EM fields of two opposite moving point charges

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I wrote and solved this problem but am having serious doubts about the answer I obtained.

Homework Statement

Two point charges $\pm q$ move along the $z$-axis with velocity $\pm v$. If they are at the origin when $t=0$, what is the electric field magnitude a distance $r$ from the $z$-axis?

Homework Equations

The only things I seemed to need for this problem are Coulomb's law, Maxwell's correction to Ampere's law, and Faraday's law:
$$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2},\qquad \mathcal{E}=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t},\qquad \oint\vec{B}\cdot\mathrm{d}\vec{s}=\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.$$

The Attempt at a Solution

The positions of the point charges are $(0,0,\pm z)$ where $z=vt$. We wish to determine the $\vec{E}$ and $\vec{B}$ fields a distance $r$ from the axis where $z=0$, i.e., at $(x,y,0)$ where $x^2+y^2=r^2$. The first contribution is that of two opposite point charges; then $E^{(0)}=-E_z^{(0)}$, where
$$E_z^{(0)}=\frac{q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{-z}{\sqrt{r^2+z^2}} +\frac{-q}{4\pi\epsilon_0}\frac{1}{r^2+z^2}\frac{z}{\sqrt{r^2+z^2}} =-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.$$
Now consider the electric flux $\Phi_E^{(0)}$ through a circle of radius $R$ at $z=0$. Then $\Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)}$, where we integrate over circular rings of thickness $\mathrm{d}r$:
$$\mathrm{d}\Phi_E^{(0)}=\vec{E}^{(0)}\cdot\mathrm{d}\vec{A}=E_z^{(0)}\,\mathrm{d}A =\left(-\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}\right)(2\pi r\,\mathrm{d}r) =-\frac{q}{2\epsilon_0}\frac{z(2r\,\mathrm{d}r)}{(r^2+z^2)^\frac{3}{2}}$$
$$\Rightarrow\Phi_E^{(0)}=\int\mathrm{d}\Phi_E^{(0)} =-\frac{qz}{2\epsilon_0}\int_0^R\frac{2r\,\mathrm{d}r}{(r^2+z^2)^\frac{3}{2}} =-\frac{qz}{2\epsilon_0}\int_{z^2}^{R^2+z^2}u^{-\frac{3}{2}}\,\mathrm{d}u =-\frac{qz}{2\epsilon_0}\big[-2u^{-\frac{1}{2}}\big]_{z^2}^{R^2+z^2}$$
where we substituted $u=r^2+z^2$, $\mathrm{d}u=2r\,\mathrm{d}r$, $u(0)=z^2$, and $u(R)=R^2+z^2$: Thus
$$\Phi_E^{(0)}(r)=-\frac{qz}{\epsilon_0}\left(\frac{1}{z}-\frac{1}{\sqrt{r^2+z^2}}\right) =\frac{q}{\epsilon_0}\left(\frac{z}{\sqrt{r^2+z^2}}-1\right).$$
Recall the Amp\`ere-Maxwell law,
$$\oint\vec{B}\cdot\mathrm{d}\vec\ell=\mu_0I_\mathrm{enc}+\mu_0\epsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}.$$
By symmetry, we have that $\vec{B}^{(0)}$ is purely azimuthal, so $\oint\vec{B}\cdot\mathrm{d}\vec\ell=(2\pi r)B_\phi^{(0)}$. Since $I_\mathrm{enc}=0$ (there is no current enclosed by the circular loop), we have
$$(2\pi r)B_\phi^{(0)}=0+\mu_0\epsilon_0\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{q}{\epsilon_0}\left(\frac{z}{\sqrt{r^2+z^2}}-1\right)\right] =\mu_0\epsilon_0\left(\frac{q}{\epsilon_0}\right)\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{z}{\sqrt{r^2+z^2}}\right]\frac{\mathrm{d}z}{\mathrm{d}t}$$
$$=\mu_0q\left(\frac{(1)(\sqrt{r^2+z^2})-(\frac{z}{\sqrt{r^2+z^2}})(z)}{r^2+z^2}\right)v =\mu_0qv\left(\frac{\frac{r^2+z^2}{\sqrt{r^2+z^2}}-\frac{z^2}{\sqrt{r^2+z^2}}}{r^2+z^2}\right) =\mu_0qv\frac{r^2}{(r^2+z^2)^\frac{3}{2}}$$
$$\Rightarrow B_\phi^{(0)}=\frac{\mu_0qv}{2\pi}\frac{r}{(r^2+z^2)^\frac{3}{2}}.$$
Now, $B^{(0)}$ itself changes with time, and a changing magnetic field induces an electric field according to Faraday's law:
$$\oint\vec{E}\cdot\mathrm{d}\vec\ell=-\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}.$$
Consider an infinitesimal rectangle of height $\mathrm{d}z$ and width $\mathrm{d}r$ about a point with $z=0$ a distance $r$ away from the $z$-axis. Then the radial components cancel, and the lengthwise components give
$$\oint\vec{E}\cdot\mathrm{d}\vec\ell=E_z^{(1)}(r)\,\mathrm{d}z-E_z^{(1)}(r+\mathrm{d}r)\,\mathrm{d}z,\qquad \frac{\mathrm{d}\Phi_B}{\mathrm{d}t}=\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}\,\mathrm{d}r\,\mathrm{d}z.$$
By symmetry, $E^{(1)}=E_z^{(1)}$. We now have
$$[E_z^{(1)}(r)-E_z^{(1)}(r+\mathrm{d}r)]\,\mathrm{d}z=-\frac{\mathrm{d}B^{(0)}}{\mathrm{d}t}\,\mathrm{d}r\,\mathrm{d}z \Rightarrow\frac{E_z^{(1)}(r+\mathrm{d}r)-E_z^{(1)}(r)}{\mathrm{d}r}=\frac{\mathrm{d}E_z^{(1)}}{\mathrm{d}r}=\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t}.$$
The right-hand side is given by
$$\frac{\mathrm{d}B_\phi^{(0)}}{\mathrm{d}t} =\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{\mu_0qv}{2\pi}\frac{r}{(r^2+z^2)^\frac{3}{2}}\right]\frac{\mathrm{d}z}{\mathrm{d}t} =\frac{\mu_0qv}{2\pi}\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{r}{(r^2+z^2)^\frac{3}{2}}\right]v =\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\frac{r(2z)}{(r^2+z^2)^\frac{5}{2}}\right)$$
$$\Rightarrow\frac{\mathrm{d}E_z^{(1)}}{\mathrm{d}r}=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\frac{r(2z)}{(r^2+z^2)^\frac{5}{2}}\right) \Rightarrow E_z^{(1)}=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\int\frac{2rz\,\mathrm{d}r}{(r^2+z^2)^\frac{5}{2}}\right)$$
$$=\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\int\frac{z\,\mathrm{d}u}{u^\frac{5}{2}}\right) =\frac{\mu_0qv^2}{2\pi}\left(-\frac{3}{2}\right)\left(-\frac{2}{3}\frac{z}{u^\frac{3}{2}}\right) =\frac{\mu_0qv^2}{2\pi}\frac{z}{(r^2+z^2)^\frac{3}{2}}.$$
Note that $1/\sqrt{\mu_0\epsilon_0}=c$, so $\mu_0=1/(\epsilon_0c^2)$ and thus
$$E_z^{(1)}=\frac{v^2}{c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.$$
We have obtained $E_z^{(1)}$ by noting that the changing electric field $\vec{E}^{(0)}$ induced a magnetic field $\vec{B}^{(0)}$, which in turn induced an electric field $\vec{E}^{(1)}$. However, the procedure does not stop there; the electric field $\vec{E}^{(1)}$ induces a magnetic field $\vec{B}^{(1)}$, which induces an electric field $\vec{E}^{(2)}$.

