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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 and , 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) .