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The questions says that two very large, conducting, parallel plates separated a distance $2a$ contain a uniform volume charge density $ρ$ between them and they both have zero potential, the permittivity between the plates is $\varepsilon$ and outside the plates is $\varepsilon_0$ For electric field between them I have supposed that the plates are at $x=a$ and $x=-a$ and written as below: $$ \overrightarrow{E}=\frac{1}{4\pi\varepsilon}\rho\int\frac{\overrightarrow{R}-\overrightarrow{R'}}{|\overrightarrow{R}-\overrightarrow{R'}|}dv' $$ $$ \overrightarrow{E}=\frac{1}{4\pi\varepsilon}\rho\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-a}^{a}\frac{(x-x')\hat{x}+(y-y')\hat{y}+(z-z')\hat{z}}{((x-x')^2+(y-y')^2+(z-z')^2)}dx'dy'dz' $$ which leads me to $$ \overrightarrow{E}=\frac{\rho a\hat{x}}{\varepsilon} $$

but if I use this equation $$ \overrightarrow \nabla \cdot \overrightarrow{E}=\frac{\rho}{\varepsilon} $$ I get: $$ \overrightarrow{E}=\frac{\rho x\hat{x}}{\varepsilon} $$ Why aren't they the same?

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    $\begingroup$ Shouldn't it be mod^3 in the denominator is E? $\endgroup$ – aneet kumar Dec 26 '20 at 9:53
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Maybe this can be a hint for you. The first formula isn't corret.
The electric field is: $$\vec{E}(x,y,z)=\frac{1}{4\pi \epsilon}\int_v\frac{\rho(x',y',z')dx'dy'dz'}{r^2}=\frac{1}{4\pi \epsilon}\int_v\frac{\rho(x',y',z')dx'dy'dz'}{[(x-x')^2+(y-y')^2+(z-z')^2]^{\frac{3}{2}}}$$ Your uniform volume charge density between them generates an electric field. Since the plates are conductors they have the role to shield (I don't know if it is the right word) the electric field. For the electric field generated, it depends also on the shape of the volume. But the first formula misses a $\frac{3}{2}$.
You can also calculate the electric field generated by your volume very easly using Gauss' law if the volume has particular symmetries, IN THIS CASE: $$\Phi (\vec{E})=\int_\Sigma \vec{E}\cdot \vec{u}_n d\Sigma=E\int_\Sigma d\Sigma=\frac{Q_{tot}}{\epsilon} \Rightarrow E\Sigma=\frac{\rho \Sigma x}{\epsilon} \Rightarrow E=\frac{\rho}{\epsilon}x$$

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    $\begingroup$ Not on vaccum, the Gauss's Maxwell ecuation reads $ \nabla \cdot \mathbf{D} = \rho$, which is equivalent to $ \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon}$. $\endgroup$ – MarcoCiafa Dec 26 '20 at 23:14
  • $\begingroup$ Yes, you are right beacause D= $\epsilon E$. Thanks! $\endgroup$ – Hitman Reborn Dec 27 '20 at 9:36
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Since the problem specifies large plates, Gauss's law can be used in the central regions. With a positive charge density the field would start at zero and point out from the center. Putting Gaussian surfaces at + and – x: 2EA = 2ρAx/$ε_o$. Suggesting that the non-conductor may be polarized would conflict with the given condition of uniform charge density.

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