Force due to an infinite line of charge I am trying to learn some basic special relativity. I am wondering if the following statement is correct. Let's say I am at rest and there is an infinite line of electrons each spaced ,say 1 inch appart. Now I accelerate to some speed v and then continue to travel at this constant velocity v parallel to the line of charges.    (Lets say I am travelling on a spaceship which has some net charge which is nonzero)
Can it be said that if we look at v's which are infinitely close to c, the force that I observe this line of charge to exert on the ship will tend to infinity?
 A: Does the spaceship have a net positive or negative charge? If positive, part of its emotion as it accelerates would be toward the line of charges. This could approach infinity by virtue of proximity, but speed. If negative, the space ship will be repelled as you move dropping off roughly as 1/r where r is the distance from line of charge.
Now you probably don't want to consider acceleration and EM force at the same time so early in relativity. So lets say you start off stationary relative to the line of charges, and you "somehow" find yourself traveling at high speed relative to those charges at a direction parallel. Somehow the EM force has had no effect on you in the mean time. 
In SR, it helps to keep track of important quantities with 4-vectors. For example you can combine charge distribution and current distribution into one object, the 4 current.
$\vec{J}=(c\rho,J_x,J_y,J_z)$
You can take its deriative with $\partial =(\frac{\partial}{c\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$
Conservation of charge is $\frac{\partial \rho}{\partial t}+\nabla\cdot\vec{J}=0$ outside of relativity. 
In SR its $\partial\vec{J}=0.$ and it is invariant under Lorentz Transformations. 
Another useful 4-vector is the 4-potential. $\vec{A}=(V/c,A_x,A_y,A_z)$ Where V is the scalar potential, and A is the vector potential of magnetism.
We usually want the Lorentz Gauge condition to hold, $\nabla\cdot\vec{A}+\frac{1}{c^2}\frac{\partial V}{\partial t}=0$
Remembering which differential equation to hold constant will help you to keep the 4 vectors straight. The $\rho$ from classic EM is multipled by c to preserve the continuity equation when expressed with the $\partial$ operator. For the same reason, the $V$ in the potential 4 vector is divided by c. 
4-vectors allow you to transform important quantities from one frame to another more readily. 
Suppose you have an infinite  line of charge along the z axis, desnity $\lambda$. 
Then $V=\frac{\lambda}{2\pi\epsilon_0}ln(r/r_0)$ where $r_0$ is designated as the zero pointof potential. Since there is no current, you only have an electric field and $\vec{A}=0$. 
Now suppose as in your scenario we find ourselves at some high speed relative to the line of charge.
Our 4-current was $(\lambda c,0,0,0)$ and  our 4-potential was $(\frac{\lambda}{2 \pi \epsilon_0c}ln(r/r_0),0,0,0)$
In the Lorentz transformation we get new quantities for these. There's scalling factor, $\gamma$ and a linear combination of the 0 component and the component parallel to the direction of relative motion. 
The first component of J represents the charge density. The other three components are the vector current densities. The first component of A is the scalar potential, the other three components are vectors of the vector potential. 
So a situation in which there was no current or vector potential is now one in which there is a current and there is a magnetic potential.
What was experienced as a static electric field willnow be experienced as an electric and magnetic field. 
So $\vec{J'}=(\gamma(-u\rho),0,0,\gamma(\rho c))$
and 
$\vec{A}=(\gamma(\frac{-u\lambda}{2\pi\epsilon_0 c}ln(r/r_0),0,0,\gamma(\frac{\lambda}{2\pi \epsilon_0 c}ln(r/r_0)))$
