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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.

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.

Place the same positive electric charge $Q$ uniformly on the sphere of radius $R$, with no external field. Every concentric interior Gaussian surface encloses zero charge; spherical symmetry then makes the interior electric field zero. Outside, it is again $Q/(4\pi {\epsilon }_0{r}^2)$. At $R$, the inner and outer limits differ by $Q/(4\pi {\epsilon }_0{R}^2)=\sigma /{\epsilon }_0$, with surface charge density $\sigma =Q/(4\pi R^2)$. The ideal surface charge makes the field discontinuous, so the plot displays one-sided limits rather than assigning a single field value on the charged sheet. An isolated charged conducting sphere in electrostatic equilibrium realizes this exterior/interior pattern.

Faraday's ice pail experiment distinguishes redistribution of electric charge by electrostatic induction from transfer of electric charge by contact. Lower a charged body into an initially neutral isolated metal pail: an externally connected electrometer responds even though the body never touches the metal. Withdraw the body without contact and the response disappears. Repeat with conductive contact before withdrawal and a residual response can remain. The striking point is that an exterior response does not establish that the pail has gained net electric charge.

Apparatus and readout. A modern implementation uses two concentric wire-mesh cylinders: a 10 cm diameter, 15 cm high inner pail on insulating rods and a grounded outer shield. Connect the electrometer's signal lead to the inner pail and its reference lead to the shield. Momentarily ground and zero the signal assembly, then isolate it; the shield remains grounded. A charged disk on an insulating handle goes into the lower half without touching. The electrometer measures electric potential difference in volts, with sign determined by the lead connections. It does not directly measure electric charge in coulombs.^[ref]

A charge ledger before contact. To separate exact reasoning from the open apparatus, first idealize the pail as a closed cavity in a conductor with one internal body of total electric charge $q$. Let the isolated metal have total electric charge $Q_0$. The metal's electric field is zero at electrostatic equilibrium.^[ref] A Gaussian surface enclosing the cavity in a conductor but lying wholly in the metal therefore has zero electric flux. Gauss's law and charge conservation giveFor $Q_0=0$ and $q>0$, the inner-wall total is $-q$ and the outer-wall total is $+q$: the metal remains neutral. The exterior electric field can therefore be nonzero without any transfer from the body. The connected electrometer and signal cable belong to the same isolated conducting assembly and share its exterior electric charge. The zero-electric field assertion concerns metal, not the occupied cavity in a conductor.

This ledger fixes totals, not the local surface charge density. Moving an off-center body changes the inner-wall distribution while its integral remains $-q$. With fixed outer geometry, fixed $Q_0+q$ and unchanged exterior sources, Uniqueness of the Dirichlet problem together with the fixed total outer electric charge determines the same exterior solution: the outside cannot reveal the body's position or detailed internal distribution. For example, internal bodies with stipulated electric charges $+2q$ and $-q$ give the same total exterior response as a single body with $+q$. Internal bodies with $+q$ and $-q$ can give a null exterior response while retaining a nonzero internal electric field. This is why a null reading cannot establish an empty cavity in a conductor.

Withdrawal is the control. Record a baseline, insert without contact, withdraw, and check return to baseline. After a fresh zero, repeat with the charged disk touching the inner wall, then withdraw and record the residual response. Finally zero the pail and reinsert the disk without contact to test remaining electric charge. Keep the operator grounded, discharge unwanted handle electric charge, and keep the unused charged disk distant. Dirty supports and humidity promote leakage. These controls separate transfer from electrostatic induction and stray coupling.^[ref]

In the ideal contact test, the internal body is itself an electrical conductor and makes a conducting connection to the wall. After charge relaxation, body and enclosure become a single equipotential; the remaining charge-free void has zero electric field. The body then carries no excess electric charge on surfaces facing that void. Withdrawal without subsequent charging leaves total metal electric charge $Q_0+q$. A coated paddle can retain electric charge on insulating patches, so merely touching it does not guarantee complete transfer. A residual reading establishes some transfer under the controls; the disk retest checks the stronger claim of complete transfer. The ideal argument assumes no leakage and no relevant contact-generated charging.

Grounding changes the ledger. With no exterior sources and zero electric potential at infinity, a closed grounded enclosure has zero exterior electric field, inner-wall electric charge $-q$ and zero outer-wall electric charge.^[ref] Consequently, disconnecting ground before withdrawal retains total metal electric charge $-q$, which moves to the exterior after withdrawal. Keeping ground connected during withdrawal instead restores zero electric charge. A lasting opposite-sign response can therefore arise through grounding without body–wall contact.

For the real open pail, no Gaussian surface lying wholly in metal closes over the mouth. Increasing insertion depth and seeking a stable response tests whether enclosure is a good approximation; mesh openings, rim proximity and nearby objects limit it. In a fixed-geometry approximation, the positive insertion response obeys $\Delta V\simeq q/C_{\mathrm{eff}}$, where $C_{\mathrm{eff}}$ is the effective capacitance of the pail, shield, cable and electrometer input. Changed geometry changes that conversion. Wait for charge relaxation and instrument settling at each step, while keeping the observation short compared with leakage. The experiment probes settled electrostatics; it does not establish shielding against arbitrary changing electromagnetic fields.

