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As we know from Maxwell's 3rd equation the magnetic field is given as

$$\nabla \times \mathbf{E} = - \frac {\partial \mathbf{B}}{\partial t}$$

Now, if we consider an electric dipole which is stationary, there will be electric field lines like this:
http://i.stack.imgur.com/z17OF.gif

I want to know what will be the curl of this electric field, if it exists. And will there be a magnetic field associated with this electric field too?

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  • $\begingroup$ Go around a circle of your choice and see how electric field flows against you, giving negative contribution and sometimes it helps you, pointing in the same direction. Try to find a loop where everytime the electric field will help you in your direction of movement. You'll soon realise, such a loop isn't possible. $\endgroup$ – Viesr Dec 20 '15 at 15:48
  • $\begingroup$ Vivek, Javier's answer is pretty good, I upvoted it. But he isn't telling you the half of it, which is that $\nabla \times \mathbf{E} = - \frac {\partial \mathbf{B}}{\partial t}$ because the electric field and the magnetic field are two aspects of the greater whole. Each particle has an electromagnetic field. The lines in your picture are lines of force. They depict the linear force that results from electromagnetic field interactions. And the arrowheads are misleading because two electrons repel, two positrons repel, and one electron and one positron attract. $\endgroup$ – John Duffield Dec 20 '15 at 16:04
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Since everything is stationary, all time derivatives are zero, and there are no currents. Now look at the relevant equations. The curl of $\mathbf{E}$ you wrote yourself: $\nabla \times \mathbf{E} = -\partial \mathbf{B} / \partial t$. And will there be a magnetic field? Well, let's look at the relevant equations:

$$\nabla \cdot \mathbf{B} = 0$$

$$\nabla \times \mathbf{B} = \mu_0 \mathbf{j} + \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t}$$

If all time derivatives and currents are zero, what happens to $\mathbf{B}$?

Edit: I want to add a couple of things based on the comments. First, while intuition is definitely a great thing to have, it can fail. It doesn't come by itself; it has to be learned. Math, however, doesn't lie. So if the equation says $\nabla \times \mathbf{E} = -\partial \mathbf{B} / \partial t$ and you know $\partial \mathbf{B} / \partial t = 0$ because everything is stationary, well, then the curl of $\mathbf{E}$ is zero and there's nothing you can do about it.

But how does this fit with the picture? A non zero curl doesn't just mean that the lines curve. Rather, it means (roughly) that the lines are going around some place, and they must be going all the way around. This is most easily seen in the magnetic field of a wire. The field has a non zero curl right at the wire, and you can see that it goes around the wire. This can be justified more rigorously by Stoke's theorem, which says that the circulation of the field around a closed curve is equal to the surface integral of the curl on a surface which has the curve as its border.

Take your picture, draw a closed curve somewhere, and sketch the electric field at different points along the curve. If you take a line integral around it, along half the curve the field points one way and along the other half it points the other way, so the contributions cancel and the integral is zero.

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  • $\begingroup$ Well,this is how I was thinking but then there will be some curl of this electric field isn't it? It means there should also be a magnetic field which confused me. $\endgroup$ – vivek Dec 20 '15 at 15:31
  • $\begingroup$ @vivek: Why do you think there will be curl? $\endgroup$ – Javier Dec 20 '15 at 15:33
  • $\begingroup$ well it is evident from the field lines I guess.. $\endgroup$ – vivek Dec 20 '15 at 15:35
  • $\begingroup$ @vivek, no it isn't. Your interpreting curl the wrong way if you think so. Curl is not that the lines curve or have a bend in them. $\endgroup$ – Aganju Dec 20 '15 at 15:54
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It would be interesting if you wrote our each vector component of the curl-E equation, and equate each of them to zero. Try to intuit what the derivative of each E-field component means with respect to your drawing.

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