Assume a capacitor where upper/lower plate has negative/positive charge. An 1D spring is attached to lower plate, and the bottom of spring is connected to ground. At the beginning everything is at the rest ($v=0$).
Let us restrict this system in two ways.
The positive charged plate and spring can not move in x direction, but they can move in y direction. We can simply achieve this by adding two walls in left/right sides of this plate.
The negative charged plate can not move in y direction, but it can move in x direction. We can restrict this plate by pinning the top and bottom of this plate (note that we don't pin the plate itself). The pins are very thin and insulator.
We know that the upper plate attracts lower plate in +y direction, so spring will be stretched in y direction, according to Hooke's law, until spring force and electric force cancel each other completely. At this point, system remains at the rest.
Now my question is as follows: What happens if we set the upper plate in motion in positive x direction, with constant velocity $v$? For the sake of simplicity, it is possible to assume that capacitor is large enough to ignore its boundaries. That is we are only interested in small part of capacitor, and figure above only shows a small part of capacitor. But this assumption is not necessary and we don't assume it.
First, let us consider observer S who is at the rest with respect to upper plate. According to this observer, the force of spring on lower plate, will be reduced by a $1/\gamma$ factor [1]. So spring will be stretched in positive y direction. From viewpoint of S, length of spring will increase. Note that if we don't assume a very large capacitor, electric field force in this part of capacitor will change over time due to the boundary conditions. However, the variation of electric field force is completely independent from spring force (i.e the force of electric field won't be reduced by $1/\gamma$ factor, it depends on other factors).
Observer S' who is at the rest with spring has another opinion. According to S', force of spring is the same as before, but the electromagnetic force of upper plate will be reduced by $1/\gamma$ factor. Let us check this out. We start from these equations [2]:
$$ E' = \gamma (E + v\times B) $$
$$B' = \gamma (B - v\times E/c^2) $$
since $E = |E| \hat{j}$ and $v = -|v| \hat{i}$ and $B = 0$ we have:
$$ E' = \gamma (|E|\hat{j}) $$
$$ B' = \gamma (|v| |E|\hat{k}/c^2) $$
and electromagnetic force will be
$$ F' = qE' + qv'\times B' $$
but since $v' = |v|\hat{i}$ we have
$$ F' = q\gamma (|E|\hat{j}) - q\gamma |v|^2 |E|/c^2 \hat{j} = q|E|/\gamma \hat{j}$$
So from viewpoint of S', electromagnetic force will be reduced by a $1/\gamma$ factor and he claims that length of spring will decrease, because force of spring will overcome electric force. Note that we could arrive at same conclusion by direct Lorentz transformation of force.
But S' and S can't be right at the same time, so what is the problem?
I appreciate any help or hints. Please do show me my errors. I believe a more complicated version of this question (which had a few problems) was asked by @MohammadJavanshiry but I was unable to find it. However I can't see those problems in my question.
Edit: For disregarding boundary conditions, you can assume this system as follows:
where gray areas are "walls" that restrict the motion of lower plate and spring in x axis. These walls also restrict the motion of upper plate in y axis. Assume Upper plate is large enough during this thought experiment.
Edit 2: I had a simple mistake in force equation. Indeed, since lower plate doesn't move in x axis, $v'=0$, and both of observer conclude that spring's length will increase, in other words
$$ F' = q\gamma (|E|\hat{j})$$
