Give an example where the electric field is zero at a point but divergence of the electric field is non zero there?
I believe that the electric field (due to that distribution) in the middle of a spherical uniform charge distribution, is zero, and the divergence is equal to the charge density divided by the permittivity.
As supplement to Malcolm White 's spherical charge distribution, I think this condition could be fulfilled on the axis of cylindrical charge distributions, either exactly midways if the cylinder is of finite length or (theoretically) on the whole axis of cylinders of infinite length.
The same seems to be true for the center plane in a layer shaped distribution, and, regarding a point, of any combination of spheres, cylinders, symmetrical cones, and planes as long as their centers coincide. Even the center point of a uniform charge distribution in the shape of a symmetrical Christmas star qualifies, for example.
One basic requirement seems to be the assumption of a continuous charge distribution.
For spherically or cylindrically symmetric charge distributions, such as those mentioned by Malcolm White and Joerg Fricke, the problem basically reduces to a one dimensional problem and then the question itself reduces to the rather trivial question whether there exist one-variable functions that vanish somewhere while their derivatives do not vanish there.
The most simple such example is f(x)=ax, with a being non-zero.
This happens in the two examples mentioned in the previous answers, since by Gauss’s law, the electric field inside a uniformly charged sphere is radial and proportional to the distance from the center of the sphere and similarly, the electric field inside a uniformly charged cylinder of infinite length is perpendicular to the axis of the cylinder and proportional to the distance from the axis.
By symmetry, the same must happen at the point of symmetry (if exists) of any three dimensional uniform charge distribution, such as at the centre of a cube, at the center of a cylinder of finite length, etc.
I agree with all previous answers, Dear Mohit Rattanpal
For all of those shapes, the Gauss theorem applies, but there is one more pure mathematical case:
We know that div rot A = 0, where A is an irrotational vector, so if vector A is not irrotational always its divergence could be different from zero.
And that probably is for those V = rot A fields that don't follow Gauss Law, am I wrong? Depend that on the gauge? Lorentz or Coulomb?
I am just curious. Thank you, interesting question. I just cannot remember despite I was taught the subject.
Dear Prof. Pedro L. Contreras E. , for any vector field A with continuous second derivatives, the divergence of its curl vanishes identically; i.e. div rot A = 0.
The proof is easy; see, for instance,
https://proofwiki.org/wiki/Divergence_of_Curl_is_Zero
Dear Spiros Konstantogiannis, thank you, I agree, it means the identity is independent of the type of gauge for the vector potential A.
I just could not remember if the vectorial identity somehow could depend on the type of one of the 2 gauges for the case of electromagnetism.
Best Regards.
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