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Do the equations $$\:\boldsymbol{\nabla}\cdot\mathbf{E}=0 \qquad \boldsymbol{\nabla}\times\mathbf{E}=\boldsymbol 0\:$$ imply that the vector field $\mathbf{E}$ is uniform? I think yes, since by Helmholtz Theorem the vector field is uniquely determined and only a uniform vector field satisfies these two conditions?

But then also: $\operatorname{rot} \mathbf{E} = 0$ implies $\mathbf{E} = \operatorname{grad}\Phi$ and $\operatorname{div}(\operatorname{grad}\Phi) = 0$ by Laplace equation. So the solution of the Laplace equation always yields a uniform electric field?

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Your second line of thinking is closer to the truth.

It is correct that the gradient of any harmonic function (i.e., a function satisfying $\nabla^2 \phi = 0$) yields a divergence-free and curl-free vector field. Moreover, the potential for any uniform field $\vec{E}$ is $\phi = - \vec{E} \cdot \vec{r}$, which (it can be shown) is a harmonic function. So it is true that all uniform vector fields can be derived from a harmonic potential.

However, there are also non-uniform vector fields that derive from harmonic potentials. For example, $\phi = \frac12 k(x^2 - y^2)$ is a harmonic function, and it leads to a vector field of $\vec{E} = k(-x \hat{\imath} + y \hat{\jmath})$, which is not uniform.

The reason this seems to be at odds with the Helmholtz Theorem is that the Helmholtz Theorem requires a set of boundary conditions to uniquely determine a vector field, along with its divergence and curl. In particular, if we require that $\vec{E} \to 0$ as $|\vec{r}| \to \infty$, then the Helmholtz theorem tells us that the only divergence- and curl-free vector field that satisfies this is $\vec{E} = 0$ everywhere.1 As we have seen above, however, there are many other divergence- and curl-free vector fields which are possible if we do not require that $\vec{E} \to 0$ at infinity.


1 Note that Laplace's equation also requires a set of boundary conditions to uniquely determine a solution. In particular, if we require that $\nabla^2 \phi = 0$ and $\phi \to \text{constant}$ as $|\vec{r}| \to \infty$, then we must have $\phi = \text{constant}$ everywhere. This is consistent with the fact that the only divergence- and curl-free vector field that vanishes at infinity is the zero vector field.

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  • $\begingroup$ Thank you very much. Now everything makes sense. $\endgroup$ Commented Apr 7, 2022 at 13:53
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Take $$\vec{E}(x,y,z) = (yz,xz,xy)^t\:.$$ It holds $\nabla \cdot \vec{E}=0$ together with $\nabla \times \vec{E}=\vec{0}$, but the field is not uniform.

The point is that $$\Delta \phi(x,y,z)=0$$ has infinitely many solutions depending on the boundary conditions.

In the case above $\phi(x,y,z) = xyz$ and $\vec{E}= \nabla \phi$ is the gradient of that $\phi$.

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Not necessarily. Consider the vector field: \begin{equation} \vec{E}(x,y,z) =\begin{pmatrix} x \\ -y \\ 0 \end{pmatrix} \end{equation} with $\vec\nabla\cdot\vec{E}=0$ and $\vec\nabla\times\vec{E}=\vec{0}$.

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