A Gaussian pillbox is a thin cylindrical Gaussian surface crossing an interface. Shrinking its height converts a divergence law into a jump condition for the field's normal component.

A Gaussian surface is a closed surface used with the integral form of Gauss's law to relate electric flux to enclosed electric charge. It need not coincide with the surface of any charged body. Choosing it to match a symmetry can make the normal component of the electric field constant on a portion of it, allowing that component to be taken outside a surface integral. A concentric sphere works for an isolated spherically symmetric charge distribution; an arbitrarily placed sphere generally does not.

Zero enclosed electric charge implies zero total electric flux. Inferring zero electric field additionally needs information such as the spherical symmetry used in the electric field of a uniformly charged spherical shell. The imagined boundary does not itself block or redirect the electric field.

Gauss's law relates the electric field on a closed surface to the net electric charge inside it. In SI units, with outward unit normal $\mathbf{n}$,Here $\rho$ describes a volume charge density, and ${\epsilon }_0$ is the vacuum permittivity; for ideal point or surface charges, use the enclosed electric charge directly rather than an ordinary volume-density integral. The left side is electric flux: the surface integral of the outward normal component of the electric field. Field directed outward contributes positively; field directed inward contributes negatively. The equality uses net signed electric charge, so positive and negative enclosed charges can cancel. The Gaussian surface is an imagined accounting boundary, not necessarily a material object. See OpenStax, explaining Gauss's law.

Where the electric field and charge density are smooth, the local statement is $\nabla \cdot \mathbf{E}=\rho /{\epsilon }_0$. The divergence theorem connects this divergence to the surface integral. This is the satisfying connection between local sources and global accounting: a surface can register how much net electric charge it encloses without revealing where that charge is located within it. A point charge requires care with the singular field at its position; the integral formulation remains useful with a boundary that avoids the charge.

The law fixes total electric flux, rather than every value of the electric field. In an uncharged region, a uniform electric field $\mathbf{E}=(E_0,0,0)$ crosses a cube of side $\ell$. Its left and right faces contribute $-E_0{\ell }^2$ and $+E_0{\ell }^2$; the other four faces contribute zero. Thus the net electric flux is zero while the electric field is nonzero everywhere. Charges outside a Gaussian surface can change the field on it without changing its net electric flux. Conversely, fields cannot generally be reconstructed from enclosed electric charge alone. Electrostatics also requires the electric field to be the negative gradient of an electric potential, with appropriate boundary information. This distinction is emphasized in Feynman lectures II.5, application of Gauss' law.

Spherical symmetry supplies the extra information needed for a particularly simple calculation. An isolated spherically symmetric charge distribution in vacuum produces a radial electric field, constant in magnitude on each concentric sphere. Its electric flux is therefore $4\pi r^2{E}_r(r)$, givingOutside the distribution, doubling $r$ quadruples the surface area and divides the radial electric field by four, while the electric flux stays $Q/{\epsilon }_0$. The inverse-square dependence is an area-dilution result. The electric field of a uniformly charged ball and electric field of a uniformly charged spherical shell show that the same total charge can give different interior fields. These symmetry arguments and their limits are developed in OpenStax, applying Gauss's law.

The outward flux of the Newtonian gravitational field through a closed surface is $-4\pi G$ times the enclosed mass. A sphere around a point mass establishes the coefficient; superposition extends the integral law to general mass distributions. The minus sign expresses attractive gravity. A symmetric Gaussian surface can turn the flux relation into a local field calculation, but the integral law alone does not imply that the field is uniform on an arbitrary surface.

A circular infinite cylinder of radius $b$ and uniform mass density $\rho$ has linear mass density $\lambda =\pi b^2\rho$. A coaxial Gaussian surface of length $L$ has inward radial gravitational field and no end-cap flux. Gauss's law for gravity gives $2\pi rL{g}_r=-4\pi G\lambda L$ outside, hence the displayed result. Inside, enclosed linear mass density is $\pi r^2\rho$, so $g_r=-2\pi G\rho r$. An exterior circular orbit has speed $\sqrt{2G\lambda }$ independent of radius. The exterior formula presupposes circular axial symmetry; area alone would not specify the near field of a noncircular rod.

For radii $a<b$ and a point charge $q$ at $r_0{\mathbf{e}}_z$, $a<r_0<b$, set $k=b/a$. The method of images gives the two families $(q{k}^n,r_0{k}^{2n}{\mathbf{e}}_z)$ and $(-qa{k}^n/r_0,a^2{k}^{2n}{\mathbf{e}}_z/r_0)$ for $n\in \mathbb{Z}$. The first family's $n=0$ term is the physical point charge. All other image charges lie outside the gap. Sum their Coulomb's law electric potentials with factor $1/(4\pi {ϵ}_0)$.

