Old exam question about potential and thermodynamics

PatF
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Homework Statement


This is a question from an old exam. The answer I have is marked wrong but I do not know why.

A conducting sphere of radius a is at potential V and sits at the center of a conducting spherical shell so large that it can be considered infinite and whose potential is zero. The whole system sits at a high temperature T, so that electrons emitted from the conductors form a dilute gas. Assume the density of these electrons is very low so that their mutal interaction can be neglected.
(a) Calculate the potential energy for an electron of charge -|e| as a function of radius r from the center of the system.

(b) In thermal equilibrium, the electrons form a gas of variable density Using the results from part (a) find the dependence of the charge density on the radial distance r.

The Attempt at a Solution



(a) The answer I have is (1/4πε0)(V/a-V/r) Is this correct?


(b) I have no idea what the questioner is asking for.
 
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I did not read the scanned problem; I only read your typed version.

For part (a) you forgot to multiply by the charge of the electron; what you have there is the potential (sans energy). There is also the ambiguity of the reference point. It looks like you chose the surface of the inner sphere to be at zero potential energy, which is generally acceptable (unless your instructor specified otherwise).

For part (b), based on the clues I would use e-βE, where E is the electrostatic potential energy (that you get from part (a)).
 
Consider this a possible plan of attack. This is not a trivial problem!

The key to part (a) is the statement to ignore mutual repulsion among the electrons. So your answer, along with the electron charge qe per what turin told you, is correct.

For (b) I would

1. use the Fermi-Dirac distribution for energies of a Fermi gas: dN(E) = f(E)dE where dN(E) is the no. of electrons with energy between E and E+dE. f(E) is of course the famous Fermi-Dirac distribution.

2. Then equate E to potential energy: p.e. = E = qe{V(a) - V(r)} = qeV(1 - a/r). Let F(r) = f(E) with E = qeV( 1- a/r) so that now
dN(r) = F(r)dr = no. of electrons sitting between r and r + dr. You now have the relative distributon of charge ranging from r = a to r = ∞.

To get the absolute distribution, realize that ∫F(r)dr from r=a to r=infinity must equal the total emitted no. of electrons per unit time. To get this number I refer you to the various formulas for thermionic emission at

http://en.wikipedia.org/wiki/Thermionic_emission.

3. You can then use the Poisson equation to directly solve for ρ(r):

del2V(r) = -ρ(r)/ε = ∂2V(r)/∂r2 + (2/r)∂V(r)/∂r.

Note that you're not solving a differential equation, you're just take the derivatives from #2 above to get ρ.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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