# Electrostatic Basic principles

(Difference between revisions)
 Revision as of 10:13, 8 October 2009 (view source)JMora (Talk | contribs)← Older edit Latest revision as of 09:05, 4 November 2009 (view source)JMora (Talk | contribs) (→References) Line 68: Line 68: [[Category: Electrostatic Application]] [[Category: Electrostatic Application]] − [[Category: Theory]] + [[Category: Electrostatic Theory]]

## Latest revision as of 09:05, 4 November 2009

Electrostatics[1][2][3] refers to the physical phenomena related to the presence of electric charges in the objects.

Electric charges[4] are the most basic sources of the electromagnetic nature. Repulsive (or attractive) electrical forces[5][6] 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[7] 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[8] ($\vec{D} \; [C/m^2])\,$.

$\oint_S \vec{D} \cdot\mathrm{d}\vec{A} = \int_V\rho\cdot\mathrm{d}V$

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:

$\vec{\nabla} \varepsilon \vec{\nabla} V + \rho_v = 0$