What would happen if charged plates are placed horizontally? My idea is placing charged conducting plates in such a way that they won't see each others' surfaces unlikely to the typical design of parallel plates. If they are placed like this, would be the force that one plate exerting to other same as in the typical design? If we say the charges on the plates are q1 and q2 how could be the force calculated? Since the electric field is perpendicular to surface of a conductor would this placement affect the electrostatic force and field?
Here is how will the plates are placed:

 A: If the two plates are made of conducting material, there is nothing preventing charges from flowing as close as possible to each other, which, in this case, means toward the edge of each plate closest to the other, right next to the insulating layer. 
If we now suppose the layer to be thin (dimension $d$) with respect to the plates' sides $L = 10\; cm$), and also that the plates are thin in the vertical direction, then charge distribution will be essentially linear, with charge density $\lambda = Q/l$ (Q the total charge on each plate). The electric field produced by this configuration, neglecting edge effects (hence the assumption $d\ll l$) is $E = 2\lambda/d$ in a direction perpendicular to the insulating layer axis. 
Hence the force per unit length $f = 2\lambda^2/d$, and the total force 
$$
F = 2 \frac{\lambda^2 l}{d} = 2 \frac{Q^2}{d\; l}
$$
is attractive and directed along the line joining the centers of the two plates, for obvious reasons of symmetry. 
If instead the two plates are made of insulating material, carrying constant surface charge density $\sigma = Q/l^2$, the force, again directed perpendicular to the insulating layer major axis, can be recovered by means of the usual quadruple integral. Any CAS (=computer algebra system) can do that. 
