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?


2 Answers 2


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.

  • $\begingroup$ I'm guessing I should've added a reminder that this is a freshman course. $\endgroup$
    – VladeKR
    Sep 29, 2014 at 22:02
  • $\begingroup$ The first answer mentioned that the plane is "infinitely" large. So, don't the considerations such as near the boundaries, or outside the region unnecessary for the scope of this question? $\endgroup$
    – VladeKR
    Sep 29, 2014 at 22:09
  • $\begingroup$ (2 comments up) well, if you haven't learned some of this stuff yet, consider it a sneak peek into the future ;-) but everything in my answer is freshman-level physics. (1 comment up) The question asks about the charge density on the emitter and collector, which are at the boundaries of the region where you're given the potential. So those considerations are very necessary. $\endgroup$
    – David Z
    Sep 29, 2014 at 22:35

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.

  • $\begingroup$ Ok.. Right, and by taking the derivative of the calculated Electric Field, you will get the charge density. My question was how do you distinguish between the "surface" density and the "volume" density as given by the question. $\endgroup$
    – VladeKR
    Sep 29, 2014 at 21:49
  • $\begingroup$ There is no volume charge density in conductors, because it would set up an electric field in the conductor, which it cannot support. If you have an insulator, you might have (and would expect it to be) a volume distribution. $\endgroup$ Sep 29, 2014 at 22:13
  • $\begingroup$ Yes I understand, but the question specifically asks for a "surface" charge density on the two conductors. But I'm not quite sure how to calculate it. $\endgroup$
    – VladeKR
    Sep 29, 2014 at 22:14
  • $\begingroup$ You have the E field between the conductors given in the problem by taking the derivative of the potential. You know the E field is zero inside the conductors. This gives you a step change as you cross the surface of the conductor, caused by the surface charge. That is where I say you should have an equation for the step change in E field caused by a surface charge. $\endgroup$ Sep 30, 2014 at 4:33
  • 1
    $\begingroup$ If you don't, consider a cylinder that crosses the surface, large in diameter compared to the height. There is no E field through the sides of the cylinder, only through the circular faces. Use Gauss' law to find the difference in E field between the two circles. $\endgroup$ Sep 30, 2014 at 4:33

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.