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For an isolated smooth conducting sphere of radius $a$, Gauss's law gives surface electric field $E_s=V/a$. Combining its capacitance $4\pi {ϵ}_{0}a$ with capacitor energy $C{V}^2/2$ gives the displayed bound when the surrounding medium has dielectric strength $E_{\mathrm{bd}}$ and approximately vacuum permittivity. Nearby conductors, surface roughness and discharge geometry can lower the practical limit.

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.

Place an infinite sheet of surface charge density $\sigma$ at $z=0$ between grounded conducting planes $z=\pm d/2$, in vacuum. Reflection symmetry and Gauss's law give $E_z=\sigma \mathrm{sgn}(z)/(2{\epsilon }_0)$ in the two gaps. Integrating the electric field with zero electric potential at the conductors gives $V(z)=\sigma (d/2-|z|)/(2{\epsilon }_0)$. Each facing conducting surface carries induced surface charge density $-\sigma /2$. For $\sigma >0$, an electron released from rest at a plate reaches the sheet with kinetic energy $e\sigma d/(4{\epsilon }_0)$; its nonrelativistic speed follows from $m_e{v}^2/2$.

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.

A spherical electrical conductor has constant electric potential throughout its connected interior in electrostatic equilibrium. For an isolated sphere in a homogeneous medium, spherical symmetry and Gauss's law give the exterior field of a point electric charge at its center.

A conductor in electrostatic equilibrium is an electrical conductor at rest whose mobile charge carriers have finished redistributing, with no maintained bulk electric current. In the ideal macroscopic model,A residual bulk electric field would drive further electric charge redistribution. The electric potential difference between any two points joined through the metal is $-\int \mathbf{E}\cdot d\mathbf{l}=0$, so each connected piece is an equipotential. Gauss's law then gives zero net bulk charge density. This is a statement about net macroscopic electric charge, not the absence of positive atomic nuclei and negative electrons. Excess electric charge is represented by surface charge density; atomic surface structure is outside this continuum model. ^[ref].

The word inside must distinguish metal from a cavity in a conductor. A completely enclosed, charge-free cavity in a conductor has constant boundary electric potential $V_c$. The constant solution of Laplace's equation and Uniqueness of the Dirichlet problem give $\Phi =V_c$ and $\mathbf{E}=0$ throughout the cavity in a conductor, regardless of its shape or external electric charges. Grounding is unnecessary for this inward electrostatic shielding; an isolated enclosure can change its overall electric potential without producing an internal electric field. Zero enclosed net electric charge alone would only give zero electric flux. A positive and negative pair inside a cavity in a conductor can have zero total electric charge and a nonzero electric field. ^[ref].

If fixed electric charges inside a cavity in a conductor sum to $q$, a Gaussian surface lying entirely in the metal and enclosing that cavity in a conductor has zero electric flux. Thus the total induced inner-wall electric charge is $-q$. For one such cavity in a conductor and an isolated enclosure with total metal electric charge $Q_c$, the outer-wall electric charge is $Q_c+q$. In particular, an initially neutral isolated enclosure generally has an exterior electric field when $q\ne 0$. For fixed internal sources, external changes only shift the constant electric potential inside an isolated enclosure, leaving the internal electric field unchanged. ^[ref].

A grounded enclosure fixes $V_c=0$ and can acquire metal electric charge $-q$; with no exterior sources and $\Phi \to 0$ at infinity, Uniqueness of the Dirichlet problem gives zero exterior electric field. With external sources present, the exterior electric field need not vanish, but it is independent of the internal sources at fixed grounded boundary electric potential. ^[ref].

Faraday's ice pail experiment makes the distinction between redistribution and transfer visible. In his letter of 4 February 1843, Faraday described an insulated pewter pail connected by wire to a gold-leaf electrometer, and a brass ball suspended on a long dry silk thread to reduce the hand's influence. He discharged the pail and charged the ball away from it. Lowering the ball without contact produced leaf divergence, which grew during entry and became approximately independent of further depth; withdrawing it collapsed the leaves. Contact with the bottom transferred electric charge, and the withdrawn ball was discharged. The readout was leaf divergence, not a direct map of the internal electric field. These observations distinguish electrostatic induction from contact transfer of electric charge. ^[ref].

