Electric field of uniformly moving charge ; misprint in Feynman textbook? (28.3) I doubt about formula (28.3) from this Feynman's lecture. 
$$\textbf{E} = - \frac{q}{4 \pi \epsilon_0 } \Big{[} \ \frac{e_{r'}}{r'^2}+ \frac{r'}{c} \frac{d}{dt} \Big( \frac{e_{r'}}{r'^2} \Big) + \frac{1}{c^2} \frac{d^2}{dt^2} e_{r'} \ \Big{]}.\tag{28.3} $$
Consider a charge moving along axis $OX$ with constant speed $v$. Let us consider field at a point on the $OX$. Then formula simplified as follows
$$\textbf{E} = - \frac{q}{4 \pi \epsilon_0 } \Big{[} \ \frac{1}{r'^2}+ \frac{r'}{c} \frac{d}{dt} \Big( \frac{1}{r'^2} \Big) \ \Big{]} e_{r'} = - \frac{q}{4 \pi \epsilon_0 r'^2} \Big{[} \ 1 \pm \frac{2v}{c}  \ \Big{]} e_{r'}$$
The sign depends on position of two points on the line. But this is not the point...
This formula seems definitely wrong. The correct formula must have factor $1 \pm \frac{v}{c}$. 
On the other hand there is no such corrections in an errata.
Questions


*

*Is it my mistake (when I substitute) or mistake in textbook?

*If it is mistake in textbook, what is correct formula? Where can I find it?
 A: Hint :
Your mistake is this
\begin{equation}
\dfrac{\mathrm dr'}{\mathrm dt}=\boldsymbol{\pm}\,\upsilon
\tag{01-wrong}
\end{equation}
The correct one is
\begin{equation}
\dfrac{\mathrm dr'}{\mathrm dt}=
\begin{cases}
\dfrac{\upsilon}{\dfrac{\upsilon}{c}-1} & \text{for charge }q \text{ to the left of the field point } \mathrm P\\
\\
\dfrac{\upsilon}{\dfrac{\upsilon}{c}+1} & \text{for charge }q \text{ to the right of the field point } \mathrm P 
\end{cases}
\tag{02-right}
\end{equation}
where $\:\upsilon(>0)\:$ the speed of the charge towards the positives of the $\:x-$axis (the velocity  being $\:\boldsymbol{\upsilon}=\upsilon\, \mathbf{i}\:$ where $\:\mathbf{i}\:$ the unit vector on the $\:x-$axis).
Note that
\begin{equation}
\mathbf{e}_{r^{'}}=
\begin{cases}
-\mathbf{i} & \text{for charge }q \text{ to the left of the field point } \mathrm P\\
\\
\hphantom{-}\mathbf{i} & \text{for charge }q \text{ to the right of the field point } \mathrm P 
\end{cases}
\tag{03}
\end{equation}
and
\begin{equation}
1-\dfrac{2\, \mathrm dr'}{c\,\mathrm dt}=\dfrac{c\pm\upsilon}{c\mp\upsilon}= 
\begin{cases}
\dfrac{c+\upsilon}{c-\upsilon} & \text{for charge }q \text{ to the left of the field point } \mathrm P\\
\\
\dfrac{c-\upsilon}{c+\upsilon}  & \text{for charge }q \text{ to the right of the field point } \mathrm P 
\end{cases}
\tag{04}
\end{equation}

Video here : Electric field of a uniformly moving point charge