To determine $E^{(2)}$, we replace $q$ with $q'=-\frac{v^2}{c^2}q$ so that $E^{(1)}=-(q'/2\pi\epsilon_0)[z/(r^2+z^2)^\frac{3}{2}]$ (which is exactly the same as $E^{(0)}$ with $q'$ instead of $q$). Therefore
$$E_z^{(2)}=-\left(\frac{v^2}{c^2}\right)^2\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}$$
It is now clear that the $n$th-order induced electric field is then
$$E_z^{(n)}=-\left(-\frac{v^2}{c^2}\right)^n\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.$$
There are infinitely many contributions, which we sum via superposition:
$$E=-E_z=-\sum_{n=0}^\infty E_z^{(n)}=\sum_{n=0}^\infty\left[\left(-\frac{v^2}{c^2}\right)^n\right]\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}} =\frac{1}{1+v^2/c^2}\frac{q}{2\pi\epsilon_0}\frac{z}{(r^2+z^2)^\frac{3}{2}}.$$
Now, I am familiar with the deep connection between relativity and electromagnetism, so when a $1/(1+v^2/c^2)$ showed up instead of a $1/(1-v^2/c^2)$, I started wondering if I did something wrong. I have checked my signs and directions of induced fields five times, and they do appear to be correct. I'm starting to think that I don't want to use Coulomb's law for point charges moving at relativistic speeds. But then what do I use?

 

[Chandra Prayaga]

You should start in the rest frame of one of the charges (call it A). In that frame, the electric field of A is the coulomb field. The electric field of B is the field of a point charge moving with constant velocity. Do you know how to get that field?