Gauss's law

Gauss's law^[9][10] states that "the total electric flux through any closed surface in free space of any shape drawn in an electric field is proportional to the total electric charge enclosed by the surface." Many numerical problems can be solved by considering a Gaussian surface around a body. Mathematically, Gauss's law takes the form of an integral equation:

${\displaystyle \Phi _{E}=\oint _{S}\mathbf {E} \cdot \mathrm {d} \mathbf {A} ={Q_{\text{enclosed}} \over \varepsilon _{0}}=\int _{V}{\rho \over \varepsilon _{0}}\mathrm {d} ^{3}r,}$

where ${\displaystyle \mathrm {d} ^{3}r=\mathrm {d} x\ \mathrm {d} y\ \mathrm {d} z}$ is a volume element. If the charge is distributed over a surface or along a line, replace ${\displaystyle \rho \,\mathrm {d} ^{3}r}$ by ${\displaystyle \sigma \,\mathrm {d} A}$ or ${\displaystyle \lambda \,\mathrm {d} \ell }$. The divergence theorem allows Gauss's Law to be written in differential form:

${\displaystyle \nabla \cdot \mathbf {E} ={\rho \over \varepsilon _{0}}.}$

where ${\displaystyle \nabla \cdot }$ is the divergence operator.

Poisson and Laplace equations

The definition of electrostatic potential, combined with the differential form of Gauss's law (above), provides a relationship between the potential Φ and the charge density ρ:

${\displaystyle {\nabla }^{2}\phi =-{\rho \over \varepsilon _{0}}.}$

This relationship is a form of Poisson's equation.^[11] In the absence of unpaired electric charge, the equation becomes Laplace's equation:

${\displaystyle {\nabla }^{2}\phi =0,}$

Electrostatic potential

As the electric field is irrotational, it is possible to express the electric field as the gradient of a scalar function, ${\displaystyle \phi }$, called the electrostatic potential (also known as the voltage). An electric field, ${\displaystyle E}$, points from regions of high electric potential to regions of low electric potential, expressed mathematically as

${\displaystyle \mathbf {E} =-\nabla \phi .}$

The gradient theorem can be used to establish that the electrostatic potential is the amount of work per unit charge required to move a charge from point ${\displaystyle a}$ to point ${\displaystyle b}$ with the following line integral:

${\displaystyle -\int _{a}^{b}{\mathbf {E} \cdot \mathrm {d} \mathbf {\ell } }=\phi (\mathbf {b} )-\phi (\mathbf {a} ).}$

From these equations, we see that the electric potential is constant in any region for which the electric field vanishes (such as occurs inside a conducting object).

Electrostatic energy

A test particle's potential energy, ${\displaystyle U_{\mathrm {E} }^{\text{single}}}$, can be calculated from a line integral of the work, ${\displaystyle q_{n}\mathbf {E} \cdot \mathrm {d} \mathbf {\ell } }$. We integrate from a point at infinity, and assume a collection of ${\displaystyle N}$ particles of charge ${\displaystyle Q_{n}}$, are already situated at the points ${\displaystyle \mathbf {r} _{i}}$. This potential energy (in Joules) is:

${\displaystyle U_{\mathrm {E} }^{\text{single}}=q\phi (\mathbf {r} )={\frac {q}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {Q_{i}}{\left\|{\mathcal {\mathbf {R} _{i}}}\right\|}}}$

where ${\displaystyle \mathbf {\mathcal {R_{i}}} =\mathbf {r} -\mathbf {r} _{i}}$ is the distance of each charge ${\displaystyle Q_{i}}$ from the test charge ${\displaystyle q}$, which situated at the point ${\displaystyle \mathbf {r} }$, and ${\displaystyle \phi (\mathbf {r} )}$ is the electric potential that would be at ${\displaystyle \mathbf {r} }$ if the test charge were not present. If only two charges are present, the potential energy is ${\displaystyle Q_{1}Q_{2}/(4\pi \varepsilon _{0}r)}$. The total electric potential energy due a collection of N charges is calculating by assembling these particles one at a time:

${\displaystyle U_{\mathrm {E} }^{\text{total}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{j=1}^{N}Q_{j}\sum _{i=1}^{j-1}{\frac {Q_{i}}{r_{ij}}}={\frac {1}{2}}\sum _{i=1}^{N}Q_{i}\phi _{i},}$

where the following sum from, j = 1 to N, excludes i = j:

${\displaystyle \phi _{i}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{\stackrel {j=1}{j\neq i}}^{N}{\frac {Q_{j}}{r_{ij}}}.}$

This electric potential, ${\displaystyle \phi _{i}}$ is what would be measured at ${\displaystyle \mathbf {r} _{i}}$ if the charge ${\displaystyle Q_{i}}$ were missing. This formula obviously excludes the (infinite) energy that would be required to assemble each point charge from a disperse cloud of charge. The sum over charges can be converted into an integral over charge density using the prescription ${\textstyle \sum (\cdots )\rightarrow \int (\cdots )\rho \,\mathrm {d} ^{3}r}$:

${\displaystyle U_{\mathrm {E} }^{\text{total}}={\frac {1}{2}}\int \rho (\mathbf {r} )\phi (\mathbf {r} )\,\mathrm {d} ^{3}r={\frac {\varepsilon _{0}}{2}}\int \left|{\mathbf {E} }\right|^{2}\,\mathrm {d} ^{3}r,}$

This second expression for electrostatic energy uses the fact that the electric field is the negative gradient of the electric potential, as well as vector calculus identities in a way that resembles integration by parts. These two integrals for electric field energy seem to indicate two mutually exclusive formulas for electrostatic energy density, namely ${\textstyle {\frac {1}{2}}\rho \phi }$ and ${\textstyle {\frac {1}{2}}\varepsilon _{0}E^{2}}$; they yield equal values for the total electrostatic energy only if both are integrated over all space.

Electrostatic pressure

On a conductor, a surface charge will experience a force in the presence of an electric field. This force is the average of the discontinuous electric field at the surface charge. This average in terms of the field just outside the surface amounts to:

${\displaystyle P={\frac {\varepsilon _{0}}{2}}E^{2},}$

This pressure tends to draw the conductor into the field, regardless of the sign of the surface charge.