Homework_5

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

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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 0 0 1 2 2 0 0 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 0 1 cos . 2 cos l l l l z P R R z Rz θ θ + = = + - Hence ( ) ( ) ( ) 1 0 0 0 1 0 0 0 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 0 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 + - + - + + + =

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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 0 0 cos 1 cos . I P x dx α α = = - The potential on the symmetry axis is ( ) ( ) ( ) ( ) 1 1 1 1 0 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 0 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 0 0 1 1 1 1 0 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 0 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 0 1 1 1 1 1 0 cos cos cos , 8 2 1 1 cos cos cos sin . 8 2 1 l r l l l l l l l l l l l Q r l E P P P R l Q r E P P P R l θ α α θ πε α α θ θ πε - - + + = - - + + = = - + = - - + At the origin, only the l = 1 term is non- zero, and so ( ) ( ) 2 2 0 2 2 0 1 1 cos cos , 24 1 1 cos sin . 24 r Q E P R Q E P R θ α θ πε α θ πε = - = - - The component parallel to the symmetry axis is ( ) 2 2 0 1 cos sin 1 cos . 24 z r Q E E E P R θ θ θ α πε = - = - The orthogonal component is identically zero, and we see that the field at the origin is parallel to the symmetry axis.

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