Homework_5 - Homework 5 solutions 3.2 First consider the...

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Unformatted text preview: Homework 5 solutions 3.2 First consider the related problem of the potential due to a spherical cap with charge density ( ) 2 4 . Q R π The potential on the symmetry axis is given by the integral ( ) ( ) ( ) 2 1 2 2 2 1 2 2 1 2 sin 4 4 2 cos sin . 8 2 cos Q R d z R R z Rz Q d R z Rz α α π θ θ πε π θ θ θ πε θ Φ = +- = +- ∫ ∫ The integrand can be written in terms of Legendre polynomials by noting that ( ) ( ) 1 2 1 2 1 cos . 2 cos l l l l z P R R z Rz θ θ ∞ + = = +- ∑ Hence ( ) ( ) ( ) 1 1 cos sin 8 cos sin . 8 l l l l l l l l Q z z P d R Q z P d R α α θ θ θ πε θ θ θ πε ∞ + = ∞ + = Φ = = ∑ ∫ ∑ ∫ The integral ( ) ( ) 1 cos cos sin l l l I P d P x dx α α θ θ θ = = ∫ ∫ can be evaluated by using the recurrence relations ( ) ( ) 1 1 1 2 1 0, l l l l P l xP lP +- +- + + = and ( ) 1 1 0. l l l dP dP x l P dx dx +-- + = Differentiating the first recurrence relation gives ( ) ( ) 1 1 1 2 1 0. l l l l dP dP dP l l P x l dx dx dx +- +- + + + = Eliminating the l dP dx from these last two equations, we obtain ( ) 1 1 2 1 . l l l dP dP l P dx dx +- + =- This is also given in Jackson equation 3.28. Hence for 1, l ≥ , ( ) ( ) ( ) ( ) 1 1 1 1 1 cos 1 1 cos cos . 2 1 2 1 l l l l l I P x P x P P l l α α α +-- + =- =- + + For l = 0, we find ( ) 1 cos 1 cos . I P x dx α α = = - ∫ The potential on the symmetry axis is ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 cos cos cos . 8 2 1 l l l l l Q z z P P R R l α α α πε ∞- + + = Φ =- +- + ∑ From this we find that the potential for r R < is ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 , 1 cos cos cos cos . 8 2 1 l l l l l l Q r r P P P R R l θ α α α θ πε ∞- + + = Φ =- +- + ∑ (a) To find the potential for this geometry, we need to subtract the above potential from the uniform potential of a complete shell. Thus for , r R < ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 1 1 , 1 cos cos cos cos 4 8 2 1 1 1 1 cos cos cos cos . 8 2 1 l a l l l l l l l l l l l Q Q r r P P P R R R l Q r P P P R R l θ α α α θ πε πε α α α θ πε ∞- + + = ∞- + + = Φ =-- +- + = +-- + ∑ ∑ The ‘exterior’ potential can be found by making use of continuity at . r R = Hence for , r R > ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 , 1 cos cos cos cos . 8 2 1 l a l l l l l Q R r P P P r r l θ α α α θ πε ∞- + + = Φ = +-- + ∑ (b) The electric field, for , r R < has components ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 1 cos cos cos , 8 2 1 1 cos cos cos sin ....
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This note was uploaded on 12/02/2011 for the course PHYS 809 taught by Professor Macdonald during the Fall '11 term at University of Delaware.

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Homework_5 - Homework 5 solutions 3.2 First consider the...

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