For a modern version of Faraday's ice pail experiment, connect an insulated inner metal pail to the signal input of a high-input-impedance electrometer, with its reference lead and an outer shield grounded. Zero and momentarily ground the pail, then isolate the signal side. Lower a charged paddle deeply without touching and record the electric potential difference; withdraw it and check return to zero. Repeat with contact, then withdrawal: a residual reading indicates transferred electric charge. Reverse the paddle's sign and repeat. The electrometer reads volts; inferring electric charge requires the effective capacitance of pail, cable and input, so a stable reading at fixed geometry is a relative measure, not automatically a value in coulombs. Keep supports clean, nearby charged objects distant, and the operator grounded; humidity can cause leakage. ^[ref].

A complementary electrostatic shielding demonstration uses a charged hollow metal enclosure, a small metal sampling ball on an insulating handle, and an electrometer. Discharge the sampling ball between trials; touch it to the exterior, withdraw it, and transfer its collected electric charge to the electrometer. Repeat by touching the inner wall without touching the exterior during withdrawal. The exterior sample produces a response, whereas the inner sample gives a null within instrument sensitivity. This compares accessible surface responses rather than assuming a perfectly spherical enclosure. ^[ref]. Useful additional controls are an uncharged enclosure baseline, a repeat exterior sample to verify sensitivity, and repeated trials at different depths. An access hole, charged handle, leakage or disturbed surface charge density limits the inference; a null bounds detectable transfer, rather than proving an exactly zero electric field. An open pail's depth plateau tests good enclosure empirically. Neither demonstration establishes shielding of arbitrary time-dependent electromagnetic fields.

Coulomb's law gives the mutual electric force between stationary point charges in vacuum. Put electric charges $q_1$ and $q_2$ at ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, and define the displacement from charge 1 to charge 2 by $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$, with separation $r=|\mathbf{r}|>0$ and unit vector $\hat{\mathbf{r}}=\mathbf{r}/r$. In SI units, the force on charge 2 due to charge 1 isHere ${\epsilon }_0$ is the vacuum permittivity. The magnitude is $k|q_1{q}_2|/r^2$. If $q_1{q}_2>0$, the force on charge 2 points away from charge 1; if $q_1{q}_2<0$, it points toward charge 1. Interchanging the charges gives ${\mathbf{F}}_{1\leftarrow 2}=-{\mathbf{F}}_{2\leftarrow 1}$, consistent with Newton's third law in this electrostatic setting. ^[ref].

For a fully computed example, stipulate $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$, ${\mathbf{r}}_1=0$, ${\mathbf{r}}_2=0.20\mathrm{m}\hat{\mathbf{x}}$ and the approximate value $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. ThenThe force is attractive, of magnitude $1.3485\times 10^{-6}\mathrm{N}$. The source electric field at charge 2 is ${\mathbf{E}}_1({\mathbf{r}}_2)=449.5N/C\hat{\mathbf{x}}$, pointing away from the positive source. Multiplication by the negative $q_2$ reverses the direction to give the attractive force. This field excludes charge 2's own self-field.

Holding the electric charges fixed while changing the separation from $R$ to $2R$ or $3R$ gives the normalized inverse-square law valuesFor the stipulated example, separations $0.20$, $0.40$ and $0.60\mathrm{m}$ give magnitudes $1.3485\times 10^{-6}$, $3.37125\times 10^{-7}$ and $1.498333\dots \times 10^{-7}\mathrm{N}$. The diagram computes these ratios and uses one common arrow-length scale for both sign choices.

The inverse-square law has a geometric explanation through Gauss's law: a source point charge gives fixed total electric flux through a concentric sphere, whose surface area is $4\pi r^2$. Dividing that flux by the area gives a radial electric field proportional to $1/r^2$. This connects a pairwise force calculation to a global conservation-like accounting of field sources. For several stationary point charges, the superposition principle adds the individual electric fields as vectors, and ${\mathbf{F}}_i=q_i{\mathbf{E}}_{\mathrm{other}}({\mathbf{r}}_i)$ uses all sources except charge $i$ itself. The Coulomb potential energy gives an alternative route when the desired quantity is work rather than instantaneous force.

