Categories: Electromagnetism, Physics.

## Electric field

The electric field $$\vb{E}$$ is a vector field that describes electric effects, and is defined as the field that correctly predicts the Lorentz force on a particle with electric charge $$q$$:

\begin{aligned} \vb{F} = q \vb{E} \end{aligned}

This definition implies that the direction of $$\vb{E}$$ is from positive to negative charges, since opposite charges attracts and like charges repel.

If two opposite point charges with magnitude $$q$$ are observed from far away, they can be treated as a single object called a dipole, which has an electric dipole moment $$\vb{p}$$ defined as follows, where $$\vb{d}$$ is the vector going from the negative to the positive charge (opposite direction of $$\vb{E}$$):

\begin{aligned} \vb{p} = q \vb{d} \end{aligned}

Alternatively, for consistency with magnetic fields, $$\vb{p}$$ can be defined from the aligning torque $$\vb{\tau}$$ experienced by the dipole when placed in an $$\vb{E}$$-field. In other words, $$\vb{p}$$ satisfies:

\begin{aligned} \vb{\tau} = \vb{p} \times \vb{E} \end{aligned}

Where $$\vb{p}$$ has units of $$\mathrm{C m}$$. The polarization density $$\vb{P}$$ is defined from $$\vb{p}$$, and roughly speaking represents the moments per unit volume:

\begin{aligned} \vb{P} \equiv \dv{\vb{p}}{V} \:\:\iff\:\: \vb{p} = \int_V \vb{P} \dd{V} \end{aligned}

If $$\vb{P}$$ has the same magnitude and direction throughout the body, then this becomes $$\vb{p} = \vb{P} V$$, where $$V$$ is the volume. Therefore, $$\vb{P}$$ has units of $$\mathrm{C / m^2}$$.

A nonzero $$\vb{P}$$ complicates things, since it contributes to the field and hence modifies $$\vb{E}$$. We thus define the “free” displacement field $$\vb{D}$$ from the “bound” field $$\vb{P}$$ and the “net” field $$\vb{E}$$:

\begin{aligned} \vb{D} \equiv \varepsilon_0 \vb{E} + \vb{P} \:\:\iff\:\: \vb{E} = \frac{1}{\varepsilon_0} (\vb{D} - \vb{P}) \end{aligned}

Where the electric permittivity of free space $$\varepsilon_0$$ is a known constant. It is important to point out some inconsistencies here: $$\vb{D}$$ and $$\vb{P}$$ contain a factor of $$\varepsilon_0$$, and therefore measure flux density, while $$\vb{E}$$ does not contain $$\varepsilon_0$$, and thus measures field intensity. Note that this convention is the opposite of the magnetic analogues $$\vb{B}$$, $$\vb{H}$$ and $$\vb{M}$$, and that $$\vb{M}$$ has the opposite sign of $$\vb{P}$$.

The polarization $$\vb{P}$$ is a function of $$\vb{E}$$. In addition to the inherent polarity of the material $$\vb{P}_0$$ (zero in most cases), there is a possibly nonlinear response to the applied $$\vb{E}$$-field:

\begin{aligned} \vb{P} = \vb{P}_0 + \varepsilon_0 \chi_e^{(1)} \vb{E} + \varepsilon_0 \chi_e^{(2)} |\vb{E}| \: \vb{E} + \varepsilon_0 \chi_e^{(3)} |\vb{E}|^2 \: \vb{E} + ... \end{aligned}

Where the $$\chi_e^{(n)}$$ are the electric susceptibilities of the medium. For simplicity, we often assume that only the $$n\!=\!1$$ term is nonzero, which is the linear response to $$\vb{E}$$. In that case, we define the relative permittivity $$\varepsilon_r \equiv 1 + \chi_e^{(1)}$$ and the absolute permittivity $$\varepsilon \equiv \varepsilon_r \varepsilon_0$$, so that:

\begin{aligned} \vb{D} = \varepsilon_0 \vb{E} + \vb{P} = \varepsilon_0 \vb{E} + \varepsilon_0 \chi_e^{(1)} \vb{E} = \varepsilon_0 \varepsilon_r \vb{E} = \varepsilon \vb{E} \end{aligned}

In reality, a material cannot respond instantly to $$\vb{E}$$, meaning that $$\chi_e^{(1)}$$ is a function of time, and that $$\vb{P}$$ is the convolution of $$\chi_e^{(1)}(t)$$ and $$\vb{E}(t)$$:

\begin{aligned} \vb{P}(t) = (\chi_e^{(1)} * \vb{E})(t) = \int_{-\infty}^\infty \chi_e^{(1)}(t - \tau) \: \vb{E}(\tau) \:d\tau \end{aligned}

Note that this definition requires $$\chi_e^{(1)}(t) = 0$$ for $$t < 0$$ in order to ensure causality, which leads to the Kramers-Kronig relations.