# A strange electric field, can it exist? [duplicate]

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?

• Why "z" for a uniform charge distribution?
– JEB
May 19 at 13:32
• 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". May 19 at 16:00
• Also see this question May 19 at 16:03

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.