HW06sol - PHY 505 Classical Electrodynamics I Fall 2010...

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1 C:\User\Teaching\505-506\F10\HW06sol.doc PHY 505 Classical Electrodynamics I Fall 2010 Homework 6 with solutions Problem 6.1 (to be graded of 10 points). An electric dipole is located above an infinite conducting plane. Calculate: (i) the distribution of the induced charge in the conductor, (ii) the force and the torque acting on the dipole, and (iii) the dipole-to-plane interaction energy. Solutions : (i) The problem may be solved by the introduction of a dipole image, at the same distance d below the plane, and with the same dipole moment magnitude p as the original dipole, but reflected in the vertical plane perpendicular to that containing the dipole moment vector (see Fig. on the right). 1 Let us prove that. The net field of these two dipoles evidently satisfies the Poisson equation in the upper half-space, so that the only thing we have to prove is that it also satisfies the boundary condition ( = 0) on the plane surface. Let us use Eq. (3.7) of the lecture notes for of a system of several dipoles – in our case, of two dipoles (let us call them p ’ and p ), with Cartesian components cos , 0 , sin ' ' ' ' ' ' ' ' ' p p p p p p p p z z y y x x , located at points d z'' z' y'' y' x'' x' , 0 , 0. (Here x is the coordinate within the vertical plane which contains vectors p and p , i.e. in the plane of our drawing, while axis y is perpendicular to the plane.) In these coordinates, Eq. (3.7) yields  . ) ( sin cos ) ( ) ( sin cos ) ( 4 ) ( ) ( 4 1 2 / 3 2 2 2 2 / 3 2 2 2 0 3 3 0 d z y x x d z d z y x x d z p " " " ' ' '  r - r p r r r - r p r r This equation shows that potential vanishes at an arbitrary point of the surface ( z = 0), thus proving our guess. Now the induced charge may be calculated as 0 0 z z , giving 1 The simplest way to understand this fact is to present the dipole in the form of two point charges (+ q ) and (- q ), slightly displaced along the direction of the dipole moment vector, and to construct the dipole image from the mirror images of these point charges in the conducting plane. However, this approach, based on a particular implementation of a dipole, and can only be used for a guess , not as a proof . d ' p '' p d x z 0
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2 C:\User\Teaching\505-506\F10\HW06sol.doc . ) ( sin 3 cos ) 2 ( 2 2 / 5 2 2 2 2 2 2 d y x dx y x d p (iii) Now, we can use Eqs. (3.16) to calculate the potential energy of interaction force between the real and imaginary dipoles (e.g., between the dipole and the plane). However, we should not forget that the image has been created by the dipole itself. Hence, following the reasoning of Sec. 1.3, we should multiply the right-hand part of Eq. (3.16) by ½: ) ( 2 1 int ' '' ' U r E p , ( * ) where E
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This note was uploaded on 09/10/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.

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HW06sol - PHY 505 Classical Electrodynamics I Fall 2010...

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