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As we know that the maxwells equation in free space are
$$\nabla\cdot \vec{E}=0$$ $$\nabla\cdot \vec{B}=0$$ $$\nabla\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \times \vec{B}= \frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$
Now if we take $\nabla \times \vec{E}=0$, that is if $ \frac{\partial \vec{B}}{\partial t}=0 $, will the $ \frac{\partial \vec{E}}{\partial t} $ be non zero. From here we can see that it can not represent a em wave equation. That's why I am curuius it will lead to $ \frac{\partial \vec{E}}{\partial t}=0 $ or it will still remain non zero. I have tried several ways to solve the equation taking $\nabla \times \vec{E}=0$ in the third equation but could not arrive at any conclution.

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    $\begingroup$ Under the condition $\:\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{E}=\boldsymbol{0}\:$ differentianting the 4th equation with respect to $\:t\:$ we have $$ \dfrac{\partial^{2} \mathbf{E}}{\partial t^{2}}=\boldsymbol{0} \quad \Longrightarrow \quad \dfrac{\partial\mathbf{E}}{\partial t}=\textbf{constant in time} \ne \boldsymbol{0} \quad \text{in general} $$ $\endgroup$ – Frobenius Dec 13 '16 at 9:10
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Consider the whole space $\mathbb R^3$ and assume that therein the electric field varies in time like this $$\vec{E}(t,x,y,z) = tE \: \bf e_z$$ for some constant $E>0$. Let us also assume that the magnetic field $\vec{B}$ is instead stationary like this $$\vec{B}(t,x,y,z) = \frac{Ex}{c^2} \: \bf e_y\:.$$ We have $$\nabla\cdot \vec{E} = \frac{\partial }{\partial z}tE = 0$$ $$\nabla\cdot \vec{B}= \frac{\partial }{\partial y}Ex = 0$$ $$\nabla\times \vec{E}=0 =-\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \times \vec{B}= \frac{1}{c^2} E {\bf e_z}=\frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$ You see that Maxwell's equations are therefore satisfied. It is interesting to notice that redefining $\vec{B}$ by replacing ${\bf e_y}$ for every other unit vector normal to ${\bf e_z}$, we would obtain another solution, since the problem is invariant under rotations around $z$. This means that $\vec B$ can only be fixed by giving boundary conditions breaking this symmetry.

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    $\begingroup$ And it's also worth noting that this solution, like any other with the required characteristics, will require an ever-growing amount of energy to sustain. $\endgroup$ – Emilio Pisanty Dec 13 '16 at 14:59
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$\frac{\partial \vec{B}}{\partial t}=0$ does not necessarily mean $\frac{\partial \vec{E}}{\partial t}=0$.

A simple counterexample is a charging capacitor. If the current remains constant, the magnetic field stays constant, while the electric field between the plates steadily increases.

Note that it costs energy to increase the electric field, so that without any source present this will not occur spontaneously.

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Why would $\nabla\times\vec{B}=0$ just because $\frac{\partial\vec{B}}{\partial{}t}=0$? Spatial and time derivatives are not the same thing. Neither side of equation (4) is necessarily zero.

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