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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 100

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online