires the continuity equation $\nabla \cdot \mathbf{J}=-\partial \rho /\partial t$, which need not vanish during charging.

Using Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the combination $\mathbf{J}+{ϵ}_0\partial \mathbf{E}/\partial t$ has zero divergence. Replacing $\mathbf{J}$ by this conserved total current density gives the Ampère-Maxwell equation:Its integral form is $\oint \mathbf{B}\cdot d\mathbf{\ell }={\mu }_0[I_{\mathrm{enc}}+{ϵ}_{0}d{\Phi }_E/dt]$. The extra term is the displacement current. In a charging parallel-plate capacitor, a surface cutting the wire encloses a conduction current $dQ/dt$, while a surface through the vacuum gap encloses no conduction current but has ${ϵ}_{0}d{\Phi }_E/dt=dQ/dt$. This modification makes the magnetic circulation independent of which spanning surface is chosen. The vacuum displacement current represents a changing electric field, not literal transport of charge across the gap.

Spherical symmetry and the vacuum Gauss's law imply $\mathbf{E}=A\hat{\mathbf{r}}/r^2$ in the gap, where $A$ is constant. Integrating the electric field between the two equipotentials givesConsequentlyThe inner charge is $Q=4\pi {ϵ}_{0}A$, which reproduces the spherical capacitor capacitance $4\pi {ϵ}_{0}ab/(b-a)$. The radial direction is outward for positive $V_1$ and inward for negative $V_1$; the formula represents the signed field, not just its magnitude.

Place eight identical cubes around the charged corner to make one larger cube, with the point charge at its centre. Each large face has electric flux $q/(6{ϵ}_0)$ and consists of four equivalent small faces. Each of the original cube's three faces opposite the charged corner therefore hasOn each of its three faces containing the corner, the electric field of the point charge obeying Coulomb's law is tangent to the face at every regular point; its outward-normal component is zero. Defining the face integral by omitting the singular corner givesA boundary point charge does not satisfy the ordinary nonsingular hypotheses for simply counting it as an interior charge in Gauss's law. The $1/8$ result expresses the octant solid angle and the specified improper surface integrals. Moving a charge towards the corner from inside a cube before integrating is a different limiting operation; its total closed-surface electric flux stays $q/{ϵ}_0$ until it reaches the boundary.

The six faces subtend equal solid angles at the central point charge. Gauss's law fixes their total outward electric flux at $q/{ϵ}_0$. By the cube's symmetry,For negative $q$ the same signed expression represents inward electric flux. The result is independent of the side length.

Define $V=\varphi (r_2)-\varphi (r_1)$ and $\eta =\ln (r_2/r_1)>0$. Cylindrical symmetry and Gauss's law imply a radial electric field proportional to $1/r$. Integrating $\mathbf{E}=-\nabla \varphi$ fixes the sign:If the inner cylinder has linear charge density ${\lambda }_q$, a coaxial Gaussian surface gives $2\pi r{E}_r={\lambda }_q/{ϵ}_0$, so ${\lambda }_q=-2\pi {ϵ}_{0}V/\eta$. Capacitance is positive regardless of the chosen terminal polarity:End effects are neglected for the long cylinders, as required by the cylindrical approximation.

The central point charge and spherical boundaries give a spherically symmetric charge distribution. A conductor in electrostatic equilibrium has zero electric field within its material. Consequently a spherical Gaussian surface inside the metal encloses zero net electric charge, so the inner surface carries $-q$. The shell's prescribed total electric charge then leaves $-2q$ on its outer surface. Gauss's law givesThe electric field is singular at the ideal central point charge. At each charged surface the normal component has distinct one-sided limits, as required by the electrostatic boundary conditions at a conductor. Spherical symmetry makes the outer surface charge density uniform:

For the uniform space charge in a grounded spherical cavity, grounding sets the metal's electric potential to zero. With no external sources and zero electric potential at infinity, the exterior electric potential is also zero by the uniqueness theorem for the electrostatic potential. The original shell electric charge is no longer fixed: electric charge can flow to or from earth.

Inside the cavity, Gauss's law gives $\mathbf{E}=\rho r\hat{\mathbf{r}}/(3{ϵ}_0)$. Integrating the electric field inward from the zero boundary electric potential givesThe inner surface carries $-Q_{\rho }$, where $Q_{\rho }=4\pi \rho r_1^3/3$, and the outer surface carries zero net electric charge. Only the cavity contributes to the final electrostatic energy:The same result follows from $U=\frac{1}{2}\int \rho V{d}^{3}r$; the conductor's surface charge contributes zero because its electric potential is zero. This is the final stored electrostatic energy, not the change from the earlier central-charge configuration, whose ideal point charge has divergent self-energy.

The current density is the moving charge density times its velocity:For a cylindrical Gaussian surface of radius $r<R$ and length $\ell$, cylindrical symmetry makes the electric field radial. Gauss's law gives $E(2\pi r\ell )=\rho \pi r^2\ell /{ϵ}_0$. A circular Ampère's circuital law contour encloses electric current $I(r)=\rho v_0\pi r^2$, giving $B(2\pi r)={\mu }_{0}I(r)$. HenceThe right-hand rule fixes the positive azimuthal direction of the magnetic field for positive electric charge moving along $+z$. Both the electric field and the magnetic field vanish on the axis. These are the steady long-cylinder electric field and magnetic field; end effects and changes of the prescribed beam distribution are excluded.

The integral and differential forms of Gauss's law in electrostatics areHere $\partial \mathcal{V}$ is a closed surface bounding a volume $\mathcal{V}$, $d\mathbf{A}=\hat{\mathbf{n}}dA$ is its outward area vector, $\mathbf{E}$ is the electric field, $\rho$ is total volume charge density, and $Q_{\mathrm{enc}}={\int }_{\mathcal{V}}\rho d^{3}r$ is enclosed electric charge. The divergence theorem converts one form to the other. For ideal sheets or point charges, interpret the differential equation with their distributional charge densities.

The continuity equation for electric charge is $\nabla \cdot J=-{\partial }_t\rho$. Taking the divergence of the magnetostatic Ampère's law would instead require $\nabla \cdot J=0$. The displacement current repairs this conflict: using Gauss's law, $\nabla \cdot ({ϵ}_0{\partial }_{t}E)={\partial }_t\rho$, so the Ampère-Maxwell equation isIn the parallel-plate capacitor, neglect fringing fields and take $E_z=V_0{e}^{-i\omega t}/d$. A circular Ampère loop of radius $s<a$ encloses displacement current $-i\omega {ϵ}_0\pi s^2{V}_0{e}^{-i\omega t}/d$. Therefore the magnetic field between circular capacitor plates has phasor and amplitudeIt is zero on the axis, grows linearly with radius, and is azimuthal by the right-hand rule. This is a quasistatic approximation: in addition to $d\ll a$, the uniform-field model requires $\omega a/c\ll 1$ and excludes the edge region. The answer gives the oscillation amplitude, rather than a time-independent instantaneous magnitude.