Learning Materials
Features
Discover
Chapter 3: Q3.6P (page 124)
A more elegant proof of the second uniqueness theorem uses Green's
identity (Prob. 1.61c), with . Supply the details.
Answer
It is proved as second uniqueness of theorem by using greens identity.
Over 30 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
Show that the average field inside a sphere of radius R, due to all the charge within the sphere, is
Where is the total dipole moment. There are several ways to prove this delightfully simple result. Here's one method:
(a) Show that the average field due to a single chargeqat point r inside thesphere is the same as the field at r due to a uniformly charged sphere with
, namely
Where r is the vector from r to
(b) The latter can be found from Gauss's law (see Prob. 2.12). Express the answerin terms of the dipole moment of q.
(c) Use the superposition principle to generalize to an arbitrary charge distribution.
(d) While you're at it, show that the average field over the volume of a sphere, dueto all the charges outside, is the same as the field they produce at the center.
a point charge located inside (same as above, in other words, only with ).(In this case, of course, Laplace's equation does not hold within the sphere.) Show that, in general,
role="math" localid="1657706668993"
where is the potential at the center due to all the external charges, and is the total enclosed charge.
For the dipole in Ex. 3.10, expand to order ,and use this
to determine the quadrupole and octo-pole terms in the potential.
Buckminsterfullerine is a molecule of 60 carbon atoms arranged
like the stitching on a soccer-ball. It may be approximated as a conducting spherical shell of radius . A nearby electron would be attracted, according to Prob. 3.9, so it is not surprising that the ion exists. (Imagine that the electron on average-smears itself out uniformly over the surface.) But how about a second electron? At large distances it would be repelled by the ion, obviously, but at a certain distance r (from the center), the net force is zero, and closer than this it would be attracted. So an electron with enough energy to get in that close should bind.
(a) Find r, in . [You'll have to do it numerically.]
(b) How much energy (in electron volts) would it take to push an electron in (from
infinity) to the point r? [Incidentally, the ion has been observed.]
Derivefrom the Rodrigues formula, and check that satisfies the angular equation (3.60) for . Check that and are orthogonal by explicit integration.
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App