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The Heaviside-Feynman Formula ( the 5fth equation in your question) is derived from the Liénard-Wiechert potentials, which in turn are derived from the first four equations of yours. But from the Liénard-Wiechert potentials you could derive the following equation which is more convenient for our case¹:

\begin{equation} \mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01} \end{equation} where

\begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{02c} \end{align} The electric field of equation (01) is given by equation 14.14 from Jackson's 'Classical Electrodynamics',3rd Edition.

For points at large distances from the charge [$R^{-2}\rightarrow 0$] and velocities $\:\upsilon\:$ of the charge always much less than c [$\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3\rightarrow 1$] the first term in the rhs of (01) tends to zero and it's ignored. For the second term with velocities $\:\upsilon\:$ of the charge always much less than c we have the following \begin{align} \beta & = \dfrac{\upsilon}{c} \ll 1 \tag{03a}\\ \mathbf{n}-\boldsymbol{\beta} & \approx \mathbf{n} \tag{03b} \end{align} so \begin{equation} \mathbf{E}(\mathbf{x},t) \approx -\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\times\mathbf{n}}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{04} \end{equation} But $\:(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\boldsymbol{\times} \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\perp \mathbf{n} }\:$ is the vectorial projection of the acceleration vector $\:\dot{\boldsymbol{\beta}}\:$ on the direction normal to $\:\mathbf{n}\:$, that is normal to $\:\mathbf{r'}$ [while the vectorial projection on the direction $\:\mathbf{n}\:$ is derived by replacing in the previous expression the outer products by the inner ones $\:(\mathbf{n}\boldsymbol{\cdot} \dot{\boldsymbol{\beta}})\cdot \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\parallel \mathbf{n} }$].

$1 \quad$ Note that \begin{equation} \mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{\boldsymbol{-}\boldsymbol{\nabla}\phi} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{\boldsymbol{-}\frac {\partial \mathbf{A}}{\partial t}} \tag{01} \end{equation}

The Heaviside-Feynman Formula ( the 5fth equation in your question) is derived from the Liénard-Wiechert potentials, which in turn are derived from the first four equations of yours. But from the Liénard-Wiechert potentials you could derive the following equation which is more convenient for our case¹:

\begin{equation} \mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01} \end{equation} where

\begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{02c} \end{align} The electric field of equation (01) is given by equation 14.14 from Jackson's 'Classical Electrodynamics',3rd Edition.

For points at large distances from the charge [$R^{-2}\rightarrow 0$] and velocities $\:\upsilon\:$ of the charge always much less than c [$\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3\rightarrow 1$] the first term in the rhs of (01) tends to zero and it's ignored. For the second term with velocities $\:\upsilon\:$ of the charge always much less than c we have the following \begin{align} \beta & = \dfrac{\upsilon}{c} \ll 1 \tag{03a}\\ \mathbf{n}-\boldsymbol{\beta} & \approx \mathbf{n} \tag{03b} \end{align} so \begin{equation} \mathbf{E}(\mathbf{x},t) \approx -\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\times\mathbf{n}}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{04} \end{equation} But $\:(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\boldsymbol{\times} \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\perp \mathbf{n} }\:$ is the vectorial projection of the acceleration vector $\:\dot{\boldsymbol{\beta}}\:$ on the direction normal to $\:\mathbf{n}\:$, that is normal to $\:\mathbf{r'}$ [while the vectorial projection on the direction $\:\mathbf{n}\:$ is derived by replacing in the previous expression the outer products by the inner ones $\:(\mathbf{n}\boldsymbol{\cdot} \dot{\boldsymbol{\beta}})\cdot \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\parallel \mathbf{n} }$].

