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As you see, this is the electric field generated by a point charge moving at constant speed v. I know that when $v$ -> 0, $E$ is just the Coloumb Law. But how do you interpret $E$ when $v$ -> $c$ ?
Can I just interpret it as the field of electromagnetic wave, because it moves at the speed of light?
The electric $\:\mathbf{E}\:$ and magnetic $\:\mathbf{B}\:$ parts of the electromagnetic field produced by a moving charge $\:q\:$ on field point $\:\mathbf{x}\:$ at time $\;t\;$are(1) \begin{align} \mathbf{E}(\mathbf{x},t) & \boldsymbol{=} \frac{q}{4\pi\epsilon_0}\left[\frac{(1\boldsymbol{-}\beta^2)(\mathbf{n}\boldsymbol{-}\boldsymbol{\beta})}{(1\boldsymbol{-} \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R^2}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \right]_{\mathrm{ret}}\!\!\!\!\!\boldsymbol{+} \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\boldsymbol{\times}\left[(\mathbf{n}\boldsymbol{-}\boldsymbol{\beta})\boldsymbol{\times} \boldsymbol{\dot{\beta}}\right]}{(1\boldsymbol{-} \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01.1}\label{eq01.1}\\ \mathbf{B}(\mathbf{x},t) & = \left[\mathbf{n}\boldsymbol{\times}\mathbf{E}\right]_{\mathrm{ret}} \tag{01.2}\label{eq01.2} \end{align} where \begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02.1}\label{eq02.1}\\ \boldsymbol{\dot{\beta}} & = \dfrac{\boldsymbol{\dot{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02.2}\label{eq02.2}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R} \tag{02.3}\label{eq02.3} \end{align} In equations \eqref{eq01.1},\eqref{eq01.2} all scalar and vector variables refer to the $^{\prime}$ret$^{\prime}$arded position and time.
Now, in case of uniform rectilinear motion of the charge, that is in case that $\;\boldsymbol{\dot{\beta}} = \boldsymbol{0}$, the second term in the rhs of equation \eqref{eq01.1} cancels out, so \begin{equation} \mathbf{E}(\mathbf{x},t) \boldsymbol{=} \frac{q}{4\pi\epsilon_0}\left[\frac{(1\boldsymbol{-}\beta^2)(\mathbf{n}\boldsymbol{-}\boldsymbol{\beta})}{(1\boldsymbol{-} \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} \quad \text{(uniform rectilinear motion : } \boldsymbol{\dot{\beta}} = \boldsymbol{0}) \tag{03}\label{eq03} \end{equation}
In this case the $^{\prime}$ret$^{\prime}$arded variable $\:\mathbf{R}\:$, so and the unit vector $\:\mathbf{n}\:$ along it, could be expressed as function of the present variables $\:\mathbf{r}\:$ and $\:\phi$, see Figure-01. Then equation \eqref{eq03} expressed by present variables is(2)
\begin{equation} \mathbf{E}\left(\mathbf{x},t\right) =\dfrac{q}{4\pi \epsilon_{0}}\dfrac{(1\boldsymbol{-}\beta^2)}{\left(1\!\boldsymbol{-}\!\beta^{2}\sin^{2}\!\phi\right)^{\frac32}}\dfrac{\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{3}}\quad \text{(uniform rectilinear motion)} \tag{04}\label{eq04} \end{equation} That is :
In case of uniform rectilinear motion of the charge the electric field is directed towards the position at the present instant and not towards the position at the retarded instant from which it comes.
