In the classic example of a charging capacitor I can see how an Amperian loop with a surface going between the capacitor's plates would feel a changing electric flux, as the electric field between the plates changes with time.  However, considering the same Amperian loop but with a flat surface crossed by the wire leading to the capacitor, there is no displacement current, only conducting current.  The fact that there is no displacement current tell us that there is no change in the electric field across that flat surface.  Can someone elaborate more to me why there is no changing electric field across the flat surface too? I can't understand it...
I'm currently studying from "Introduction to Electromagnetics" by D.J. Griffiths. In the book the significance of the displacement current term is explained by looking a non-steady capacitor circuit (shown below).  Using the integral equation and balloon-shaped surface, it is said that $I_{enc}=0$ but $\int \partial\vec{E}/\partial t \cdot d\vec{a}=I/\epsilon_0$. This statement makes sense to me - there is no physical current flowing between the plates but there is a changing electric flux through the balloon surface.  My misunderstanding is with the flat Amperian loop surface where it is mentioned that $\vec{E}=0$ and $I_{enc}=I$ in this case. Obviously there is a current flow $I_{enc}$ in the Amperian loop, but why is $\vec{E}=0$? Just looking at the equations directly, I would have thought there would be a combination of both current terms (including displacement current due to changing electric field in the wire) in a capacitor charging/discharging situation.  I've attempted to seek other explanations on this from other texts,  and elsewhere in this stackexchange site, but this has made me more confused. Clarification on this would be appreciated.     Differential form of Ampere's-Maxwell's equation: $$\nabla \times \vec{B}=\mu_0\vec{J}+\mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$$  Integral form of Ampere's-Maxwell's equation: $$ \oint\vec{B}\cdot d\vec{l} = \mu_0I_{enc}+\mu_0\epsilon_0 \int \frac{\partial \vec{E}}{\partial t} \cdot d\vec{a} $$
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