Bastion said:
Does that mean that when q1 and q2 start moving relative to x, that relative to x, the force on q1 and q2 increases while relative to q1 and q2 the force does not increase?
As you aren't interested in the acceleration phase we can ignore it and just consider the "before" and "after" inertial phases. That way we don't have to mess with Lienard-Wiechert potentials and can use Coulomb's law and the Lorentz transforms.
Working in the rest frame of the charges we place ##q_1## at the origin and ##q_2## at ##x=R## (not ##r## because I'm going to use that symbol for the generic distance). The electric field due to ##q_1## is then ##(x,y,z)^T(E/r^3)## where ##r^2=x^2+y^2+z^2## and ##E=k_eq_1## is equal to the electric field strength at ##r=1##. Work in units where ##c=1## and plug this into the Faraday tensor:$$\begin{eqnarray*}
F&=&
\left(\begin{array}{cccc}
0&-E_x&-E_y&-E_z\\
E_x&0&-B_z&B_y\\
E_y&B_z&0&-B_x\\
E_z&-B_y&B_x&0
\end{array}\right)\\
&=&
\left(\begin{array}{cccc}
0&-Ex/r^3&-Ey/r^3&-Ez/r^3\\
Ex/r^3&0&0&0\\
Ey/r^3&0&0&0\\
Ez/r^3&0&0&0
\end{array}\right)
\end{eqnarray*}$$Now we can write the Lorentz transforms as matrices:$$\Lambda=
\left(\begin{array}{cccc}
\gamma&-\gamma v&0&0\\
-\gamma v&\gamma&0&0\\
0&0&1&0\\
0&0&0&1
\end{array}\right)$$where ##\gamma=1/\sqrt{1-v^2}##. Now we can easily write the electromagnetic fields in a frame where the charges are moving by transforming the Faraday tensor: $$\begin{eqnarray*}
F'&=&\Lambda.F.\Lambda^T\\
&=&\left(\begin{array}{cccc}
0&-Ex/r^3&-\gamma Ey/r^3&-\gamma Ez/r^3\\
Ex/r^3&0&\gamma v Ey/r^3&\gamma vEz/r^3\\
\gamma Ey/r^3&\gamma v Ey/r^3&0&0\\
\gamma Ez/r^3&\gamma v Ez/r^3&0&0
\end{array}\right)
\end{eqnarray*}$$Note that I've only transformed the field components and not the coordinates. Formally I would need to also transform coordinates and there are risks in not doing so, but in this case it doesn't matter because I'm not going to do anything except transform ##x=R## and ##y=z=0## and plug that in, and there are no issues of simultaneity because the field components don't change at ##q_2##'s location. So I can just plug in ##x=R## and ##y=z=0## and read off that the transformed electric field is the same as the untransformed one.
So in this case there is no change. There will be changes if the line joining the charges is not parallel to the direction of motion.