For $u=\log (r/a)$ and $\Phi =\beta q\varphi -2u$, a counterion-only Poisson-Boltzmann equation reduces to ${\Phi }^{\prime \prime }=-C{e}^{-\Phi }$. Conditions $\Phi (0)=0$, $\Phi \to \infty$ and ${\Phi }^{\prime }\to 0$ give $\Phi =2\log (1+bu)$, $b=\sqrt{C/2}>0$. Matching surface Gauss's law gives $b=\eta -1$, so this branch requires the Manning parameter $\eta >1$. Its number density is $n(r)=(\eta -1{)}^2(a/r{)}^2/[2\pi {\ell }_B{a}^2(1+(\eta -1)\log (r/a){)}^2]$. It leaves residual line-charge magnitude $q/{\ell }_B$.

The positive mobile charge per unit axial length accumulated from cylinder radius $a$ to $r$ is $Q(r)=2\pi {\int }_a^{r}s\rho (s)ds$. For fixed negative line charge $-\lambda$, Gauss's law gives signed radial electric field $E_r=2[-\lambda +Q(r)]/(ϵr)$ in Gaussian units. Its absolute value is the field magnitude. An integral over a local infinite-dilution profile measures condensed charge, rather than every ion in a finite neutral cell.

For a uniformly charged sphere of radius $a$, total electric charge $Q$, and an exterior dilute electrolyte of permittivity $\epsilon$, the Debye–Hückel approximation gives $(r^2{\varphi }^{\prime }{)}^{\prime }={\kappa }^2{r}^2\varphi$. Decay at infinity and Gauss's law, $-\epsilon {\varphi }^{\prime }(a)=Q/(4\pi a^2)$, select the displayed electric potential. Substituting $u=r\varphi$ reduces the proof to $u^{\prime \prime }={\kappa }^{2}u$. The screening parameter is ${\kappa }^2={\sum }_i{z}_i^2{e}^2{n}_{i,\infty }/(\epsilon k_{B}T)$, and the ion profiles are $n_i\simeq n_{i,\infty }(1-z_{i}e\varphi /(k_{B}T))$. The validity condition is $|z_{i}e\varphi (a)|\ll k_{B}T$ for every relevant species. Integrating the diffuse charge density using the radial equation gives total screening charge $-Q$.

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.

Let the spherically symmetric volume charge density be $\rho (r)=A{r}^p$, with $p>-3$ so the origin contains finite charge. Integration gives $Q(r)=4\pi A{r}^{p+3}/(p+3)$. Gauss's law yields the radial electric field $\mathbf{E}=A{r}^{p+1}\hat{\mathbf{r}}/[{ϵ}_0(p+3)]$ for $r>0$. The field tends to zero at the centre if $p>-1$; finite enclosed charge alone does not guarantee a regular central field.

For spherical charge density ${\rho }_e=-e{e}^{-2r/a_0}/(\pi a_0^3)$, integrate $4\pi s^2{\rho }_e(s)$ from zero to $r$ to obtain the displayed enclosed charge. Gauss's law then gives $E_e=Q_e(r)/(4\pi {ϵ}_0{r}^2)$, directed inward. Near the origin $E_e\simeq -er/(3\pi {ϵ}_0{a}_0^3)$; at large radius it approaches the field of charge $-e$.

Adding a point nucleus of charge $+e$ leaves the unperturbed total fieldIt is exponentially small rather than identically zero, because the electron cloud has no compact boundary. This frozen-density field alone does not describe the dominant long-range response to another charge: charge-induced dipole attraction requires allowing the atom to polarize.

Distribute a positive electric charge $Q$ uniformly through a closed ball of radius $R$ in vacuum, with no external field. Its volume charge density is $3Q/(4\pi R^3)$. For $0<r<R$, the enclosed fraction is $(r/R{)}^3$. Applying Gauss's law on a concentric Gaussian surface givesThe electric field grows linearly inside and decays as an inverse square outside. With $E_R=Q/(4\pi {\epsilon }_0{R}^2)$, its values at $r/R=1/2,1,2$ are $E_R/2,E_R,E_R/4$. The field vanishes at the centre by spherical symmetry and is continuous at the surface because this ideal volume distribution has no singular surface layer. Its interior behavior differs from a conductor in electrostatic equilibrium, where mobile charge redistributes.

