Chapter3

# Chapter3 - Chapter 3 Boundary Value Problems, Introduction...

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Chapter 3 Boundary Value Problems, Introduction 3.1 The method of images There are some problems in electrostatics for which a solution can be obtained by adding ‘image’ charges outside the region of interest. The most obvious one is the potential of a charge above a conducting, in f nite plane. This is solved by noting that the plane which is the perpendicular bisector of the line between charges of opposite sign but the same magnitude is an equipotential. In fact if the potential at in f nity is zero, the potential of the plane will equal the potential at an in f nite distance form the charge. 3.1.1 The potential due to a charge above an in f nite plane Let the plane in this system lie at z =0 and let the charge q lie at ( x 0 ,y 0 ,z 0 ) . The potential in the half space z> 0 satis f es 2 φ ( r )= 4 πk 1 ρ ( r 0 4 1 (3) ( r r 0 ) (3.1) = 4 πqδ (3) ( r r 0 ) .k 1 =1 in Gaussian units = q ± 0 δ (3) ( r r 0 ) k 1 = 1 4 π± 0 in SI units The general solution to Poisson’s equation in the upper half plane is φ ( r k 1 ZZZ ρ ( r 0 ) | r r 0 | d 3 x 0 + E 0 · r + V 0 + V ( r ) (3.2) = qk 1 q ( x x 0 ) 2 +( y y 0 ) 2 z z 0 ) 2 (3.2b) + E 0 · r + V 0 + V ( r ) where E 0 · r , V 0 and V ( r ) represent terms, F i ( r ) , which satisfy 2 F i ( r )=0 for 0 and which are included only if they satisfy the boundary conditions. The V ( r ) is the potential of any charge distribution which would lie in the region z< 0 . Note that Eq. 3.1 holds only for 0 . If the potential on the plane is zero relative to a point at in f nity then φ ( r φ 1 ( r ) with φ 1 ( r 1 q ( x x 0 ) 2 y y 0 ) 2 z z 0 ) 2 (3.3) 1 q ( x x 0 ) 2 y y 0 ) 2 z + z 0 ) 2 Here the second term is V ( r ) for a point charge q at z = z 0 . φ 1 ( r ) is the potential of a ‘dipole’ with charge separation 2 z 0 . The requirement that the normal component of the electric f eldvanishat z provides another special solution for the problem with charge a q in the upper half plane. In this case φ ( r φ 2 ( r ) : φ 2 ( r 1 q ( x x 0 ) 2 y y 0 ) 2 z z 0 ) 2 (3.4) + 1 q ( x x 0 ) 2 y y 0 ) 2 z + z 0 ) 2 . 56

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lim z > 0 [ ˆ z · φ 2 ( r )] = qk 1 [( x x 0 ) 2 +( y y 0 ) 2 + z 2 0 ] 3 / 2 lim z > 0 [( z z 0 )+( z + z 0 )] = 0 (3.5) Note that the normal component of the electric f eld, E 1 = φ 1 ( r ) is not zero at z =0 . lim z > 0 [ ˆ z · φ 1 ( r )] = 1 [( x x 0 ) 2 y y 0 ) 2 + z 2 0 ] 3 / 2 lim z > 0 [( z z 0 ) ( z + z 0 )] (3.6) = 2 1 z 0 [( x x 0 ) 2 y y 0 ) 2 + z 2 0 ] 3 / 2 = q 1 z 0 2 π± 0 [( x x 0 ) 2 y y 0 ) 2 + z 2 0 ] 3 / 2 But the tangential component of E 1 = φ 1 ( r ) vanishes at z . lim z > 0 [ ˆ z × φ 1 ( r )] = 1 [( x x 0 ) 2 y y 0 ) 2 + z 2 0 ] 3 / 2 lim z > 0 ˆ z × [ ¡ r r + 0 ¢ ¡ r r 0 ¢ ] (3.7) = 1 [( x x 0 ) 2 y y 0 ) 2 + z 2 0 ] 3 / 2 [ ˆy [( x x 0 ) ( x x 0 )] ˆx [( y y 0 ) ( y y 0 )] 3.1.2 Attraction between a charged particle and a conducting plane The tangential component of the electric f eld at the surface of a conductor is zero. It follows that φ 1 ( r ) provides a solution for the potential of the isolated charge - plane system. The electric f eld in the z> 0 half-plane is the sum of the electric f eld of the point charge and the electric f eld of the surface charge on the plane. The surface charge on the plane satis f es σ 1 ( x, y )= ± 0 [
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## This note was uploaded on 12/19/2010 for the course PHYS 411 taught by Professor G during the Spring '10 term at Missouri S&T.

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Chapter3 - Chapter 3 Boundary Value Problems, Introduction...

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