$1 \quad$ Note that \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{\boldsymbol{-}\boldsymbol{\nabla}\phi} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{\boldsymbol{-}\frac {\partial \mathbf{A}}{\partial t}} \tag{01-false} \end{equation}

EDIT : Thanks to a comment by @verdelite : "Your partitioning of $\mathbf{E}(\mathbf{x},t)$ into the two parts as in footnote 1 is incorrect..." the correct one is \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{\boldsymbol{-}\boldsymbol{\nabla}\phi\boldsymbol{+}\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{\boldsymbol{-}\frac {\partial \mathbf{A}}{\partial t}\boldsymbol{-}\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)} \tag{01} \end{equation} where $\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)$ a vector function of $\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}$.

The Heaviside-Feynman Formula ( the 5fth equation in your question) is derived from the Liénard-Wiechert potentials, which in turn are derived from the first four equations of yours. But from the Liénard-Wiechert potentials you could derive the following equation which is more convenient for our case¹:

\begin{equation} \mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01} \end{equation} where

\begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{02c} \end{align} The electric field of equation (01) is given by equation 14.14 from Jackson's 'Classical Electrodynamics',3rd Edition.

For points at large distances from the charge [$R^{-2}\rightarrow 0$] and velocities $\:\upsilon\:$ of the charge always much less than c [$\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3\rightarrow 1$] the first term in the rhs of (01) tends to zero and it's ignored. For the second term with velocities $\:\upsilon\:$ of the charge always much less than c we have the following \begin{align} \beta & = \dfrac{\upsilon}{c} \ll 1 \tag{03a}\\ \mathbf{n}-\boldsymbol{\beta} & \approx \mathbf{n} \tag{03b} \end{align} so \begin{equation} \mathbf{E}(\mathbf{x},t) \approx -\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\times\mathbf{n}}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{04} \end{equation} But $\:(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\boldsymbol{\times} \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\perp \mathbf{n} }\:$ is the vectorial projection of the acceleration vector $\:\dot{\boldsymbol{\beta}}\:$ on the direction normal to $\:\mathbf{n}\:$, that is normal to $\:\mathbf{r'}$ [while the vectorial projection on the direction $\:\mathbf{n}\:$ is derived by replacing in the previous expression the outer products by the inner ones $\:(\mathbf{n}\boldsymbol{\cdot} \dot{\boldsymbol{\beta}})\cdot \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\parallel \mathbf{n} }$].

$1 \quad$ Note that \begin{equation} \mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{-\boldsymbol{\nabla}\phi} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{-\frac {\partial \mathbf{A}}{\partial t}} \tag{01} \end{equation}

The Heaviside-Feynman Formula ( the 5fth equation in your question) is derived from the Liénard-Wiechert potentials, which in turn are derived from the first four equations of yours. But from the Liénard-Wiechert potentials you could derive the following equation which is more convenient for our case¹:

\begin{equation} \mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01} \end{equation} where

\begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{02c} \end{align} The electric field of equation (01) is given by equation 14.14 from Jackson's 'Classical Electrodynamics',3rd Edition.

For points at large distances from the charge [$R^{-2}\rightarrow 0$] and velocities $\:\upsilon\:$ of the charge always much less than c [$\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3\rightarrow 1$] the first term in the rhs of (01) tends to zero and it's ignored. For the second term with velocities $\:\upsilon\:$ of the charge always much less than c we have the following \begin{align} \beta & = \dfrac{\upsilon}{c} \ll 1 \tag{03a}\\ \mathbf{n}-\boldsymbol{\beta} & \approx \mathbf{n} \tag{03b} \end{align} so \begin{equation} \mathbf{E}(\mathbf{x},t) \approx -\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\times\mathbf{n}}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{04} \end{equation} But $\:(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\boldsymbol{\times} \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\perp \mathbf{n} }\:$ is the vectorial projection of the acceleration vector $\:\dot{\boldsymbol{\beta}}\:$ on the direction normal to $\:\mathbf{n}\:$, that is normal to $\:\mathbf{r'}$ [while the vectorial projection on the direction $\:\mathbf{n}\:$ is derived by replacing in the previous expression the outer products by the inner ones $\:(\mathbf{n}\boldsymbol{\cdot} \dot{\boldsymbol{\beta}})\cdot \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\parallel \mathbf{n} }$].

