Starting with this generalized formula for a electric field created by a point charge: $$\vec{E} =\frac{q}{4 \pi \epsilon_0} \frac{r}{\left(\vec{r}\cdot\vec{u}\right)^3} [(c^2 − v^2)\vec{u} + \vec{r} × (\vec{u} × \vec{a})] \tag1$$
I was able to show that the electric field of point charge $q$ is constrained to move along the $x$ axis with velocity $v$ is:
$$\vec{E} = \frac{q}{4 \pi \epsilon_0} \frac{1}{r^2} \left(\frac{c + v}{c - v}\right)\tag2$$
However, according to what I deducted from "Introduction to Electrodynamics" by David J. Griffiths, when a point charge $q$ moves along $x$ axis with velocity $v$, without being constrained the electric field is:
$$\vec{E} = \frac{q}{4 \pi \epsilon_0} \frac{1 - \frac{v^2}{c^2}}{\left(1-\frac{v^2 \sin^2\left (\theta \right)}{c^2}\right)^{3/2}} \frac{\hat{R}}{R^2}\tag3$$
$\theta$ is the angle between $\vec{v}$ and $\vec{R}$, if $\theta = 0$, wouldn't it be the case where a point charge is contrained? If so, wouldn't give the exact same electric field?
When I plug $\theta = 0$, I get $$\vec{E} = \frac{q}{4 \pi \epsilon_0} \frac{1 - \frac{v^2}{c^2}}{R^2}\hat{R}\tag4,$$ which clearly isn't the same as (2).
Is my reasoning wrong or is it supposed to be different?
