Electrostatic Basic principles

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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:

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

References

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)
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	[[Category: Electrostatic Application]]		[[Category: Electrostatic Application]]
−	[[Category: Theory]]	+	[[Category: Electrostatic Theory]]

Electrostatic Basic principles

Latest revision as of 09:05, 4 November 2009

References

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