Question: From Griffiths, Introduction to Electrodynamics:

[![enter image description here](https://i.stack.imgur.com/s11hm.png)](https://i.stack.imgur.com/s11hm.png)

I don't understand why the author says that this problem depends on $x$. The strip is sitting on the left edge. It has a negligible thickness, so we can that $dx$ is the distance it covers in the $x$ direction. Clearly, this problem doesn't depend on $z$ and it most certainly depends on $y$, but why $x$?

Answer: 
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> I don't understand why the author says that this problem depends on x.
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> 
> 

In general, one should assume the potential depends on $x,y,z$ and then, if possible, eliminate one or more coordinates due to, e.g., a symmetry.

The potential is given on two surfaces of constant $y$ and one surface of constant $x$ *and* the potential $V$ must satisfy Laplace's equation:

$$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} + \frac{\partial^2 V}{\partial z^2} = 0$$

It is easy to see that $V$ is not a function of $z$. Why? A translation $z \rightarrow z + Z$ doesn't change the system (the system 'looks' the same regardless of where one chooses to set $z = 0$).

So $V$ must satisfy

$$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0$$

Now, if $V$ did not depend on $x$, we would have

$$\frac{\partial^2 V}{\partial y^2} = 0$$

which implies that $V = V\_{y=0} + \alpha y$.

However, the boundary condition at $y=0$ implies $V\_{y=0} = 0$ and the boundary condition at $y=a$ implies that $\alpha = 0$ and thus $V = 0$.

But this is inconsistent with the boundary condition at $x = 0$ where $V = V\_0(y)$ which is arbitrary.

Thus, $V$ *must* depend on $x$.
It depends on x because the potential will vary with x - the distance away from the edge plate at x=0, where the potential is Vo.