# Gauss’s Law for Electric Fields¶

Fig. 33 Charge enclosed

Gauss’s law for the electric field describes the static electric field generated by a distribution of electric charges. It states that the electric flux through any closed surface is proportional to the total electric charge enclosed by this surface. By convention, a positive electric charge generates a positive electric field. The law was published posthumously in 1867 as part of a collection of work by the famous German mathematician Carl Friedrich Gauss.

## Integral Equation¶

Gauss’s law in integral form is given below:

(34)$\int_V \boldsymbol{\nabla} \cdot \mathbf{e} \; ~dv =\oint_{S} \mathbf{e} \cdot \hat{\mathbf{n}} \; ~da = \frac{Q}{ \varepsilon_{0} }\;,$

where:

• $$\mathbf{e}$$ is the electric field

• $$Q$$ is the enclosed electric charge

• $$\varepsilon_0$$ is the electric permittivity of free space

• $$\hat{\mathbf{n}}$$ is the outward pointing unit-normal

Flux is a measure of the strength of a field passing through a surface. Electric flux is defined in general as

(35)$\boldsymbol{\Phi} = \int_S \mathbf{e} \cdot \hat{\mathbf{n}} \, \mathrm{d}a.$

We can think of electric field as flux density. Gauss’s law tells us that the net electric flux through any closed surface is zero unless the volume bounded by that surface contains a net charge.

## Differential Form¶

When considering a spatially extended charged body, we can think of its charge as being continously distributed throughout the body with density $$\rho$$. The total charge is then given by the integral of the charge density over the volume of the body.

(36)$Q = \int_V \rho \; \mathrm{d}v\;.$

Using this definition and applying the divergence theorem to the left hand side of Gauss’s law (34), we can rewrite the law as:

(37)$\int_V \boldsymbol{\nabla} \cdot \mathbf{e} \; \mathrm{d}v = \int_V \frac{\rho}{\varepsilon_0} \; \mathrm{d}v \;.$

Since this equation must hold for any volume $$V$$ , we can equate the integrands, giving the differential form of Gauss’s law:

(38)$\boldsymbol{\nabla} \cdot \mathbf{e} = \frac{\rho}{\varepsilon_0}.$

It can be shown that Gauss’ law for electric fields is equivalent to Coulomb’s law (see Equivalence of Gauss’ Law for Electric Fields to Coulomb’s Law)

## Gauss’s Law in Matter¶

Gauss’s law for electric fields is most easily understood by neglecting electric displacement ($$\mathbf{d}$$). In matter, the dielectric permittivity may not be equal to the permittivity of free-space (i.e. $$\varepsilon \neq \varepsilon_0$$). In matter, the density of electric charges can be separated into a “free” charge density ($$\rho_f$$) and a “bounded” charge density ($$\rho_b$$), such that:

(39)$\rho = \rho_f + \rho_b$

The free-charge density refers to charges which flow freely under the application of an electric field; i.e. they produce a current which is divergence-free. The bounded-charge density refers to electrical charges attributed to electrical polarization ($$\mathbf{p}$$). By combining Eqs. (38) and (39) with our definition for electrical polarization, we find that:

(40)$\nabla \cdot \mathbf{d} - \nabla \cdot \mathbf{p} = \rho_f + \rho_b$

By using the constitutive relationship $$\mathbf{d} = \varepsilon \mathbf{e}$$ and separating the previous equation into bounded and free contributions, we find that:

(41)$-\nabla \cdot \mathbf{p} = \rho_b$

and

(42)$\nabla \cdot \mathbf{d} = \rho_f$

The above equation is the differential form of Gauss’s equation in matter. Meanwhile, the integral form of Gauss’s equations in matter is given by:

$\int_V \nabla \cdot \mathbf{d} \; dV = \oint_S \mathbf{d} \cdot \mathbf{\hat n} \; da = Q_f$

where $$Q_f$$ is the total enclosed free charge.

## Units¶

 Surface area $$\text{S}$$ m $$\! ^{2}$$ Square meter Volume $$V$$ m $$\! ^{3}$$ Cubic meter Electric charge $$q, Q, Q_f$$ C Coulomb Electric charge density $$\rho, \rho_f, \rho_b$$ C/m $$\! ^{3}$$ Coulomb per cubic meter Electric field $$\mathbf{e}$$ V/m Volt per meter Electric displacement $$\mathbf{d}$$ A/m $$\! ^{2}$$ Volt per meter Dielectric permittivity $$\varepsilon$$ F/m Farad per meter

Conversions

$\varepsilon_0 = \frac{\text{F}}{\text{m}} = \frac{\text{C}}{\text{V} \cdot \text{m}}.$