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[tex]\vec{\nabla} \cdot \vec{E}= \frac{\rho}{\epsilon_0}[/tex]

where [tex]\rho[/tex] is the volume charge density.

I know that if I can show that the net flux of the electric field (in three directions xyz) out of the a small gaussian surface in the shape of a cube with faces parallel to the xy, xz and yz planes is:

[tex]\phi_n_e_t = ( \displaystyle{\frac{\delta E_x}{\delta x}} + \displaystyle{\frac{\delta E_y}{\delta y}} + \displaystyle{\frac{\delta E_z}{\delta z}} ) \Delta V[/tex] = [tex](\vec{\nabla} \cdot \vec{E}) \Delta V[/tex]

with [tex]\Delta V[/tex] is the volume enclosed by the gaussian surface.

then according to

[tex]\phi_n_e_t = \displaystyle{\frac{\Delta q}{\epsilon_0}} = \displaystyle{\frac{\rho \Delta V}{\epsilon_0}}= (\vec{\nabla} \cdot \vec{E}) \Delta V[/tex] the two [tex]\Delta V[/tex] cancel, leaving:

[tex]\vec{\nabla} \cdot \vec{E}= \frac{\rho}{\epsilon_0}[/tex]

The only question is: how do I show that [tex](\vec{\nabla} \cdot \vec{E}) \Delta V[/tex] is a correct formula?

Perhaps it would be easier to show that if the electric field was only along the x axis the equation would be:

[tex]\phi_n_e_t = \displaystyle{\frac{\delta E_x}{\delta x}} \Delta V[/tex]

but how will I do that?

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# Help needed with (the point form of) Gauss's law & divergence of E and

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