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Show that the electric field $$\mathbf{\vec{E}}=\begin{pmatrix}0 \\ E_0x \\ 0\end{pmatrix}$$ where $E_0$ is a constant, cannot be generated by any static distribution of charges.

I understand that an electrostatic field is irrotational and the divergence of the electric field is the charge density over epsilon 0 but I have no idea why that makes the answer.

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Calculate curl of the given field and confront the result with the sentence in your post. –  Ján Lalinský Jan 11 at 12:20
    
Ah yes, I get it now, thanks –  user37187 Jan 11 at 12:22
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1 Answer 1

As one commenter has already noted, you could proceed by showing that the curl of the electric field in question is not identically zero.

You can also show this by way of contradiction. If you assume that the electric field $\mathbf{E}$ can be generated by some static charge distribution, you can use the differential form of Gauss's Law to find out what that distribution is:

$$\rho=\epsilon_0\nabla\cdot\mathbf{E}=\epsilon_0\nabla\cdot\left(0,E_0x,0\right)=0.$$

But it follows immediately from Coulomb's Law that if the charge distribution is identically zero, then the electric field is identically zero. Thus, if $E_0\neq0$ then we have a contradiction, and so the initial assumption must be false.

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