Physics 7701: Problem Set #10
Here are some hints, suggestions, and comments on the assignment.
Recent changes to this page:
 16Nov2013  original version.
 18Nov2013  additional comments for 14
 19Nov2013  still more comments for 14
 Potential in cylinder.
In general, we follow the expansion method in class for a finite cylinder.
The difference here is that the boundary conditions are, in a sense, reversed:
the end caps are grounded and we specify the potential on the cylindrical
surface.
 We have the full range of φ dependence so we apply the condition
for singlevalued functions of φ. Don't forget to consider
the possibility of a constant term.

The boundary conditions in z means that the separation constant kappa^{2}
will be the opposite sign from what we considered in class, so sinhs and coshs
become sines and cosines. Do both contribute?
 You can think of the change in sign of kappa^{2}
as meaning that k becomes i*k.
This in turn makes the Bessel functions become modified Bessel functions.
Look up the properties of modified Bessel functions in a text or online.
You will find the asymptotic forms for I(x) and K(x) (which includes what
happens at small values of x); which of these functions
do you use (or both?)
for the interior scalar potential? Why?
 Can you think of limiting cases to check your answer?
 Cylinder again.

The first part is just to carry out the integrals to find the coefficients
given the particular potential in equation (1).

Use the asymptotic (small ρ) limits of the modified Bessel function.
[An open question is whether this is really justified, given that the
sums go up to infinity so the argument of these Bessel functions will
eventually not be small in the sum for fixed b/L. That the sum converges
is not in question because it is the ratio that appears.]
You will need to do sums, which you can do using Mathematica, but you will
still have some manipulations to do to make it equal to equation (2).
In class we'll discuss how to do sums like this "by hand" (you can apply
the power series for arctan twice!). A helpful identity is
tan(A+B) = (tan A + tan B)/(1  tan A tan B), remembering that
tan(tan^{1}(x))=x.
 Concentric spheres.
 Use the general expansion for a φindependent scalar potential.
Remember that restrictions on how the potential behaves at the origin
won't be relevant here.
You'll determine the coefficients from the boundary conditions on the
hemispheres.
 The large b limit is a bit unexpected, because the potential
is not zero at infinity because of the boundary conditions.
 Spherical surfaces.
Note that we have a spherical surface of charge here and not a conductor.
So we want to find a potential given a charge distribution, which suggests
using a Green function.
 What is the charge density written with appropriate theta and delta
functions? What is the relevant Green function (again remembering that
we are not counting the sphere as a conductor) to integrate over to
get the scalar potential. You can also write an expansion for the potential
on the inside of the sphere. Equate at a simple choice for
theta and find the coefficients. You'll need to use recursion relations
to get the form requested.
 Same procedure as for inside the sphere.
 Use the standard formula for the electric field given the potential.
 What happens to α in this limit?
 Freespace Green function in polar coordinates.
Here the only variables are ρ and φ (all functions are independent
of z). Some comments:
 Please note the use of Zangwill's conventions. This means that factors
of ε_{0} and 4π (or 2π in this case) appear in the
Green function. This doesn't change anything significant in your solution.
 The idea is to do separation of variables and solve
for the angular part in a complete expansion. Remember that we need
this part to be single valued and don't forget the possibility of a
constant term. Use the completeness relation (the sum that gives a delta
function) in the φ functions.
 Don't forget to exploit the idea that the Green function should be symmetric
under interchange of the unprimed and primed variables.
 When you solve the onedimensional equation by integrating across the
delta function, make sure you first make any rearrangement needed so that
you get the difference of the first derivative of the Green function at
the matching point.
 Note that the units seem to be wrong for the ln term (if we associate
ρ with a unit of length). You can think of there being a constant there
that is equal to one in our units. Where does this enter into your solution
for the ρ dependence?
 Green function inequalities.
 What would the Green function be in the absence of boundary conditions
on the surface of the volume V? What are those boundary conditions for
a Dirichlet Green function? Could the opposite inequality to the one
stated in equation (8) be possible if these boundary conditions are to
hold? Consider the physical interpretation of the second term
in equation (7).
 Earnshaw's theorem says that the scalar potential in a chargefree region
takes both its minimum and maximum value on the boundary of that region.
Suppose G_{D} were less than zero; what would that mean about
the contribution of the second term in equation (7)? If we considered
just this term, what would it imply for the scalar potential in V?
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Physics 7701: Assignment #10 hints.
Last modified: 08:48 am, November 20, 2013.
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