because E=-\nabla\varphi.
from upper equation \varphi=C1Z+C2;
so E=-C1=E0, it is independence with the position. So, I can said when Z=0 or somewhere z<1/2a, E=E0.
by the way, maybe it could be \epsilonE0 because of the continuity at the boundary.
So sorry, lots of confusion.