Solution of Griffiths Problem 3.26

Solution of Griffiths Problem 3.26
D. Sober
Revised 9 February 2011
Sphere of radius R with charge density
To find: an approximate expression for the potential at
points on the z-axis, far away from the sphere
Note: For a field point on z-axis,
, so 2N = (angle of
relative to z axis) = 2. [In the general case, 2N in (3.96)
is the angle between
, which is more complicated.]
Strategy: Evaluate the integrals of Eq. (3.96) to find the
lowest multipole terms in the expansion of V(r), and identify the first non-zero contribution.
Monopole term:
(dJN = a ring element of radius rN sin 2N, width rN d2N and thickness drN )
Dipole term:
Calculating the n = 1 term of (3.95-3.96) and letting r 6 z,
Note that the 2 integral must vanish because the integrand is odd about B/2.
We can also show that the dipole term vanishes by calculating the dipole moment, which
involves the same integrals because
Quadrupole term:
(letting r 6 z, 2N6 2, rN6r in (3.96) , and using
[Schaum 18.30])
The quadrupole term is the lowest-order non-vanishing term, so, for large z (z o R), V(z) . V2 (z) .