0
$\begingroup$

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 $z=0$?

Griffiths

$\endgroup$
2
$\begingroup$

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 $z$.

$\endgroup$
1
$\begingroup$

The configuration doesn't depend of z axis because there is no limit its axial. Otherwise those variable $y$ and $x$ does.

Pain attention about this figure:

enter image description here

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – phyundergrad Jun 24 '15 at 3:11
  • $\begingroup$ 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. $\endgroup$ – Ultima Jun 24 '15 at 3:54
  • $\begingroup$ 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, $\endgroup$ – phyundergrad Jun 24 '15 at 4:39
  • $\begingroup$ 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. $\endgroup$ – miguel747 Jun 24 '15 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.