- #1
center o bass
- 560
- 2
The differential form of Gauss' law states that
[tex]\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}[/tex].
So the divergence of the electric field is the chargedensity divided by epsilon zero.
I just wondered.. since divergence is a local or "point" property. Is the chargedensity in this law also the charge density at that point? That is to say for some x,y,z is
[tex]\nabla \cdot \vec{E}(x,y,z) = \frac{\rho (x,y,z) }{\epsilon_0}[/tex]?
[tex]\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}[/tex].
So the divergence of the electric field is the chargedensity divided by epsilon zero.
I just wondered.. since divergence is a local or "point" property. Is the chargedensity in this law also the charge density at that point? That is to say for some x,y,z is
[tex]\nabla \cdot \vec{E}(x,y,z) = \frac{\rho (x,y,z) }{\epsilon_0}[/tex]?