In electrostatic electric field in a system is always irrotational ∇×E=0. And divergence of electric field is non zero ∇.E=ρ/ε but in some cases divergence of electric field is also zero ∇.E=0 such as in case of dipole I had calculated and got that ∇.E=0 for a dipole
So in case of this dipole divergence and curl both are zero So what does it mean when a vector fieLd do not diverge and not rotational at all So what kind of nature it has?? ∇×E=0 , ∇.E=0.
So it means the electric field is both solenoidal and irrotational ,but how can these two conditions satisfy simultaneously? If a vector field is solenoidal then it has to rotate ,must have some curliness

But in pic of a dipole I can see that the electric field is bending or rotating enter image description here Then what does it mean about zero curl (∇×E=0)? I can see the electric field is rotational

  • $\begingroup$ Start with a simpler problem, calculate $\nabla\cdot \mathbf{E}$ and $\nabla \times \mathbf{E}$ for a point charge. Also, $\nabla \times \mathbf{E}=0$ isn't about the field not bending, it's about the field not shearing past itself. So while the field lines of a dipole are bent, the balance of field strength cancels. To see an example of what's going on, calculate the curl of $$\mathbf{V} = x \hat{j}.$$ $\endgroup$ – Sean E. Lake Nov 4 '16 at 4:07
  • $\begingroup$ I didn't understand why ∇×E=0 is not about the field not bending? $\endgroup$ – user101134 Nov 4 '16 at 4:58
  • $\begingroup$ Graph that field, $\mathbf{V}$, I gave you. The field lines are all parallel to each other, with density changing as you move in the $x$-direction. Calculate the curl, and you'll find it's the unit vector in the $z$-direction, $\hat{k}$, constant everywhere. Curl is about shear, not bending. $\endgroup$ – Sean E. Lake Nov 4 '16 at 5:05

div E does not vanish everywhere for the dipole, you should get delta-functions in the points where there are charges.

  • $\begingroup$ But otherwise curl is zero but I pic I can see they are bending so how can it be irrotational? $\endgroup$ – user101134 Nov 4 '16 at 4:59
  • $\begingroup$ @user101134: Who said irrotational field cannot bend (whatever that means)? If curl vanishes, it means the field is typically a gradient of some scalar field, if, in addition div vanishes, that means the Laplacian of the scalar field vanishes. $\endgroup$ – akhmeteli Nov 4 '16 at 7:51
  • $\begingroup$ Then what does it mean by ''irrotational'' if ∇×E =0 ? I used to think that as it is called irrotational that means there will be no rotation. Am I wrong in this sense? $\endgroup$ – user101134 Nov 5 '16 at 14:46
  • $\begingroup$ @user101134: This is correct in some sense. If the curl of a vector field vanishes, an integral of the vector field over any closed curve vanishes (according to a relevant theorem). Let us imagine (to make it more intuitive) that the vector field is a field of velocities of a fluid. If there is a rotational motion of a fluid along some closed curve, the velocity will be directed clockwise (or counterclockwise) along the entire curve and the integral of the velocity vector along the curve will not vanish. However, if the vector field is irrotational, it does not mean it cannot "bend" anywhere. $\endgroup$ – akhmeteli Nov 5 '16 at 15:03

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