For an infinite insulating cylinder of radius $R$ and uniform fixed charge density $\rho$, Gauss's law gives interior electric field $\rho \mathbf{r}/(2{ϵ}_0)$ and exterior electric field $\rho R^2\mathbf{r}/(2{ϵ}_0|\mathbf{r}{|}^2)$, where $\mathbf{r}$ is transverse displacement from its axis. The electric field is continuous at $|\mathbf{r}|=R$ when there is no separate surface charge density. The interior linear dependence makes displaced-cylinder superposition particularly simple.

Infinite sheets at $x=0$ and $y=0$ with surface charge density $+\sigma$ and $-\sigma$ have piecewise uniform electric fields. Gauss's law and superposition give the displayed result outside the sheets. The electric field lines bisect each quadrant and connect the positive sheet to the negative one. Its magnitude is $\sigma /(\sqrt{2}{ϵ}_0)$.

Electric flux measures how much electric field crosses a chosen oriented surface, with a sign distinguishing the two crossing directions. A strong electric field running along a surface contributes nothing to its electric flux: only the normal component counts. This lets Gauss's law turn a potentially complicated electric field into one number recording the net enclosed electric charge.

For a surface $S$ with chosen unit normal $\mathbf{n}$, the electric flux is the surface integralHere $dA$ is positive surface area, while $d\mathbf{S}$ is the vector area element. The dot product makes electric flux a signed scalar, rather than a vector like the electric field. Its SI unit is $\mathrm{N}{\mathrm{m}}^2/C$: electric field measured using the newton per coulomb, multiplied by surface area in square metres. Reversing the orientation of a surface replaces $\mathbf{n}$ by $-\mathbf{n}$ and reverses the electric flux. Calling it electric flux does not imply that electric charge is moving through the surface. ^[ref].

For an open patch in a plane, with surface area $A$ and a constant electric field, the unit normal is constant and the surface integral simplifies toThe angle $\theta$ is between the electric field and the chosen unit normal, not between the electric field and the plane. The cosine describes the signed projected surface area seen by the electric field. Thus electric flux is positive for $0\le \theta <90^{\circ }$, zero at $90^{\circ }$, and negative for $90^{\circ }<\theta \le 180^{\circ }$. ^[ref].

For example, take $E=100N/C$ and $A=0.02{\mathrm{m}}^2$, so $EA=2\mathrm{N}{\mathrm{m}}^2/C$. Three choices of angle giveKeeping the $60^{\circ }$ patch fixed and reversing its unit normal gives $120^{\circ }$ and changes $+1$ to $-1\mathrm{N}{\mathrm{m}}^2/C$. The electric field and physical patch have not changed. If the electric field is instead tangent to the patch, its dot product with the unit normal vanishes: electric flux is zero even though $E=100N/C$.

For a closed surface enclosing a volume $V$, choose the outward unit normal. Gauss's law givesHere ${\epsilon }_0$ is the vacuum permittivity, and $Q_{\mathrm{enclosed}}$ is the signed total electric charge inside. The electric field in the surface integral includes all sources, including external electric charges. Assume no ideal point charge lies on the integration surface. Zero enclosed electric charge implies zero total electric flux; it does not imply zero electric field everywhere on the closed surface or inside it. Positive and negative contributions to the surface integral can cancel. ^[ref].

To see the cancellation with sources inside, place point charges $+q$ at $(-a,0,0)$ and $-q$ at $(a,0,0)$, where $q>0$, $a=0.30\mathrm{m}$, and both lie inside the sphere of radius $R=1\mathrm{m}$ centered at the origin. This separated pair has the electric dipole geometry. Its enclosed electric charge is $q-q=0$, so its total outward electric flux is zero. At the point $P=(0,0,R)$ on the sphere, Coulomb's law and the superposition principle give the individual electric fieldsThe $z$ components cancel and the positive $x$ components add:For the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $q=1\mathrm{nC}$, this is about $4.740N/C$ along positive $x$. At $P$, the outward unit normal is $\hat{\mathbf{z}}$, so this nonzero electric field is tangent to the sphere. At other points of the sphere, its normal component can be positive or negative. The source-field formula applies away from the two point charges, where the electric field has zero divergence; it is singular at the sources. ^[ref].

