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Lisa...
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I need to show that
[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?
[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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