Mathematical Methods for Physics and Engineering - Matematica.NET

21.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONSexample, in solving Poisson’s equation in two dimensions in the half-space x>0we again require just one image charge, of strength q 1 = −1, at a position r 1 thatis the reflection of r 0 in the line x = 0. Since we require G(r, r 0 )=0whenr lieson x = 0, the constant in (21.93) must equal zero, and so the Dirichlet Green’sfunction isG(r, r 0 )= 1 (ln |r − r0 |−ln |r − r 1 | ) .2πClearly G(r, r 0 ) tends to zero as |r| →∞. If, however, we wish to solve the twodimensionalPoisson equation in the quarter space x>0, y>0, then more imagepoints are required.◮A line charge in the z-direction of charge density λ is placed at some position r 0 in thequarter-space x>0, y>0. Calculate the force per unit length on the line charge due tothe presence of thin earthed plates along x =0and y =0.Here we wish to solve Poisson’s equation,∇ 2 u = − λ ɛ 0δ(r − r 0 ),in the quarter space x>0, y>0. It is clear that we require three image line chargeswith positions and strengths as shown in figure 21.13 (all of which lie outside the regionin which we seek a solution). The boundary condition that the electrostatic potential u iszero on x =0andy = 0 (shown as the ‘curve’ C in figure 21.13) is then automaticallysatisfied, and so this system of image charges is directly equivalent to the original situationof a single line charge in the presence of the earthed plates along x =0andy = 0. Thusthe electrostatic potential is simply equal to the Dirichlet Green’s functionu(r) =G(r, r 0 )=−λ (ln |r − r0 |−ln |r − r 1 | +ln|r − r 2 |−ln |r − r 3 | ) ,2πɛ 0which equals zero on C and on the ‘surface’ at infinity.The force on the line charge at r 0 , therefore, is simply that due to the three line chargesat r 1 , r 2 and r 3 . The elecrostatic potential due to a line charge at r i , i = 1, 2 or 3, is givenby the fundamental solutionu i (r) =∓λ ln |r − r i | + c,2πɛ 0the upper or lower sign being taken according to whether the line charge is positive ornegative, respectively. Therefore the force per unit length on the line charge at r 0 , due tothe one at r i ,isgivenby−λ∇u i (r)∣ = ± λ2 r 0 − r ir=r02πɛ 0 |r 0 − r i | . 2Adding the contributions from the three image charges shown in figure 21.13, the totalforce experienced by the line charge at r 0 is given by(F =λ2− r 0 − r 12πɛ 0 |r 0 − r 1 | + r 0 − r 22 |r 0 − r 2 | − r )0 − r 3,2 |r 0 − r 3 | 2where, from the figure, r 0 − r 1 =2y 0 j, r 0 − r 2 =2x 0 i +2y 0 j and r 0 − r 3 =2x 0 i. Thus, in761