F07_hw03a - Physics 505 Fall 2007 Homework Assignment#3 —...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 two-dimensional 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 two-dimensional 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 Φ-π 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 ) = ∞...
View Full Document

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.

Page1 / 13

F07_hw03a - Physics 505 Fall 2007 Homework Assignment#3 —...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online