# Why can't electrostatic field be rotational?

I have studied that curl of electrostatic field is zero Or $\nabla \times \mathbf E =\mathbf 0$ and hence we can say that E has a electrostatic potential $(v)$ but why the curl of this electrostatic field is zero? Rather $\nabla\times \mathbf E=\mathbf 0$ implies that the electric field is irrotational.

But why can't electric field be a rotational vector ?

Can I any how make an electric field rotational so that $\nabla\times \mathbf E\ne \mathbf 0$ like $\mathbf E= (q/4\pi\varepsilon)(y~\mathbf i+x~\mathbf j)\,?$

Here it has non-zero curl

On the other hand $\nabla\cdot \mathbf E=\rho/\varepsilon\ne 0$

So this implies that electric field has non zero divergence.

So in the nature of electric field vector something is coming out or coming in.

So I can't understand why the curl of electric field is zero and divergence of electric field is non zero, in the sense of physical significance of curl and divergence.

• Experiment tells us that the electrostatic force is conservative, so the curl has to be zero. You can probably formulate a theory with a non-zero curl, but it wouldn't match experiment. Oct 1, 2016 at 7:35

The Faraday-Maxwell law says that $$\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$$
You are confused regarding the conservative electrostatic field and the non conservative electric field. For any conservative field $\vec E$ , it can be written as $\vec E = -\nabla V$ for some scalar potential V. Now, you can easily prove using basic concepts of curl,grad and div that the curl of a grad is always zero. In other words : $\nabla \times (-\nabla V$) = 0 always. However, the non conservative electric field cannot be written as $\vec E = -\nabla V$. Moreover, Maxwell's law specifies that $\nabla \times \vec E$ = -$\frac{\partial\vec B}{\partial t}$, which shows thats here the curl may be non zero, based on how the magnetic field in space varies. This reduces to the electrostatic condition when the right side of the last equation vanishes.