Is a given charge density a surface charge density or volume charge density?

Question

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?

Answer 1

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.

Answer 2

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.