For the magnitude of the electric field \begin{equation} \Vert\mathbf{E}\Vert =\dfrac{\vert q \vert}{4\pi \epsilon_{0}}\dfrac{(1\boldsymbol{-}\beta^2)}{\left(1\!\boldsymbol{-}\!\beta^{2}\sin^{2}\!\phi\right)^{\frac32}r^{2}} \tag{05}\label{eq05} \end{equation} Without loss of generality let the charge be positive ($\;q>0\;$) and instantly at the origin $\;\rm O$, see Figure-02. Then \begin{equation} r^{2} = x^{2}+y^{2}\,,\quad \sin^{2}\!\phi = \dfrac{y^2}{x^{2}+y^{2}} \tag{06}\label{eq06} \end{equation} so \begin{equation} \Vert\mathbf{E}\Vert =\dfrac{q}{4\pi \epsilon_{0}}\left(1\boldsymbol{-}\beta^2\right)\left(x^{2}\!\boldsymbol{+}y^{2}\right)^{\boldsymbol{\frac12}}\left[x^{2}\!\boldsymbol{+}\left(1\!\boldsymbol{-}\beta^{2}\right)y^{2}\right]^{\boldsymbol{-\frac32}} \tag{07}\label{eq07} \end{equation}
For given magnitude $\;\Vert\mathbf{E}\Vert\;$ equation \eqref{eq07} is represented by the closed curve shown in Figure-02. More exactly the set of points with this magnitude of the electric field is the surface generated by a complete revolution of this curve around the $\;x-$axis.
Now, let the charge that has been moving with constant velocity reaches the origin $\;\rm O\;$ at $\;t_{0}=0$, is abruptly stopped, and remains at rest thereafter. At a later instant $\;t>t_{0}=0\;$ the Coulomb field (from the charge at rest on the origin $\;\rm O$) has been expanded till a circle (sphere) of radius $\;\rho=ct$. Outside this sphere the field lines are like the charge continues to move uniformly to a point $\;\rm O'\;$, so being at a distance $\;\upsilon t =(\upsilon/c)c t=\beta\rho\;$ inside the Coulomb sphere as shown in Figure-03 (this Figure is produced with $\beta=\upsilon/c=0.60$). Note that the closed oval curves (surfaces) refer to constant magnitude $\;\Vert\mathbf{E}\Vert\;$ and must not be confused with the equipotential ones.
In case that the charge stops abruptly the second term in the rhs of equation \eqref{eq01.1} dominates the first one. Furthermore since the velocity $\;\boldsymbol{\beta}\;$ and the acceleration $\;\boldsymbol{\dot{\beta}}\;$ are collinear we have $\;\boldsymbol{\beta}\boldsymbol{\times}\boldsymbol{\dot{\beta}}=\boldsymbol{0}\;$ so \begin{equation} \mathbf{n}\boldsymbol{\times}\left[(\mathbf{n}\boldsymbol{-}\boldsymbol{\beta})\boldsymbol{\times} \boldsymbol{\dot{\beta}}\right]=\mathbf{n}\boldsymbol{\times}(\mathbf{n}\boldsymbol{\times}\boldsymbol{\dot{\beta}})=\boldsymbol{-}\boldsymbol{\dot{\beta}}_{\boldsymbol{\perp}\mathbf{n}} \tag{08}\label{eq08} \end{equation} that is the projection of the acceleration on a direction normal to $\;\mathbf{n}$, see Figure-06. But don't forget that this unit vector is the one on the line connecting the field point with the retarded position. But the retarded position in the time period of an abrupt deceleration and a velocity very close to zero is very close to the rest point. So it's reasonable a field line inside the Coulomb sphere to continue as a circular arc on the Coulomb sphere and then to a field line outside the sphere as shown in Figure-04.
To find the correspondence $^{\prime}$inside line-circular arc-outside line$^{\prime}$ we apply Gauss Law on the closed surface $\:\rm ABCDEF\:$ shown in Figure-05. Don't forget that we mean the closed surface generated by a complete revolution of this polyline around the $\;x-$axis. The electric flux through the surface $\:\rm BCDE\:$ is zero since the field is tangent to it. So application of Gauss Law means to equate the electric flux through the spherical cap $\:\rm AB\:$ to that of the electric flux through the spherical cap $\:\rm EF$. The final result of this application derived analytically is a relation(3) between the angles $\:\phi, \theta\:$ as shown in Figure-04 or Figure-05 \begin{equation} \tan\!\phi=\gamma \tan\!\theta=\left(1-\beta^2\right)^{\boldsymbol{-}\frac12}\tan\!\theta \tag{09}\label{eq09} \end{equation}
(1) From J.D.Jackson's $^{\prime}$Classical Electrodynamics$^{\prime}$, 3rd Edition, equations (14.14) and (14.13) respectively.
