Jackson's Electrodynamics 3.2 Solutions
Homer Reid's Solutions to Jackson Problems: Chapter 3 2 and at r = b , V Z 0 - 1 P l ( x ) dx = 2 2 l + 1 h A l b l + B l b - ( l +1) i . The integral from 0 to 1 vanishes for l even, and is given in the text for l odd: Z 1 0 P l ( x ) dx = ( - 1 2 ) ( l - 1) / 2 ( l - 2)!! 2 ( l +1 2 ) ! . The integral from -1 to 0 also vanishes for l even, and is just the above result inverted for l odd. This gives V ( - 1 2 ) ( l - 1) / 2 ( l - 2)!! 2 ( l +1 2 ) ! = 2 2 l + 1 h A l a l + B l a - ( l +1) i - V ( - 1 2 ) ( l - 1) / 2 ( l - 2)!! 2 ( l +1 2 ) ! = 2 2 l + 1 h A l b l + B l b - ( l +1) i . or α l = A l a l + B l a - ( l +1) - α l = A l b l + B l b - ( l +1) with α l = V ( - 1 2 ) a ( l - 1) / 2 (2 l + 1)( l - 2)!! 4 ( l +1 2 ) ! . The solution is A l = α l b l +1 + a l +1 a 2 l +1 - b 2 l +1 B l = - α l a l +1 b l +1 ( b l + a l ) a 2 l +1 - b 2 l +1 The first few terms of (1) are Φ( r, θ ) = 3 4 V ( a 2 + b 2 ) r a 3 - b 3 - a 2 b 2 ( a + b ) r 2 ( a 3 - b 3 ) P 1 (cos θ ) - 7 16 ( a 4 + b 4 ) r 3 a 7 - b 7 - a 4 b 4 ( a 3 + b 3 ) r 4 ( a 7 - b 7 ) P 3 (cos θ )+ · · · In the limit as b → ∞ , the problem reduces to the exterior problem treated in Section 2.7 of the text. In that limit, the above expression becomes Φ( r, θ ) → 3 4 V a r 2 P 1 (cos θ ) - 7 16 V a r 4 P 3 (cos θ ) + · · · in agreement with (2.27) with half the potential spacing. When a → 0, the problem goes over to the interior version of the same problem, as treated in section 3.3 of the text. In that limit the above expression goes to Φ( r, θ ) → - 3 4 V r b P 1 (cos θ ) + 7 16 V r b 3 P 3 (cos θ ) + · · · This agrees with equation (3.36) in the text, with the sign of V flipped, because here the more positive potential is on the lower hemisphere.
Jackson's Electrodynamics 3.2 Solutions
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