8/13/2019 Jackson 2 12 Homework Solution
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Jackson 2.12 Homework Problem Solution
Dr. Christopher S. BairdUniversity of Massachusetts Lowell
PROBLEM:
Starting with the series solution (2.7! for the two"di#ensional
potential pro$le# with the potentialspecified on the surface of a
cylinder of radius b% evaluate the coefficients for#ally%
su$stitute the#
into the series% and su# it to o$tain the potential inside the
cylinder in the for# of &oisson's integral
(,)=
2)
2
(b , ') b
22
b2+22bcos ( ')
d '
*hat #odification is necessary if the potential is desired in
the region of space $ounded $y the cylinder
and infinity+
SOLU!O":
*hen the potentials and charges are unifor# in the z"direction%
as is the case in this pro$le# with thecylinder% the syste# reduces
down to a two"di#ensional pro$le#. ,he circular cross"section of
the
cylinder dictates that polar coordinates are the #ost fitting
two"di#ensional coordinate syste# to use.
,here are not charges involved% so we need to solve the Laplace
e-uation. ,he series solution to theLaplace e-uation in polar
coordinates was found in ac/son to $e (0-. 2.7!
(,)=a)+b) ln+n=
annsin (n n)+
n=
bnn
sin(n+n)
Because this is a solution to a two"di#ensional second order
differential e-uation (the Laplacee-uation!% there #ust $e four
undeter#ined coefficients (or sets of coefficients for series
solutions!. *e
therefore need four $oundary conditions1 or a $oundary condition
at the #ai#u# and #ini#u# of
each di#ension. 3or this pro$le# the $oundary conditions are
(! angular #ini#u#( (=))=(=2)(2! angular #ai#u# (=2)=(=))(4!
radial #ini#u# (=))=finite(5! radial #ai#u#( (=b)=V()
Boundary conditions (! and (2! are the sa#e $oundary conditions
ensuring periodicity $ecause the fullangular sweep is included in
the region of interest. 6pplying $oundary conditions (! and (2!
leads to
the coefficient #ultiplied against phi $eco#ing an integer.
,hese $oundary conditions have already
$een applied in the solution given in ac/son 0-. 2.7. *ith two
$oundary conditions left% we should
only have two undeter#ined sets of coefficients. But 0-. 2.7
see#s to contain four independentundeter#ined coefficients. n
reality% they are not all independent and will $e ta/en care of as
we
proceed.
*hen we apply $oundary condition (4! we see right away that
8/13/2019 Jackson 2 12 Homework Solution
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bn=) and b)=)
so that the solution $eco#es
(,)=a)+n=
annsin (nn)
Because anand nare ar$itrary at this point% we can redefine the#
as we want to get this general
solution into a #ore useful for#. ,a/e a b-nout of the
anconstants.
(,)=a)+
n=
an(b)
n
sin (nn)
Use sin=(e iei)/(2 i)
(,)=a)+
n=
an
2i(b )
n
[ein ei n ei n ei n ]
8edefine constants
(,)=a)+n=
(b)n
[cn ei n+dne
i n ]
Both ter#s (as well as a)! can $e co#$ined $y letting the
su##ation inde etend to negative nu#$ers
(,)=n=
cn(b )n
ei n
6pply $oundary condition (5! to find
V()=n=
cn ei n
Multiple $oth sides $y a co#ple eponential and integrate
)
2
V() ei n 'd= n=
cn)
2
ei (nn ')
d
9ow recogni:e the integral on the right as the state#ent of
orthogonality for co#ple eponentials sothat
)
2
V()ei n 'd=n=
cn 2n n '
