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.

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    $\begingroup$ 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. $\endgroup$ Oct 1, 2016 at 7:35

2 Answers 2


The Faraday-Maxwell law says that $$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$$

So, if the curl of the electric field is non-zero, then this implies a changing magnetic field. But if the magnetic field is changing then this "produces" (or rather must co-exist with) a changing electric field and is thus inconsistent with an electrostatic field.

The divergence of an electrostatic field can be zero (where there is no actual charge), but it cannot be zero everywhere. This because a conservative field (which is one that has a curl of zero) cannot form closed field lines; the field lines have to begin and end (on charges). Otherwise, you could move a test charge along a closed electric field line and return to the same point in space with the charge having gained kinetic energy because it was acted on by an electric force all the way around the loop in the direction of motion. But this is the antithesis of a conservative field.


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.

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    $\begingroup$ So can a electrostatic field be non conservative? $\endgroup$
    – user101134
    Oct 6, 2016 at 14:47
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    $\begingroup$ No. For electrostatic conditions, curl(E) = 0 . We know that curl of any div is zero. Hence we can only write E = -div(V) for an electrostatic field , which makes it conservative. Only a non electrostatic field can be non conservative. $\endgroup$
    – Lelouch
    Oct 6, 2016 at 14:50

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