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$\nabla\times E = 0/(r^2\sin\theta)$ where $\theta$ is the polar angle. Clearly $\nabla \times E = 0$ for all $r$ except $r=0$. But how do we conclude that $\nabla\times E$ at $r=0$?

One can surely argue that work done by the electrostatic field around any closed path enclosing the origin is zero and hence as the curl is zero everywhere except origin, and as the surface integral of curl is zero so $\nabla\times E = 0$ at r=0 using Stokes' theorem.

But how can we even apply Stokes' theorem in this case as the vector field is not defined at all at the origin ? We need a vector field which is differentiable everywhere on the surface on which we compute the surface integral of curl in Stokes'.

Similarly $\nabla\cdot B$ everywhere except $r=0$.

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  • $\begingroup$ $\nabla \times F$ and $\nabla \cdot F$ $\endgroup$
    – R. Emery
    Jan 13 at 11:20
  • $\begingroup$ $\nabla \times \mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \boldsymbol{\hat\imath} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \boldsymbol{\hat\jmath} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \boldsymbol{\hat k} = \begin{bmatrix}\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \\ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \\ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\end{bmatrix}$ $\endgroup$
    – R. Emery
    Jan 13 at 11:28
  • $\begingroup$ $\operatorname{div} \mathbf{F} = \nabla\cdot\mathbf{F} = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \cdot (F_x,F_y,F_z) = \frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}$ $\endgroup$
    – R. Emery
    Jan 13 at 11:29
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Note that for a vector field, as the electric $\mathbf{E}$, an equivalent definition of the curl(1) at a point $\mathrm P$ (based on its properties) is \begin{equation} \left(\boldsymbol{\nabla} \boldsymbol{\times}\mathbf{E}\right)_{\mathrm P}\boldsymbol{\cdot} \mathbf{n}\stackrel{\texttt{def}}{\boldsymbol{=\!=}}\lim\limits_{A \boldsymbol{\rightarrow}0}\dfrac{1}{\boldsymbol{\vert}A\boldsymbol{\vert}}\oint_{C}\mathbf{E}\boldsymbol{\cdot}\mathrm d\mathbf{r} \tag{01}\label{01} \end{equation} where the line integral is calculated along the boundary $C$ of the area $A$ in question, $\boldsymbol{\vert}A\boldsymbol{\vert}$ being the magnitude of the area. This equation defines the projection of the curl of $\mathbf{E}$ onto $ \mathbf{n}$ . The infinitesimal surfaces bounded by $C$ have $ \mathbf{n}$ as their normal. $C$ is oriented via the right-hand rule. If $C$ is always a circle with center the point charge in the line integral of the rhs of equation \eqref{01}, which is zero, the value of $\mathbf{E}_{\mathrm q}$ at the point charge nowhere and never is used.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

Similarly for a vector field, as the magnetic $\mathbf{B}$, an equivalent definition of the divergence(2) at a point $\mathrm P$ (based on its properties) is \begin{equation} \left(\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{B} \right)_{\mathrm P} \stackrel{\texttt{def}}{\boldsymbol{=\!=}}\lim\limits_{V \boldsymbol{\rightarrow}0}\dfrac{1}{\boldsymbol{\vert}V\boldsymbol{\vert}}\iint_{S(V)}\mathbf{B}\boldsymbol{\cdot}\mathbf{n}\,\mathrm dS \tag{02}\label{02} \end{equation} where $\boldsymbol{\vert}V\boldsymbol{\vert}$ is the volume of $V$, $S(V)$ is the boundary of $V$, and $\mathbf{n}$ is the outward unit normal to that surface. It can be shown that the above limit always converges to the same value for any sequence of volumes that contain $\mathrm P$ and approach zero volume. The result, $\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{B}$, is a scalar function of the coordinates of $\mathrm P$. If $V$ is always a right cylinder with axis around the wire at point $\mathrm P$ then surface integral in the rhs of equation \eqref{02} is zero.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

(1) Curl (mathematics)

(2) Divergence

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  • $\begingroup$ Yes, so since in the limit in RHS, the numerator is always zero, so, Curl E = 0 everywhere, including the origin. Right ? $\endgroup$
    – user285600
    Jan 14 at 12:03
  • $\begingroup$ @Avinandan Mondal ph20b005 :Yes, precisely. $\endgroup$
    – Frobenius
    Jan 14 at 12:15
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Curl E is zero for finite r. Also a contour integral is zero for a contour avoiding the origin. Hence the surface integral over any surface of curl E is zero. Now consider a surface through the origin bounded by this contour. We know that all points but the origin contribute zero and the total is strictly zero. Then the origin must also contribute zero.

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  • $\begingroup$ So are we forcefully defining Curl E =0 at origin in order to make Stokes' theorem applicable ? $\endgroup$
    – user285600
    Jan 13 at 22:06
  • $\begingroup$ Expect trouble when discussing point particles. $\endgroup$
    – my2cts
    Jan 13 at 22:23

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