Additional electric field due to changing magnetic field around a moving charge Imagine a point charge moving along x-axis in empty space at a constant velocity $v$. Since the charge is moving, the electric and magnetic field around it should be surmisable from this image (courtesy: brilliant.org)

I suppose one can explain the shape of the above magnetic field from the following Maxwell's Equation:
$\nabla\times \boldsymbol{B} = \frac{1}{c^2} \frac{\partial \boldsymbol{E}}{\partial t}$
The current component $\boldsymbol{J}$ is not needed since we have a lone charge moving in space.
If I understand the above equation correctly, the change w.r.t. time in $\boldsymbol{E}$ at any point in space is accompanied by $\boldsymbol{B}$ curling around it. Since the moving charge in our frame of reference concedes $\boldsymbol{E}$ at some points in space, and bolsters $\boldsymbol{E}$ at others, we see a concomitant curling $\boldsymbol{B}$ in proportion to the change in $\boldsymbol{E}$.
However, just like the diverging $\boldsymbol{E}$, the curling $\boldsymbol{B}$ is also changing w.r.t. time. Where is the accompanying $\boldsymbol{E}$ field that should arise due to the equation below?
$\nabla\times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t}$
Why don't the authors usually comment on $\boldsymbol{E}$ accompanying the curling $\boldsymbol{B}$? I am assuming that there is no additional $\boldsymbol{E}$. Maybe, it's just that the curl of the already curling $\boldsymbol{B}$ is the original diverging $\boldsymbol{E}$, although moving in one direction. Am I correct? Where can I learn more about this?
 A: 
I suppose one can explain the shape of the above magnetic field from the following Maxwell's Equation:
$\nabla\times \boldsymbol{B} = \frac{1}{c^2} \frac{\partial \boldsymbol{E}}{\partial t}$
The current component $\boldsymbol{J}$ is not needed since we have a lone charge moving in space.

Actually it's the opposite. Without $\mathbf J$ due to moving charge, there would not be such electric and magnetic field, moving together with the charged body, concentrated around it. The Ampere-Maxwell equation relating magnetic and electric field has also another term $\mu_0 \mathbf J$ and only then it is valid everywhere.
Outside the charged body, the equation can be written without the $\mu_0 \mathbf J$ term, but then it alone does not imply magnitude and direction of these fields; this simplified equation (mathematically, "homogeneous equation") allows for infinity of different fields, including such that both fields vanish. The fact they do not vanish is due to charge and current being present at the point where this simplified equation does not hold.
There is no additional electric field, the field of a moving charge is compressed in direction of motion (Lorentz contraction) and thus has non-zero curl.
A: Suppose a small neighbourhood without charges: $\delta x$ $\delta y$ $\delta z $ around a point $P$, and during a small time interval $\delta t$. If $\mathbf E$ changes during $\delta t$, then $\mathbf B$ is not constant over the neighbourhood, because its spatial derivatives changes (curl not zero).
On the other hand, it is possible to imagine that $\mathbf E$ only changes with time, but it is spatially constant around $P$. In this case $\mathbf B$ doesn't change during $\delta t$.
Of course, the most general case (as that resulting from the OP moving charge) is both fields changing with time. The homogeneous Maxwell equations tell that they also must change in space. The mutual relation between fields in that small neighbourhood obeys the homogeneous Maxwell equations.
A: Changing B (E) fields don't cause E (B) fields. They are both generated by charges, currents, and their time derivatives such that the curl of E (B) is proportional to the time derivative of B (E).
To analyze the above problem, you can consider a a charge:
$$ q(t, \vec r) = q\delta(x-vt) $$
and a current density
$$ j(t, \vec r) = q(t, \vec r)v\hat x $$
but that is hard. Just take a Coulomb point charge field, B=0,  and Lorentz transform it.
