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Unformatted text preview: PHY501 Lecture IIPart 1 Boundary Value Problems V1.3 Boundary Value Problems Statement of the Problem: Find the solution to Poissons (Laplaces) Equation 2 = ( x ) (1) in a volume V bounded by a surface (or surfaces) S. The boundary conditions have two basic forms. Dirichlet Boundary Conditions (DBC): Neumann Boundary Conditions (NBC): ( x ) specified for x on S (fixed voltages on surfaces) ( x ) n S specified for x on S (fixed charges on surfaces) Note that one cant in general arbitrarily specify both DBCs and NBCs at the same time, since this would overdetermine the solution to Eq. 1. Note also that the solution to Eq. 1 for a given set of boundary conditions is unique (to within an overall added constant for NBC). Proof of this assertion can be found in J1.9 Method of Images In situations possessing a high degree of symmetry, it is sometimes possible to deduce the form of by introducing f ctitious image charges, which, when acting in concert with the real charges of the problem, result in potentials that satisfy the BCs. = Thus A point charge +q at z=+d is placed above a conducting plane held at ground potential . = 0 d d x x q y x z +q ( x ) DBC : Intuitively we know that by placing q at z=d we can produce a potential such that . ( x , y ,0) = ( x ) = q 4 1 x x ' 1 x x '' where x ' = + d z x '' = d z (2) R = r 2 + d 2 2 rd cos R ' = r 2 + d 2 + 2 rd cos The resulting potential is easily expressed in spherical polar coordinates, using the Law of Cosines r d d z x +q q r sin d r cos R R Method of Images Thus Thus z = 0 = 2 and r , 2 ( ) = Note that for ( r , ) = q 4 1 R 1 R ' = q 4 1 r 2 + d 2 2 rd cos 1 r 2 + d 2 + 2 rd cos (3) E z = z = + 1 r 2 = ( x , y ) (4) The charge induced on the conducting plane is given by = r z = 2 = q 4 r rd sin r 2 + d 2 2 rd cos ( ) 3 / 2 rd sin r 2 + d 2 + 2 rd cos ( ) 3 / 2 = 2 = q 2 d r 2 + d 2 ( ) 3 / 2 (4) r/d (r) 1 3 2 4 2 1 3 Method of Images The total charge induced on the conducting plane...
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 '11
 MARLOW
 Magnetism

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