The fiber's metal coating is a conductor, so its charges will distribute so that the electric field at its surface is perpendicular to that surface. Imagine the electric field bending towards the fiber. Unfortunately, I don't have anything quantitative at hand.
Update:
It turns out there is a standard problem (Schwartz, section 2-11) of a conducting rod of radius $a$ placed in an electric field which approaches a uniform field $E_0 \, \boldsymbol{\hat{x}}$ far from the rod, in other words that approaches an ideal capacitor field (in your problem, $ 2V/d \text{ , with } d \approx 200 \mu m$).
Solving in cylindrical coordinates $r,\theta,z$, with $r=0$ at the center of the rod and $x=r \cos \theta$, one gets a simple form for the potential:
$$ \phi = E_0 \left[ \left(\frac{a}{r}\right)^2 - 1 \right] r \cos \theta
\text{ , with } E_0 = \frac{2V}{d} $$
For $ r >> a$, the potential is constant on the surfaces $r \cos \theta =$ constant, in particular:
$$ \phi \approx \left\{ \begin{array}{cc} +V & r \cos \theta = x = -d/2 \\ -V & r \cos \theta = x = d/2 \end{array} \right. $$
On the rod surface ($r=a$), the potential is constant ($0$). The electric field there is
$$ E_{rod} = - \left. \frac{\partial \phi}{\partial r} \right|_a = 2 E_0 \cos \theta$$
The maximum field intensification factor is 2.
