Electric potential between two infinite plates

Below is a solved exercise from Griffiths' Electrodynamics.

I don't understand why it's directly assumed that the configuration is independent of $z$. Shouldn't there be a contribution from it since the plates span all over the $xz$-plane? Maybe I'm getting confused by the picture, I don't know, does the author mean the electric potential between the plates at $z=0$? Let's maybe clear up a (possible) misunderstanding first.

Mathematically speaking, to say that $V$ has no dependence on $z$ implies the following: $V(x_1,y_1,z_1) = V(x_1,y_1,z_2)$.

The simple and intuitive way to see this is to observe the symmetry along $z$: There is no distinguishable difference between the points ${x_1,y_1,z_1}$ and ${x_1,y_1,z_2}$; if you were to, say, place a point charge at either point, they should move from their point of origin in exactly the same way.

A more physical explanation is to consider the field lines. You know from $V= - \int \mathbf E \cdot \mathbf {dl}$ that for the potential to change, you must move along the field line. So: should $\mathbf E(x,y,z)$ have a component in the positive or negative $z$ direction? The answer is that it can't have either, by the arguments given above, so the potential must not change as we move along $z$.

The configuration doesn't depend of z axis because there is no limit its axial. Otherwise those variable $y$ and $x$ does. In $z = 0$ is just a $x-y$ plane and Its too complicated to say what the potential all this plane. So because of this the Laplace equation must be solve with boundary conditions.
• Could you please elaborate on how it's independent of $z$ because it's "axial"? There are no limits on $z$ so it's independent of it? What if the given potential were $V_0(y,z)$ instead? – phyundergrad Jun 24 '15 at 3:11
• Physically, it makes no sense for $V(x,y)$ to be a function of $z$. The plate extends to infinity in the $x$-$z$ plane and there is nothing to break the $z$-symmetry. That is, the boundary conditions are invariant under translations of the form $z \to z + a$. If you want further proof, you can solve the system assuming $V = V(x,y)$. This function satisfies Laplace's equation and the boundary conditions, so by the uniqueness theorem it must be the correct potential. – Ultima Jun 24 '15 at 3:54
• Well, of course it doesn't make sense once you specify $V$ as a function of $x$ and $y$, and this is exactly what I'm having trouble visualizing, how you can straight forward assume $V=V(x,y)$, is it because there were no limits stated for the $z$ variable, so it's like we're evaluating the potential solely within any plane parallel to the $yz$-plane? Like, $V(x,y,z_1)=V(x,y,z_2)=V(x,y,z_n)=V(x,y)$? That figure really doesn't do it justice, – phyundergrad Jun 24 '15 at 4:39
• Your observation is wrong because you need to choose a specific point along the plane $xy$ and move for any point of z plane. So the potential $V(x_0,y_0,z)$ is equal wherever along the z (vertically or horizontally). One more time, if you choose one point the potential is equal even you change the point z. – miguel747 Jun 24 '15 at 19:55