Metal rod between a capacitor In my experiment, I am filling liquid crystal between two glass plates. Nn AC voltage is applied on them. The thickness of the liquid crystal cell is around $200 \mu m$. One glass plate is with +V and another with -V. Liquid crystal is aligned perpendicular to the electric field direction. Now I am inserting a metal coated fiber in the liquid crystal cell (between the glass plate filled with liquid crystal). 


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*Is this metal coated fiber has any effect on electric field?  

*If yes, then what happens to the electric field orientation and strength. 

*Is it still transverse? 

 A: 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.
