Electric field at plane of non-conducting uniformly charged thin disk Will field at plane of a disk excluding the disk itself be zero? Here basis of above conclusion is that the field lines will originate perpendicularly to the plane, and from both faces therefore they will never be able to reach plane of disk at finite distance.
From above the potential at edge will also be zero, if a test charge is bought from infinity through the path passing through plane to the edge, which is obviously not true.
Therefore, tell me where I'm wrong and please provide the field line distribution due to the disk. 
 A: Your assumption that field lines come out perpendicularly of the disc is incorrect.
Imagine the disc seen from a very large distance: it's indistinguishable (I'm short-sighted) from a point particle, which has radially symmetric lines. This gives you an approximation for the long-distance behaviour of the field, which is not zero on the disc plane.
I can't think of an easy way to find the field at all points in space. However this link shows how to compute the field on the vertical of the disc centre.
A: Assuming the disk is charged +ve, the electric field will point away from the disk in all directions, including in the plane of the disk. It will not be zero at any point outside of the disk.

In the plane of the disk, by symmetry the electric field points radially away from the centre. If the potential at infinity is zero, the potential at the edge of the disk will not be zero.
Out of the plane of the disk the field lines will not be straight except along the axis of the disk - which is a line of symmetry. Very close to the surface the field will not necessarily be perpendicular to the faces, because the disk is not a conductor. 
As Bzazz says, at large distances the field is the same as that from a point charge.
