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If the electric field strength is $E_x=x, E_y=E_z=0$, then by $\nabla\cdot E=\frac1{\epsilon_0}\rho_e$ where $\rho_e$ is the density of charge, we get $\rho_e=-1$ for any point in the space.

But if $\rho_e=-1$ for any point in the space, then the distribution of charge in space is completely symmetric, so we shouldn't get a $\vec{E}$ which only have $E_x$ component.

I am really confused. Can you explain it for me?

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  • $\begingroup$ Why "z" for a uniform charge distribution? $\endgroup$
    – JEB
    May 19 at 13:32
  • $\begingroup$ Maxwell's equations alone don't determine the electric field. You also need boundary conditions. The equation $\vec\nabla\cdot \vec E =-1$ has many solutions, and you need to define what happens at the boundary before you know which one is "real". $\endgroup$ May 19 at 16:00
  • $\begingroup$ Also see this question $\endgroup$ May 19 at 16:03
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Think about the case where $\rho = 1$ between $x = -x_0$ and $x = x_0$, and two charged planes of $\sigma = -x_0$ at $x = \pm x_0$ . You get $E_x = x$ with this setup inside the interval, and zero field outside. Now let $x_0 \to \infty$, and you get the desired field strength, with some strange charge setup at infinity.

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It cannot exist, as it requires an infinite amount of charge and an infinite energy to concentrate this. It also conflicts with average charge neutrality.

Perhaps it is instructive to consider a uniformly charged spherical shell. This is an two dimensional implementation of your idea. The component of the electric field parallel to the surface is indeed zero.

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This makes a perfect sense if you consider a plane volume of a finite thickness and the variables inside it.

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    $\begingroup$ This is not an answer. $\endgroup$
    – garyp
    May 19 at 11:39
  • $\begingroup$ It is an answer, because you need a location for x = 0 where E is zero. It does require a positive charge density. $\endgroup$
    – R.W. Bird
    May 19 at 14:09

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