,
using the orthogonality relation (3.68). That is, we obtain 2 if l=0, and 0 for l;SPMgt;0.
-- in spherical coordinates with azimuthal symmetry (no 
-dependence) -- for the boundary condition
constant potential on a spherical shell at radius R. As shown in the text, the general eigensolution to Laplace's equation in spherical coordinates with azimuthal symmetry is
for all integer l, where 
is a Legendre polynomial.
We need to find solutions to Laplace's equation with our boundary
condition valid first inside the shell, and then outside, and then
patch the two together. Here the patching will be automatic, as our
boundary condition at r=R will automatically make both solutions
equal at r=R, where both give potential 
.
First, inside the shell, we have two implicit boundary conditions:
We note condition (2.1) is because the potential can diverge, physically, only at a charge source. Imposing (2.1) means we must discard the 
solution, giving
At r=R, condition (2.2) implies
This is an expansion of the constant function in terms of Legendre polynomials, with coefficients 
. As noted in Example 3.6, the orthogonality relation (3.68) of the Legendre polynomials implies that the coefficients obey
But this integral is just times the integral 
evaluated in problem 1, and gives as the only nonzero coefficient
Thus our solution inside the sphere is
Outside the shell, we also have two implicit boundary conditions:
We note condition (2.3) is again because the potential can diverge, physically, only at a charge source. Imposing (2.3) means we must discard the 
solution, giving
At r=R, condition (2.4) implies
This is an expansion of the constant function in terms of Legendre polynomials, with coefficients 
. Again the coefficients -- here -- obey
(This is the same integral as above.) We thus have for our only nonzero coefficient
Thus our solution outside the sphere is
Our full solution is thus given by
constant inside the sphere, and falling as 1/r outside the sphere. (This potential could be physically realized by a charged spherical shell at radius R, or equivalently a charged conducting sphere of radius R, with charge 
.)
as the sum 
, noting that only terms with 
are required to represent a third order polynomial. We find the coefficients 
by
term comes only from 
. Since has coefficient 4, 
must be 8/5. So now we have
This remainder has no second order term, so 
must be zero. Since 
, so we have exactly 
, giving
then
Since is of degree 3, 
vanishes for l;SPMgt;3, from
We thus must find 
and by integration. They are
Here, in integrating 
and , we have used the fact that the integral of an odd integrand over the interval [-1, 1] vanishes. We 've also noted that in evaluating
odd terms in f(x) give twice the contribution from x=1.
We must find the potential inside and outside a sphere which has
on its surface at r=R. We must then find the charge density 
on the spherical shell, assuming no charge density elsewhere.
As in problem 2, we need to find solutions to Laplace's equation with
our boundary condition valid first inside the shell, and then outside,
and then patch the two together. Again the patching will be automatic,
as our boundary condition at r=R will automatically make both
solutions equal at r=R, where both give potential 
.
The general eigensolution to Laplace's equation in spherical coordinates with azimuthal symmetry is again
for all integer l, where is a Legendre polynomial.
First, inside the shell, we have two implicit boundary conditions:
Imposing (4.1) means we must discard the solution, giving
At r=R, condition (4.2) implies
This looks simpler in terms of the variable 
:
This is an expansion of the polynomial 
in terms of Legendre polynomials, with coefficients . Fortunately, we've worked out this expansion in problem 3, finding
Thus we identify coefficients
Thus our solution inside the sphere is
using
Note that at r=R this indeed gives
Outside the shell, we also have two implicit boundary conditions:
Imposing (4.3) means we must discard the solution, giving
At r=R, condition (4.4) implies
or again in terms of ,
This is again an expansion of the polynomial in terms of Legendre polynomials, now with coefficients . Again from problem 3,
allowing us to identify coefficients
We might have gotten this directly from Griffiths' equation (3.81), 
, forcing both solutions to agree at r=R.
Thus our solution outside the sphere is
using
Note that at r=R this indeed gives
Our full solution is thus given by
Finally, we find on the r=R shell by using the boundary condition
which Griffiths shows via some quick algebra to be equal to
equation (3.83). Here we have
yielding
using
Multiplying by 
, then separating variables 
and dividing by V yields
Each term is thus a constant, and we must choose the signs of the constants so that
so that the eigensolutions 
are 
-periodic in . This gives as the s-equation
which is soluble by a polynomial S. Assuming 
and plugging in gives
which is solved by 
. Thus our general eigensolution is
(A slight complication arises when n=0. Here the equation is solved by 
, of which we can only keep the periodic constant b. The s equation,
requires that 
be constant, which occurs for eigensolution 
. ) Thus the general solution to Laplace's equation for cylindrical coordinates with z-independence is
Now back to our specific problem. The entire conducting pipe is an equipotential, whose potential we choose to set to zero (by making the origin our reference point for calculating potentials). Thus
is our first boundary condition. Because the pipe is a conductor,
V(s) = 0 on the equipotential inside the pipe, at 
. Outside the pipe, the
distortion the pipe causes to the electric field 
must vanish asymptotically far away, so
is our second boundary condition. (In principal, there could be an additional constant piece to this unperturbed V, but it vanishes by making the unperturbed potential agree with our convention that V=0 at the origin.)
Applying condition (5-2) to our general solution, only terms with
n=1 give the proper -dependence. We discard the 
solution to give, from condition (5-2),
As 
, this reproduces the asymptotic behavior
so we can identify 
from condition (5-2).
Now, applying condition (5-1) at s=R gives
This gives for our solution for V outside the pipe
You might wonder whether our boundary condition (5-2) at infinity
truly enabled us to set 
for 
, since those terms go
as 
and vanish asymptotically anyway. The answer is no,
really, condition (5-2) alone doesn't fix for ,
though it does force 
. However, the term we must
keep, 
, makes a contribution to our boundary condition
at s=R which can only be canceled by another term varying as
. Thus, cancellation at the second boundary excludes all but
n=1 terms.
Finally, we must calculate
since V=0 inside. This gives
, Laplace's equation in cylindrical coordinates with no z-dependence. This time we must solve Laplace's equation inside and outside our pipe, a cylindrical surface at s=R with charge density
Thus we must solve Laplace's equation inside, with boundary condition
and outside, with boundary condition
We must then match our two solutions via boundary conditions
As before, inside the pipe we have general solution
Requiring finiteness at s=0, condition (6.1), means that b and
must all vanish. Thus we have
Outside the pipe we have general solution
Requiring finiteness as , condition (6.2), means that b' and
must all vanish. Thus we have
Matching these solutions at s=R, condition (6.3), yields
or
Thus our final boundary condition,
On the right, we have a Fourier expansion for the function 
over the interval 
. Here the coefficients are trivial: we keep the term, and discard all other orthogonal ones. Our only nonzero coefficient is 
, which obeys
Plugging this back in to our inside and outside solutions, we have
