Is a given charge density a surface charge density or volume charge density? The exact question goes like this:

In a certain electronic tube, electrons are emitted from a hot plane metal surface, and collected by a plane metal plane parallel to the emitter, at a distance $d$ away. (The distance $d$ is small compared with the lateral dimensions of the plates)
The electric potential between the plates is given by $\varphi =kx^{4/3} $ where $x$ is the distance from the emitter.
a) What is the surface charge density $\sigma $ on the emitter? On the collector?
b) What is the volume charge density $\rho(x) $ for $0 < x < d$?

Now I tried to do just do a Laplacian on the electric potential so that it would give me a charge density. But I'm confused.
By taking Laplacian on the electric potential, it would give me :
$$\nabla^{2}\varphi = -\frac{\rho}{\varepsilon_{0}}$$
where $\rho$ is charge density. How do I know if it's surface charge density, or volume charge density?
 A: This equation will always give you a volume charge density. One way to see this is that surface charge density and volume charge density have different units - $\mathrm{C/m^2}$ and $\mathrm{C/m^3}$ respectively - and in order for the units to be consistent, $\rho$ has to be the latter. The fact that the equation is written with $\rho$ is a helpful reminder that it is a volume charge density.
Of course, keep in mind that the potential is not $kx^{4/3}$ everywhere. That function only describes the potential within a certain region. You also have to think about what's happening outside that region, and on the boundaries of the region.
If you try solving Poisson's equation $\nabla^2\varphi = -\rho/\epsilon_0$ in region where the potential is not so nicely behaved (as you have to do here, if you think about the boundaries), you might get a solution that involves a delta function. Just to pull an example out of thin air, something like
$$\rho(x, y, z) = \delta(x - L) e^{-y^2 - z^2}$$
That is the signature of a surface charge density being expressed as a volume charge density. $\sigma$ is the part other than the delta function; in general:
$$\rho(x, y, z) = \delta(x - a)\sigma(y, z)$$
so in this purely hypothetical example you could discern that $\sigma(y, z) = e^{-y^2 - z^2}$.
This is consistent with the statement that surface charge densities correspond to discontinuities in the electric field, because remember you can write Poisson's equation as
$$\vec\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}$$
When $\vec{E}$ is discontinuous, its derivative is "infinite", and therefore $\rho$ needs to be represented as a product involving a delta function.
A: As this is a conductor, the charges are on the surface.  The statement that $d$ is small indicates that you are expected to model this as an infinite plane surface.  You can compute the electric field just outside the conductor by taking $\frac {d\varphi}{dx}$  The electric field inside the conductor is zero because it is a conductor.  You should have a formula that gives the change in field when crossing a surface charge density.
