# Show that the electric field, $\mathbf{\vec{E}}$ cannot be generated by any static distribution of charges

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.

• Calculate curl of the given field and confront the result with the sentence in your post. – Ján Lalinský Jan 11 '14 at 12:20

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.