Driver in electrostatic balance

A conducting electric in electrostatic balance is a driver for which the free loads are on average fixed, consequently there is absence of current inside this one. Let us suppose this ohmic driver i.e which it checks the local Loi of Ohm, therefore $\backslash \; vec\; \{J\}\; =\; \backslash \; sigma\; \backslash \; vec\; \{E\}\; \_\; \{int\}$ thus (since $\backslash \; vec\; \{J\}\; =\; \backslash \; vec\; \{0\}$):

$\backslash \; vec\; \{E\}\; \_\; \{int\}\; =\; \backslash \; vec\; \{0\}$

Consequently the potential $V$ is uniform inside the driver. The Poisson's equation ($\backslash \; vec\; \{\backslash \; nabla\}\; ^\; \{2\}\; V+\; \backslash \; frac\; \{\backslash \; rho\}\; \{\backslash \; epsilon\_0\}\; =0$) thus gives us since $\backslash \; vec\; \{\backslash \; nabla\}\; ^\; \{2\}\; V=0$:

$\backslash \; rho=0$.

Consequently if a driver in electrostatic balance is then charged it is charged surfaciquement .

See too

• conducting

• Vector density of current
• Law of Ohm

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