The diagram shows streamlines of the three-dimensional electric field in the section $y=0$. The circle is a cross-section of the sphere, not the integration surface: Gauss's law uses the entire three-dimensional surface integral. A two-dimensional line integral of outward field around that circle would be a different quantity, with units $\mathrm{N}m/C$ rather than the electric flux units $\mathrm{N}{\mathrm{m}}^2/C$.

In source-free vacuum the electric field and magnetic field obeyA brief derivation takes the curl of Faraday's law and substitutes the Ampère-Maxwell equation. The vector calculus identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, together with source-free Gauss's law, leaves ${\nabla }^2\mathbf{E}={\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$. The same argument gives the magnetic field equation. The operator ${\nabla }^2$ is the Laplacian, applied componentwise.

The electrostatic boundary conditions at a conductor relate the one-sided total electric fields at a smooth metal–vacuum interface. Choose a unit normal $\mathbf{n}$ from metal into the adjacent vacuum, let ${\mathbf{E}}_{\mathrm{metal}}$ and ${\mathbf{E}}_{\mathrm{vac}}$ be the two limiting electric fields, and let $\sigma$ be the signed surface charge density. A thin Gaussian pillbox with cap area $A$ has outward cap normals $+\mathbf{n}$ on the vacuum side and $-\mathbf{n}$ on the metal side. As its height tends to zero, its side electric flux tends to zero for bounded local electric fields, leavingThe minus sign comes from the second outward cap normal, not from the sign of the electric charge. A shrinking rectangular loop across the interface and the zero closed line integral in electrostatics give continuity of the tangential component:For a conductor in electrostatic equilibrium, ${\mathbf{E}}_{\mathrm{metal}}=0$, henceThus positive surface charge density gives an electric field away from metal and negative surface charge density gives an electric field toward metal. The derivative uses the same chosen unit normal, from the vacuum side. The ideal surface layer has no unique electric field value on the layer itself; these are limiting values. Ordinary surface electric charge preserves continuity of electric potential; an ideal electric dipole sheet would require a different model. ^[ref].

Here “exterior” means just outside the metal, including the vacuum within a cavity in a conductor. On a spherical inner wall of radius $a$, its unit normal from metal into the cavity in a conductor is $-\hat{\mathbf{r}}$. A central positive point charge $q$ gives ${\mathbf{E}}_{\mathrm{vac}}=q\hat{\mathbf{r}}/(4\pi {ϵ}_0{a}^2)$ there, soThe outward normal of the cavity in a conductor as a volume is instead $+\hat{\mathbf{r}}$; confusing these two unit normals reverses the inferred sign. For a stipulated spherical enclosure with inner radius $a$, outer radius $b$, no external sources and total isolated metal electric charge $Q_c$, spherical symmetry and Gauss's law give the analytical modelIts outer surface charge density is $(Q_c+q)/(4\pi b^2)$. Grounding to $V_c=0$ with zero electric potential at infinity instead gives $Q_c=-q$ and zero exterior electric field, while preserving the same nonzero electric field inside the charged cavity in a conductor.

As a numerical sign check, stipulate $q=1\mathrm{nC}$, $a=0.10\mathrm{m}$, $b=0.20\mathrm{m}$ and $1/(4\pi {ϵ}_0)\simeq 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. The inner electric field is approximately $899N/C$ along $+\hat{\mathbf{r}}$, but ${\mathbf{E}}_{\mathrm{vac}}\cdot \mathbf{n}=-899N/C$ and ${\sigma }_{\mathrm{inner}}\simeq -7.96nC/{\mathrm{m}}^2$. An isolated neutral enclosure has $+1\mathrm{nC}$ on the outer wall, approximately $+1.99nC/{\mathrm{m}}^2$ and $225N/C$ just outside. Multiplying each surface charge density by its spherical surface area recovers the signed wall electric charges.

The local result uses the total electric field, including all induced and imposed sources. The factor $\sigma /{ϵ}_0$ follows from the zero metal-side electric field; a symmetric isolated sheet in vacuum instead has magnitude $|\sigma |/(2{ϵ}_0)$ on each side. ^[ref]. At sharp corners, limiting electric fields can be singular and the smooth local pillbox argument needs care. In material outside the metal, use the normal free-charge jump of the electric displacement field and the material's permittivity, as in electrostatic interface boundary conditions. ^[ref].