Gauss's law (${\oint }_{\partial V}\mathbf{E}\cdot \mathbf{n}dS=Q_{\mathrm{enclosed}}/{\epsilon }_0$)

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.

Place a fixed sheet of charge density $-\sigma$ at $z=0$ and a uniform fixed volume charge $\rho$ in $-d<z<0$. With no imposed uniform background field, the above-sheet and below-slab fields are $\pm (\rho d-\sigma )/(2{ϵ}_0)$. The interior field is the title expression, obtained by adding the slab's linear field to the sheet's constant field. If $0<\sigma <\rho d$, it vanishes at $z=-(d+\sigma /\rho )/2$, inside the lower half of the slab. At the sheet it jumps down by $\sigma /{ϵ}_0$, while it remains continuous at the slab's uncharged lower face.

For a point charge exactly at a cube corner, each of the three opposite faces has outward electric flux $q/(24{ϵ}_0)$. Reflecting the cube in its three incident faces makes a larger cube centred on the charge and proves the result by symmetry. The three incident faces have zero normal electric field away from their singular corner. Their improper integrals vanish; the total is $q/(8{ϵ}_0)$. This boundary convention differs from moving a charge to the corner from the interior after computing a closed-surface integral.

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.

Remove an axial cylindrical hole, wholly contained in an infinite insulating cylinder with uniform fixed charge density $\rho$. Let $\mathbf{a}$ point from the original axis to the hole axis. The superposition principle replaces the removed material by a cylinder of charge density $-\rho$. At any point $\mathbf{r}$ inside the hole, the two interior electric fields add to $\rho [\mathbf{r}-(\mathbf{r}-\mathbf{a})]/(2{ϵ}_0)=\rho \mathbf{a}/(2{ϵ}_0)$. The result is uniform throughout the hole, even when the hole is not concentric; it does not apply to a hole in an electrical conductor whose electric charges can redistribute.

Outside a bounded spherically symmetric charge distribution in vacuum, the electric field is that of a central point charge with the same total electric charge. With zero electric potential at infinity, the exterior electric potential is $Q/(4\pi {ϵ}_{0}r)$. A uniform surface charge on a sphere has zero electric field inside.

For a spherically symmetric electric field in vacuum, integrate Gauss's law over a sphere. Its electric flux is $4\pi r^2{E}_r(r)$, so taking the enclosing radius to infinity gives the displayed total charge, whenever that limit and total charge exist. In particular, a radial field decaying faster than $r^{-2}$ can be nonzero locally while its total source charge is zero. The origin must also be checked: a point charge contributes a nonzero limiting flux as $r\to 0$.

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.

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.

In static time times round three-dimensional hyperbolic space of curvature radius $R$, a geodesic sphere has area $4\pi R^2{\sinh }^2(\xi /R)$. The Gauss law therefore gives the displayed radial field. It approaches the flat-space inverse-square field near the charge and falls exponentially at large geodesic distance. The curvature changes the area of a sphere, not the conserved flux.

For uniform charge density ${\rho }_e=3Q/(4\pi R^3)$, Gauss's law gives the interior electric field $E={\rho }_{e}r/(3{\epsilon }_0)$. Integrate its vertical force density ${\rho }_e^{2}z/(3{\epsilon }_0)$ over the upper hemisphere: $\int zdV=\pi R^4/4$. Internal pair forces in that hemisphere cancel, so its net force is exactly the force from the lower hemisphere. The result is upward and has magnitude $3Q^2/(64\pi {\epsilon }_0{R}^2)$.

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.

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.