(2) From W.Rindler's $^{\prime}$Relativity-Special, General, and Cosmological$^{\prime}$, 2nd Edition. Equation \eqref{eq04} here is identical to (7.66) therein.
(3) From E.Purcell, D.Morin $^{\prime}$Electricity and Magnetism$^{\prime}$, 3rd Edition 2013, Cambridge University Press. The derivation is given as Exercise 5.20 and equation \eqref{eq09} here is identical to (5.16) therein.
Equation \eqref{eq09} is proved by equating the electric flux through the spherical cap $\:\rm AB\:$
to that of the electric flux through the spherical cap $\:\rm EF$, see Figure-05 and discussion in main section.
$\boldsymbol{\S a.}$ Spherical cap $\:\rm AB\:$ of angle $\:\theta$
Let $\;\:\mathrm{O'F}=r\;$ the radius of the cap. The flux of the electric field through the cap is
\begin{equation}
\Phi_{\rm EF}=\iint\limits_{\rm EF}\mathbf{E}\boldsymbol{\cdot}\mathrm d\mathbf{S}
\tag{p-05}\label{eqp-05}
\end{equation}
The field, of variable magnitude as in equation \eqref{eq05} of the main section
\begin{equation}
\mathrm E\left(r,\psi\right)=\dfrac{q}{4\pi \epsilon_{0}}\dfrac{(1\boldsymbol{-}\beta^2)}{\left(1\!\boldsymbol{-}\!\beta^{2}\sin^{2}\!\psi\right)^{\frac32}r^{2}}
\tag{p-06}\label{eqp-06}
\end{equation}
is everywhere normal to the spherical surface. So taking the infinitesimal ring formed between angles $\;\psi\;$ and $\;\psi\boldsymbol{+}\mathrm d \psi\;$ we have for its infinitesimal area
\begin{equation}
\mathrm dS=\underbrace{\left(2\pi r \sin\psi\right)}_{length}\underbrace{\left(r\mathrm d \psi\right)}_{width}=2\pi r^2 \sin\psi \mathrm d \psi
\tag{p-07}\label{eqp-07}
\end{equation}
and
\begin{align}
\Phi_{\rm EF} & =\int\limits_{\psi=0}^{\psi=\phi}\mathrm E\left(r,\psi\right)\mathrm dS=\dfrac{q(1\boldsymbol{-}\beta^2)}{2\epsilon_{0}}\int\limits_{\psi=0}^{\psi=\phi}\dfrac{\sin\psi \mathrm d \psi}{\left(1\!\boldsymbol{-}\!\beta^{2}\sin^{2}\!\psi\right)^{\frac32}}
\nonumber\\
& \stackrel{z\boldsymbol{=}\cos\psi}{=\!=\!=\!=\!=}\boldsymbol{-}\dfrac{q(1\boldsymbol{-}\beta^2)}{2\epsilon_{0}}\int\limits_{z=1}^{z=\cos\phi}\dfrac{\mathrm d z}{\left(1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} z^2\right)^{\frac32}}
\tag{p-08}\label{eqp-08}
\end{align}
From the indefinite integral
\begin{equation}
\int\dfrac{\mathrm d z}{\left(1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} z^2\right)^{\frac32}}=\dfrac{z}{\left(1\!\boldsymbol{-}\!\beta^{2}\right)\left(1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} z^2\right)^{\frac12}}+\text{constant}
\tag{p-09}\label{eqp-09}
\end{equation}
equation \eqref{eqp-08} yields
\begin{equation}
\Phi_{\rm EF} = \boldsymbol{-}\dfrac{q}{2\epsilon_{0}}\Biggl[\dfrac{ z}{\left(1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} z^2\right)^{\frac12}}\Biggr]_{z=1}^{z=\cos\phi}
\nonumber
\end{equation}
so
\begin{equation}
\boxed{\:\:\Phi_{\rm EF} = \dfrac{q}{2\epsilon_{0}}\Biggl[1\boldsymbol{-}\dfrac{ \cos\phi}{\sqrt{\left(1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} \cos^2\phi\right)}}\Biggr]\:\:}
