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Unformatted text preview: Physics 505 Fall 2007 Homework Assignment #3 — Solutions Textbook problems: Ch. 2: 2.14, 2.15, 2.22, 2.23 2.14 A variant of the preceeding twodimensional problem is a long hollow conducting cylinder of radius b that is divided into equal quarters, alternate segments being held at potential + V and V . a ) Solve by means of the series solution (2.71) and show that the potential inside the cylinder is Φ( ρ,φ ) = 4 V π ∞ X n =0 ρ b 4 n +2 sin[(4 n + 2) φ ] 2 n + 1 The general series solution for the twodimensional problem in polar coordinates is given by (2.71) Φ( ρ,φ ) = a + b log ρ + ∞ X n =1 a n ρ n sin( nφ + α n ) + b n ρ n sin( nφ + β n ) Since we are interested in the interior solution, we demand that the potential remains finite at ρ = 0. This indicates that the b n coefficients must all vanish. We are thus left with Φ( ρ,φ ) = a + ∞ X n =1 a n ρ n sin( nφ + α n ) which we may choose to rewrite as Φ( ρ,φ ) = A 2 + ∞ X k =1 [ A k ρ k cos( kφ ) + B k ρ k sin( kφ )] (1) This form of the series is supposed to be reminiscent of a Fourier series. The boundary condition for this problem is that the potential at ρ = b is either + V or V , depending on which quadrant we are in V V V V b This can be plotted as a function of φ Φπ π V V ϕ ) , ϕ b ( It should be obvious that Φ( b,φ ) is an odd function of φ . As a result, we im mediately deduce that the A k Fourier coefficients in (1) must vanish, leaving us with Φ( ρ,φ ) = ∞ X k =1 B k ρ k sin( kφ ) (2) On the interior surface of the conducting cylinder, this reads Φ( b,φ ) = ∞ X k =1 B k b k sin( kφ ) where Φ( b,φ ) is given by the figure above. In particular, we see that the quantities B k b k are explicitly the Fourier expansion coefficients of a square wave with period π (which is half the usual 2 π period). As a result, we may simply look up the standard Fourier expansion of the square wave and map it to this present problem. Alternatively, it is straightforward to calculate the coefficients directly B k b k = 1 π Z π π Φ( b,φ )sin( kφ ) dφ = V π Z π/ 2 π Z π/ 2 + Z π/ 2 Z π π/ 2 ! sin( kφ ) dφ = V kπ cos( kφ ) π/ 2 π + cos( kφ ) π/ 2 cos( kφ ) π/ 2 cos( kφ ) π π/ 2 = 2 V kπ 1 2cos kπ 2 + cos( kπ ) = 8 V kπ k = 2 , 6 , 10 , 14 ,... ( ie k = 4 n + 2) Substituting B k = 8 V/kπb k into (2) and usiing k = 4 n + 2 then gives Φ( ρ,φ ) = 4 V π ∞ X n =0 ρ b 4 n +2 sin[(4 n + 2) φ ] 2 n + 1 (3) b ) Sum the series and show that Φ( ρ,φ ) = 2 V π tan 1 2 ρ 2 b 2 sin2 φ b 4 ρ 4 This series is easy to sum if we work with complex variables. Since sin θ is the imaginary part of e iθ , we write (3) as Φ( ρ,φ ) = 4 V π = ∞ X n =0 ( ρ/b ) 4 n +2 e (4 n +2) iφ 2 n + 1 = 4 V π = ∞ X n =0 z 2 n +1 2 n + 1 = 4 V π = ( z + 1 3 z 3 + 1 5 z 5 + ··· ) (4) where z ≡ ρ 2 b 2 e 2 iφ (5) Now recall that the Taylor series expansion for log(1 + z ) is given by log(1 + z ) = ∞...
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This note was uploaded on 11/21/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Stephens,P
 Magnetism, Work

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