$1 \quad$ Note that \begin{equation} \mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{\boldsymbol{-}\boldsymbol{\nabla}\phi} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{\boldsymbol{-}\frac {\partial \mathbf{A}}{\partial t}} \tag{01} \end{equation}

The Heaviside-Feynman Formula ( the 5fth equation in your question) is derived from the Liénard-Wiechert potentials, which in turn are derived from the first four equations of yours. But from the Liénard-Wiechert potentials you could derive the following equation which is more convenient for our case  :

\begin{equation} \mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01} \end{equation} where

\begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{02c} \end{align} The electric field of equation (01) is given by equation 14.14 from Jackson's 'Classical Electrodynamics',3rd Edition.

For points at large distances from the charge [$R^{-2}\rightarrow 0$] and velocities $\:\upsilon\:$ of the charge always much less than c [$\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3\rightarrow 1$] the first term in the rhs of (01) tends to zero and it's ignored. For the second term with velocities $\:\upsilon\:$ of the charge always much less than c we have the following \begin{align} \beta & = \dfrac{\upsilon}{c} \ll 1 \tag{03a}\\ \mathbf{n}-\boldsymbol{\beta} & \approx \mathbf{n} \tag{03b} \end{align} so \begin{equation} \mathbf{E}(\mathbf{x},t) \approx -\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\times\mathbf{n}}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{04} \end{equation} But $\:(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\boldsymbol{\times} \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\perp \mathbf{n} }\:$ is the vectorial projection of the acceleration vector $\:\dot{\boldsymbol{\beta}}\:$ on the direction normal to $\:\mathbf{n}\:$, that is normal to $\:\mathbf{r'}$ [while the vectorial projection on the direction $\:\mathbf{n}\:$ is derived by replacing in the previous expression the outer product by the inner one $\:(\mathbf{n}\boldsymbol{\cdot} \dot{\boldsymbol{\beta}})\cdot \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\parallel \mathbf{n} }$].

The Heaviside-Feynman Formula ( the 5fth equation in your question) is derived from the Liénard-Wiechert potentials, which in turn are derived from the first four equations of yours. But from the Liénard-Wiechert potentials you could derive the following equation which is more convenient for our case¹:

\begin{equation} \mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01} \end{equation} where

\begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{02c} \end{align} The electric field of equation (01) is given by equation 14.14 from Jackson's 'Classical Electrodynamics',3rd Edition.

For points at large distances from the charge [$R^{-2}\rightarrow 0$] and velocities $\:\upsilon\:$ of the charge always much less than c [$\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3\rightarrow 1$] the first term in the rhs of (01) tends to zero and it's ignored. For the second term with velocities $\:\upsilon\:$ of the charge always much less than c we have the following \begin{align} \beta & = \dfrac{\upsilon}{c} \ll 1 \tag{03a}\\ \mathbf{n}-\boldsymbol{\beta} & \approx \mathbf{n} \tag{03b} \end{align} so \begin{equation} \mathbf{E}(\mathbf{x},t) \approx -\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\times\mathbf{n}}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{04} \end{equation} But $\:(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\boldsymbol{\times} \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\perp \mathbf{n} }\:$ is the vectorial projection of the acceleration vector $\:\dot{\boldsymbol{\beta}}\:$ on the direction normal to $\:\mathbf{n}\:$, that is normal to $\:\mathbf{r'}$ [while the vectorial projection on the direction $\:\mathbf{n}\:$ is derived by replacing in the previous expression the outer products by the inner ones $\:(\mathbf{n}\boldsymbol{\cdot} \dot{\boldsymbol{\beta}})\cdot \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\parallel \mathbf{n} }$].

$1 \quad$ Note that \begin{equation} \mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{-\boldsymbol{\nabla}\phi} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{-\frac {\partial \mathbf{A}}{\partial t}} \tag{01} \end{equation}