Talk:Gauss's law

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It doesn't make sense to speak of a path integral over a surface. A path or curve is by definition one-dimensional; a surface is by definition two-dimensional. Michael Hardy 01:24 Mar 21, 2003 (UTC)

indeed, must we still use the \oint symbol then?

yes, the \oint symbol is still appropriate. It's not just notation for one-dimensional closed-paths, it's also used for the integral taken over any closed manifold, a closed path is just the one-dimensional case, here, it means a closed 2-mainfold, or surface. Revolver

What we really need is the symbol for an integral over a closed surface, which is two integral signs with a loop through both of them: sort of an \oiint, which doesn't exist in standard TeX and LaTeX. (It's in Unicode: U+222F (insert pitch for MathML here).) Mike 21:11, 29 March 2006 (UTC)

1 Differential Form
- 1.1 Partial form
2 Please vet
3 Symbols

[edit] Differential Form

Shouldn't we include it in differential form too? I am too afraid to make mistakes myself to do it.

[edit] Partial form

$\nabla \cdot \mathbf{D} = \rho$

$\nabla \cdot$ is the divergence

D is the electric displacement field (in units of C/m²).

ρ is the free electric charge density (in units of C/m³), not including dipole charges bound in a material

Andries 09:29, 4 Sep 2004 (UTC)

[edit] Please vet

Today's edits by User:128.6.83.17 should be carefully vetted. --Wetman 00:12, 10 Jun 2005 (UTC)

2.000? o_O what's the point --210.6.198.242 15:34, 16 Jun 2005 (UTC)

Reply: I believe that there are certain theoretical reasons for insisting that the square be exactly 2, and not just approximately 2. One, I think, has to do with the central nature of the electric force which does not show up with 1/(r^n) type dependence if n is not equal to 2. I could be wrong, so take this with a chunk of salt.

--24.84.203.193 28 June 2005 06:45 (UTC)

solve gauss's law for a sherical charge its just 2, with no reason to insist on a decimal Cpl.Luke 00:04, 13 July 2005 (UTC)