Electric field

The electric field $\mathbf{E}$ at a point in space is the force per unit positive test charge placed at that point:

$\mathbf{E}={\lim }_{q^{\prime }\to 0}\frac{\mathbf{F}}{q^{\prime }}\text{(V/m)}.$

A charge $q^{\prime }$ placed in the field experiences a force $\mathbf{F}=q^{\prime }\mathbf{E}$. Taking the limit $q^{\prime }\to 0$ is the standard idealization — the test charge has to be small enough that it doesn’t disturb the field it’s measuring.

The electric field is a Vector field: at every point in space it assigns a vector (magnitude and direction). It’s the carrier of electrostatic interaction — charges create $\mathbf{E}$, and other charges respond to the local $\mathbf{E}$. This decouples the question “what does charge $q_1$ feel from charge $q_2$” into two steps: $q_2$ sets up a field, and $q_1$ feels the local field where it sits.

Field of a point charge

For an isolated charge $q$ in a medium with permittivity $ϵ$, the field at distance $R$ is

$\mathbf{E}=\hat{\mathbf{R}}\frac{q}{4\pi ϵR^2}\text{(V/m)},$

pointing radially outward for $q>0$, inward for $q<0$. This is Coulomb’s law written in field form.

Radial electric field lines from a positive point charge.

In free space, $ϵ={ϵ}_0=8.854\times 10^{-12}$ F/m. In a Dielectric medium, $ϵ={ϵ}_r{ϵ}_0$ where ${ϵ}_r$ is the relative permittivity.

Electric flux density

In a medium, it’s convenient to define a second vector field, the electric flux density $\mathbf{D}$:

$\mathbf{D}=ϵ\mathbf{E}{\text{(C/m}}^2\text{)}.$

$\mathbf{D}$ has the advantage of being directly tied to free charge via Gauss’s law: $\nabla \cdot \mathbf{D}={\rho }_v$. The complication of the medium (its ${ϵ}_r$) gets absorbed into the relationship $\mathbf{D}=ϵ\mathbf{E}$, leaving $\mathbf{D}$‘s sources as just the free charges.

Superposition

The field obeys the Superposition principle: the field due to multiple charges is the vector sum of the fields of each charge taken individually. For $N$ point charges,

$\mathbf{E}(\mathbf{R})={\sum }_{i=1}^N\frac{q_i(\mathbf{R}-{\mathbf{R}}_i)}{4\pi ϵ|\mathbf{R}-{\mathbf{R}}_i{|}^3},$

where ${\mathbf{R}}_i$ is the Position vector of charge $q_i$ and $\mathbf{R}$ is the observation point.

Two-charge superposition: $\mathbf{E}$ at $P$ is the vector sum of ${\mathbf{E}}_1$ and ${\mathbf{E}}_2$.

For continuous distributions, the sum becomes an integral over the charge density. See Charge density.

Field from distributed charge

If charge is spread out, the field at observation point $\mathbf{R}$ is

$\mathbf{E}(\mathbf{R})=\frac{1}{4\pi ϵ}\int \frac{{\rho }_v({\mathbf{R}}^{\prime })(\mathbf{R}-{\mathbf{R}}^{\prime })}{|\mathbf{R}-{\mathbf{R}}^{\prime }{|}^3}d{v}^{\prime },$

where ${\mathbf{R}}^{\prime }$ ranges over the source region. Same form for surface (${\rho }_{s}d{s}^{\prime }$) and line (${\rho }_{l}d{l}^{\prime }$) distributions.

In practice you’d rather use Gauss’s law when symmetry permits — direct integration only when symmetry is absent.

Relationship to potential

The electric field is the negative Gradient of the Electric potential:

$\mathbf{E}=-\nabla V.$

This is the “easy way” to find $\mathbf{E}$ when $V$ is known. Equivalently, $V$ is recovered from $\mathbf{E}$ by line integration: $V_{21}=V_2-V_1=-{\int }_{P_1}^{P_2}\mathbf{E}\cdot d\mathbf{l}$.

The sign convention: $\mathbf{E}$ points in the direction of decreasing potential. A positive charge released in an $\mathbf{E}$ field rolls “downhill” toward lower potential.

In electrostatics vs time-varying fields

In electrostatics, $\nabla \times \mathbf{E}=0$ — the electric field is irrotational, and a scalar potential exists. In time-varying problems, Faraday’s law gives $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}\ne 0$ in general, so the line integral of $\mathbf{E}$ around a closed loop is no longer zero. The scalar potential alone is insufficient; you need both a scalar and a vector potential.