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Electromagnetism

by: Miss Lucienne Hamill

24

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2

Electromagnetism PHY 712

Miss Lucienne Hamill
WFU
GPA 3.91

Natalie Holzwarth

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Natalie Holzwarth
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KARMA
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This 2 page Class Notes was uploaded by Miss Lucienne Hamill on Wednesday October 28, 2015. The Class Notes belongs to PHY 712 at Wake Forest University taught by Natalie Holzwarth in Fall. Since its upload, it has received 24 views. For similar materials see /class/230730/phy-712-wake-forest-university in Physics 2 at Wake Forest University.

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Date Created: 10/28/15
January 22 2002 Notes for Lecture 2 Examples of solutions of the onedimensional Poisson equation Consider the following one dimensional charge distribution 0 formlt a pu for altalt0 pu for 0 lt lt a 1 for gt a We want to nd the electrostatic potential such that 12 p 12 l E l 2 J 1 with the boundary condition oo 0 The solution to the differential equation is given by U for lt a PittHIV for 1ltFIIltU 25 12 pua2su for 0 lt lt a 39 7311 for gt a The electrostatic eld is given by U for lt a E for altatlt0 4 for0ltarlta U formgta The electrostatic potential can be determined by piecewise solution within each of the four regions or by use of the Green s function Cm 47rarlt where 1 47mg ltIgtltargt ammar39wmt a In the expression for Garar arlt should be taken the smaller of and ar It can be shown that Eq 5 gives the identical result for given in Eq 3 Notes on the onedimensional Green s functions The Green s function for the Poisson equation can be de ned a solution to the equation V2Gat 47r6a7 6 Here the factor of 47r is not really necessary but ensures consistency with your text s treat ment of the 3dimensional case The meaning of this expression is that 37 is held xed while taking the derivative with respect to It is easily shown that with this de nition of the Green s function 6 Eq 5 nds the electrostatic potential for an arbitrary charge density In order to nd the Green s function which satis es Eq 6 we notice that we can use two independent solutions to the homogeneous equation Wad o 739 where i 1 or 2 to form 4 GW39 73 1ltmltgt 2ltxgtgt 8 This notation means that a7lt should be taken the smaller of and 37 and a7gt should be taken the larger In this expression IV is the VVronskian 03 2 IV E 96 1at We can check that this recipe works by noting that for 35 37 Eq 8 satis es the de ning equation 6 by virtue of the fact that it is equal to a product of solutions to the homogeneous equation 7 The de ning equation is singular at ar but integrating 6 over in the neighborhood of 37 6 lt lt 37 6 gives the result dGm dGm I 7 1 l b 6 l e 4 0 In our present case we can choose 1at and 12 1 so that W 1 and the Green s function is given above For this piecewise continuous form of the Green s function the integration 5 can be evaluated 1 47161 cm Gm m pm dm x Ga m pm dm 11 which becomes 1 at x j m39pm dat pat da 12 cu 730 L Evaluating this expression we nd that we obtain the same result given in Eq In general the Green s function Cm solution 5 depends upon the boundary conditions of the problem well as on the charge density In this example the solution is valid for all neutral charge densities that is 13 0

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