For a short proof, reflection of $(Q,z)$ in a grounded sphere of radius $R$ gives $(-RQ/z,R^2/z)$. On $r=a$ this pairs first-family index $n$ with second-family index $-n$; on $r=b$ it pairs $n$ with $1-n$. Each pair has zero boundary electric potential. The negative-index electric potential tails are $O(k^n)$ and the positive-index tails are $O(k^{-n})$, uniformly on the closed gap after excluding the physical singularity. These geometric series establish absolute convergence, boundary cancellation, and termwise differentiation away from the physical point charge. Uniqueness of Poisson equation identifies the solution. A spherical Gaussian surface just outside $a$ encloses the first family with $n\le -1$ and the second with $n\le 0$, so Gauss's law and the geometric series give $Q_a=-qa(b-r_0)/(r_0(b-a))$.

Choose a spherical Gaussian surface of radius $s$ with $a<s<r_0$. Its physical enclosed electric charge is the total induced electric charge $Q_a$ on the inner sphere. The method of images produces the same electric field on that Gaussian surface, so Gauss's law identifies $Q_a$ with the sum of the enclosed image charges. These are the first family with $n\le -1$ and the second family with $n\le 0$:Summing these geometric series givesThe electric field in the empty central cavity vanishes by electrostatic shielding, so it contributes no additional inner-face surface charge density. The formula tends to $-q$ as $r_0\downarrow a$ and to zero as $r_0\uparrow b$, as expected when the point charge approaches one grounded boundary or the other.

Translate the concentric spheres so that their centre is $(0,0,-a)$, take $b=a+d$, and keep the point charge fixed at $(0,0,h)$. Its distance from the centre is $r_0=a+h$. As $a\to \infty$, the upper portions of the inner and outer spheres approach the planes $z=0$ and $z=d$. The preceding total induced electric charge becomesThe outer sphere has total induced electric charge $Q_b=-q-Q_a$: a Gaussian surface within its conducting material has zero electric flux, and the grounded exterior is field-free by uniqueness of Poisson equation. Therefore $Q_b\to -qh/d$.

One can justify the total-charge limit without assuming that local convergence of electric fields preserves their surface integrals. Use induced charge as harmonic measure: if a harmonic function $w$ is one on the selected grounded boundary and zero on the other, Green's second identity with the electric potential givesIndeed, ${\nabla }^2\varphi =-q{\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)/{ϵ}_0$, $\varphi =0$ on the boundaries, and their unit normal outward from the gap satisfies $\sigma ={ϵ}_0{\partial }_n\varphi$. For the spheres the weighting harmonic function is $w_a(r)=a(b-r)/(r(b-a))$, exactly the factor already obtained. For the slab the two weighting harmonic functions are $w_0=1-z/d$ and $w_d=z/d$. Applying the same identity to the decaying grounded-slab electric potential, whose boundary term at lateral infinity vanishes, establishes directlyTheir sum is $-q$, and their relative sizes measure how close the point charge lies to each grounded plane.

By cylindrical symmetry the electric field of an infinite line charge is radial, independent of $z$ and angle. Use a coaxial cylindrical Gaussian surface of radius $r$ and length $L$. The end caps have zero flux, so Gauss's law gives $2\pi rL{E}_r=\mu L/{\epsilon }_0$. Hence the field and electrostatic potential areHere $r_0>0$ is an arbitrary reference length, and $\mathbf{E}=-\nabla \varphi$. One cannot choose zero electrostatic potential at infinity for a single infinite line charge, because its logarithmic potential diverges there; the arbitrary reference handles the additive constant.

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.

Introduce cylinder length $\ell$ and line charge density $\lambda =Q/\ell$; only charge per length matters in the long-cylinder approximation. A coaxial cylindrical Gaussian surface gives $E_{r}2\pi r\ell =Q/{ϵ}_0$. Choosing the outer cylinder's electric potential to be zero,The capacitance per unit length follows from $C=Q/[V(a)-V(b)]$:If $Q$ is intended as a line charge rather than a finite total charge, replace $\lambda$ by that $Q$. Neglecting end effects, equal opposite cylinder charges make the exterior electric field vanish; the potential difference between the conductors remains the quantity defining their capacitance.

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:

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.

Put electric charge $+Q$ on the inner electrical conductor and $-Q$ on the outer one. In the gap, spherical symmetry makes the electric field radial and constant on a spherical Gaussian surface. Gauss's law givesThe inner-to-outer electric potential difference is thereforeThus the capacitance of the spherical capacitor isAs $b\to \infty$ this gives the capacitance of an isolated conducting sphere, $4\pi {ϵ}_{0}a$; for $b-a\ll a$ it approaches the parallel-plate capacitor result with area $4\pi a^2$.

Use opposite electric charges $\pm Q$ over a common length $L$, and write the inner linear charge density as $\lambda =Q/L$. Away from the ends, axisymmetry and Gauss's law on a coaxial cylindrical Gaussian surface of length $\ell$ giveThe electric potential difference and capacitance of the coaxial cylindrical capacitor areThis neglects the fringing field of a capacitor at the ends, as justified when $L\gg b$.