# Electrostatic Basic principles

Electrostatics refers to the physical phenomena related to the presence of electric charges in the objects.

Electric charges are the most basic sources of the electromagnetic nature. Repulsive (or attractive) electrical forces are obtained by joining two charges of the same (or different) sign (Coulomb's law). $F = \frac{Q_1Q_2}{4\pi\varepsilon_0 r^2} \,$ $[Newton] \,$

where:

• $Q_1 \,$ and $Q_2 \,$ $[Coulomb]\,$ are two electric charges in interaction ;
• $r \,$ $[meter]\,$ is the distance between the charges;
• $\varepsilon_0 \,$   is a electric constant defined as $\varepsilon_0 \ = 8.854\ \times 10^{-12} [Farad \; · m^{-1}] \; or \; [C^2·N^{-1}·m^{-2}] \,$ The image shows the force lines (in blue) produced by two charges, that are the same that those which define the electric field ( $\vec{E} \; [Volt/m] \,$, the electric force per unit of charge associated to each electric charge). $\vec{F} = q\vec{E}\,$ $E = \frac{Q}{4\pi\varepsilon_0 r^2} \,$

For closed surfaces, the total electric flux ( $\Phi_E \,$) is is proportional to the total charge enclosed within the surface (Gauss' law). The electric flux over a surface S is given by the surface integral: $\Phi_E = \int_S \vec{E} \cdot d\vec{A}$

This expression can be written in terms of the electric displacement field ( $\vec{D} \; [C/m^2])\,$. $\oint_S \vec{D} \cdot\mathrm{d}\vec{A} = \int_V\rho\cdot\mathrm{d}V$

with $\rho \,$ the volume charge density $[C/m^3] \,$.

The equation becomes in differential form to our first Maxwell's equation: $\vec{\nabla}\cdot\vec{D} = \rho$

where: $\vec{D} = \varepsilon\vec{E} = \varepsilon_0\varepsilon_r\vec{E}$

with: $\varepsilon, \varepsilon_0, \varepsilon_r$ the permittivity of the medium, of the free space and relative permittivity, respectively.

The electric field is a conservative vector field. That means that is the gradient of a scalar potential, called electrostatic potential ( $V [Volts]\,$) and that the line integral from one point to another is path independent. $\vec{E}=-\vec{\nabla} V$

Therefore, the electric field is also irrotational (our second Maxwell's equation). $\vec{\nabla}\times\vec{E} = 0$

By using the first Maxwell's equation ( $\vec{\nabla}\cdot\vec{D} = \rho$) and this definition of electrostatic potential ( $\vec{E}=-\vec{\nabla} V$), we can easily can obtain the Poisson's equation: $\vec{\nabla} \varepsilon \vec{\nabla} V + \rho_v = 0$