F10HW03 - Problem 2.7 gonsider potentil prolem in the...

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Problem 2.7 Consider a potential problem in the half-space de ned by z 0, with Dirichlet boundary conditions on the plane z = 0 (and at in nity). 2.7.a. Write down the appropriate Green function G ( ~x; ~x 0 ) . G D ( 0 ) = 1 p ( x 1   x 0 1 ) 2 + ( x 2   x 0 2 ) 2 + ( x 3   x 0 3 ) 2   1 p ( x 1   x 0 1 ) 2 + ( x 2   x 0 2 ) 2 + ( x 3 + x 0 3 ) 2 where x 1 , x 2 , and x 3 denote the x , y , and z coordinates, respectively. 2.7.b. If the potential on the plane z = 0 is speci ed to be ¨ = V inside a circle of radius a centered at the origin, and ¨ = 0 outside that circle, nd an integral expression for the potential at the point P speci ed in terms of cylindrical coordinates ( ; '; z ). We're going to use equation 1.44 from Jackson: ¨( ~x ) =   1 4 I S ¨( 0 ) @G D @n 0 da 0 (1) Note that ¨( 0 ) = V inside the circle of radius a centered at the origin. Let's convert G D ( 0 ) to cylindrical coordinates: G D ( 0 ) = 1 p ( cos '   0 cos ' 0 ) 2 + ( sin '   0 sin ' 0 ) 2 + ( z   z 0 ) 2   1 p ( cos '   0 cos ' 0 ) 2 + ( sin '   0 sin ' 0 ) 2 + ( z + z 0 ) 2 = 1 s 2 + 0 2   0 (cos ' cos ' 0 + sin ' sin ' 0 | {z } cos( '   ' 0 ) ) + ( z   z 0 ) 2   1 s 2 + 0 2   0 (cos ' cos ' 0 + sin ' sin ' 0 | {z } cos( '   ' 0 ) ) + ( z + z 0 ) 2 = 1 p 2 + 0 2   0 cos( '   ' 0 ) + ( z   z 0 ) 2   1 p 2 + 0 2   0 cos( '   ' 0 ) + ( z + z 0 ) 2 1
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We need to nd the normal derivative of G D . Note that the normal points away from the region of interest { since we're considering z 0, let ^ n =   ^ z 0 : @G D @n 0 = r G D ( ~x; ~x 0 ) ¡ (   ^ z 0 ) =   2 6 4 ¨ ¨ ¨ ¨ ¨ ¨ ¨¨* 0 1 0 @ @ 0 ( 0 G D ) + > 0 1 0 D @' 0 + D @z 0 z 0 =0 3 7 5 =   "   1 2 2( z   z 0 )(   1) ( 2 + 0 2   0 cos( '   ' 0 ) + ( z   z 0 ) 2 ) 3 = 2 + 1 2 2( z + z 0 ) ( 2 + 0 2   0 cos( '   ' 0 ) + ( z + z 0 ) 2 ) 3 = 2 # z 0 =0 =   z ( 2 + 0 2   0 cos( '   ' 0 ) + z 2 ) 3 = 2   z ( 2 + 0 2   0 cos( '   ' 0 ) + z 2 ) 3 = 2 =   2 z ( 2 + 0 2   0 cos( '   ' 0 ) + z 2 ) 3 = 2 Note that the terms which \cancel to zero" do so because the derivatives of the two terms of G D sum to zero when evaluated at z 0 = 0. Now, plug this into equation (1). Note that we only need to integrate over the circle which has potential V because the integrand is zero elsewhere. ¨( ~x ) =   1 4 Z 2 ' 0 =0 Z a 0 =0 ( V )   2 z ( 2 + 0 2   0 cos( '   ' 0 ) + z 2 ) 3 = 2 ! 0 d 0 d' 0 = V z 2 Z 2 ' 0 =0 Z a 0 =0 0 ( 2 + 0 2   0 cos( '   ' 0 ) + z 2 ) 3 = 2 d 0 d' 0 2.7.c. Show that, along the axis of the circle ( = 0 ), the potential is ¨ = V 1   z p a 2 + z 2 . Letting = 0: ¨( ) = V z 2 Z 2 ' 0 =0 Z a 0 =0 0 ( 0 2 + z 2 ) 3 = 2 d 0 d' 0 2
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Using the substitution u = 0 2 + z 2 and du = 2 0 : ¨( ~x ) = V z 2 Z 2 ' 0 =0 " Z a 2 + z 2 u = z 2 1 2 du u 3 = 2 # d' 0 = V z 2 Z 2 ' 0 =0 (   2) 1 2 u 1 = 2 a 2 + z 2 u = z 2 d' 0 =   V z 2 Z 2 ' 0 =0 1 p a 2 + z 2   1 p z 2 d' 0 =   V z 2 2 1 p a 2 + z 2   1 z = V 1   z p a 2 + z 2 2.7.d.
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F10HW03 - Problem 2.7 gonsider potentil prolem in the...

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