\tag{p-10}\label{eqp-10}
\end{equation}
Equating the two fluxes,
\begin{equation}
\Phi_{\rm EF} = \Phi_{\rm AB} \quad \stackrel{\eqref{eqp-04},\eqref{eqp-10}}{=\!=\!=\!=\!=\!\Longrightarrow}\quad \cos\theta=\dfrac{ \cos\phi}{\sqrt{\left(1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} \cos^2\phi\right)}}
\tag{p-11}\label{eqp-11}
\end{equation}
Squaring and inverting \eqref{eqp-11} we have
\begin{equation}
\cos^2\theta=\dfrac{\cos^2\phi}{1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} \cos^2\phi}\quad\Longrightarrow\quad 1+\tan^2\theta =\beta^{2}+\left(1\!\boldsymbol{-}\!\beta^{2}\right)\left(1+\tan^2\phi\right)
\tag{p-12}\label{eqp-12}
\end{equation}
and
\begin{equation}
\boldsymbol{\vert}\tan\phi\,\boldsymbol{\vert}=\left(1\!\boldsymbol{-}\!\beta^{2}\right)^{\boldsymbol{-}\frac12}\boldsymbol{\vert}\tan\theta\,\boldsymbol{\vert}
\tag{p-13}\label{eqp-13}
\end{equation}
Now $\;\theta,\phi \in [0,\pi]\;$ so $\;\sin\theta,\sin\phi \in [0,1]\;$ while from \eqref{eqp-11} $\;\cos\theta\cdot\cos\phi \ge 0\;$ so $\;\tan\theta\cdot\tan\phi \ge 0\;$ and finally
\begin{equation}
\boxed{\:\:\tan\!\phi=\gamma \tan\!\theta=\left(1-\beta^2\right)^{\boldsymbol{-}\frac12}\tan\!\theta\:\:\vphantom{\dfrac{a}{b}}}
\tag{p-14}\label{eqp-14}
\end{equation}
Note that for $\;\theta=\pi=\phi\;$ equations \eqref{eqp-04},\eqref{eqp-10} give as expected
\begin{equation}
\Phi_{\rm AB} =\dfrac{q}{\epsilon_{0}}= \Phi_{\rm EF}
\tag{p-15}\label{eqp-15}
\end{equation}
while from equation \eqref{eqp-14} we have also
\begin{equation}
\theta =\dfrac{\pi}{2} \quad \Longrightarrow \quad \phi =\dfrac{\pi}{2}
\tag{p-16}\label{eqp-16}
\end{equation}
as shown in Figure-08.
There is no "going" going on in field-line diagrams. The direction of the field lines indicates, by convention, the direction of the electrostatic force experienced by a positive test charge at that location.
Field lines do not indicate the 'flow' of any physical quantity, and there is nothing being 'generated'; instead, all you have is a force field, and ways to study and analyze it. This extends to the concept of electric flux (i.e. for a given surface $S$, the integral $\iint_S\mathbf E\cdot\mathrm d\mathbf S$): we call it 'flux' by analogy, but there's nothing at all actually 'flowing'; instead, it is just one more tool to understand and analyze the force field and the laws that govern it.
For more on field lines, see Why does the density of electric field lines make sense, if there is a field line through every point?.
Best Answer
Moving charges don't emit electromagnetic waves unless they are accelerating. A charge moving at a constant speed will just have a field that moves with it, but not a propagating field (i.e., electromagnetic waves).
Sources: Introduction to Electrodynamics by David J Griffiths 4th edition, section 10.3.2 and example 10.4 (pages 459-461), or the Wikipedia page for the Liénard-Wichart Potential.