# 2_7 - K 2 Ý J 2 2_J cos S Þà 3/2 or after expanding d Ý...

This preview shows pages 1–2. Sign up to view the full content.

PHY 5346 HW Set 3 Solutions – Kimel 5. 2.7 The system is described by a) The Green’s function, which vanishes on the surface is obviously G Ý x 3 , x 3 v Þ = 1 | x 3 ? x 3 v | ? 1 | x 3 ? x 3 I v | where x 3 v = x v î + y v c ! + z v k ! , x 3 I v = x v î + y v c ! ? z v k ! b) There is no free charge distribution, so the potential everywhere is determined by the potential on the surface. From Eq. (1.44) d Ý x 3 Þ = ? 1 4 ^ X S d Ý x 3 v Þ / G Ý x 3 , x 3 v Þ / n v da v Note that n ! v is in the ? z direction, so / / n v G Ý x 3 , x 3 v Þ | z v = 0 = ? / / z G Ý x 3 , x 3 v Þ | z v = 0 = ? 2 z Ý x ? x v Þ 2 + Ý y ? y v Þ 2 + z 2 3/2 So d Ý x 3 Þ = z 2 ^ V X 0 a X 0 2 ^ _ v d _ v d d v Ý x ? x v Þ 2 + Ý y ? y v Þ 2 + z 2 3/2 where x v = _ v cos d v , y v = _ v sin d v . c) If _ = 0, or equivalently x = y = 0, d Ý z Þ = z 2 ^ V X 0 a X 0 2 ^ _ v d _ v d d v ß _ v 2 + z 2 à 3/2 = zV X 0 a _ d _ ß _ 2 + z 2 à 3/2 d Ý z Þ = zV ? z ? Ý a 2 + z 2 Þ z Ý a 2 + z 2 Þ = V 1 ? z Ý a 2 + z 2 Þ d) d Ý x 3 Þ = z 2 ^ V X 0 a X 0 2 ^ _ v d _ v d d v Ý _ 3 ? _ 3 v Þ 2 + z 2 3/2

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

View Full Document
In the integration choose the x ? axis parallel to _ 3 , then _ 3 6 _ 3 v = __ v cos d v d Ý x 3 Þ = z 2 ^ V X 0 a X 0 2 ^ _ v d _ v d d v ß _ 2 + _ v 2 ? 2 _ 3 6 _ 3 v + z 2 à 3/2 Let r 2 = _ 2 + z 2 , so d Ý x 3 Þ = z 2 ^ V r 3 X 0 a X 0 2 ^ _ v d _ v d d v 1 + _ v 2 ? 2 _ 3 6 _ 3 v r 2 3/2 We expand the denominator up to factors of O Ý 1/ r 4 Þ , ( and change notation d v S , _ v J , r 2 1 K 2 Þ d Ý x 3 Þ = z 2 ^ V r 3 X 0 a X 0 2 ^ J d J d S 1 + J 2 ? 2 _ 3 6 J 3 r 2 3/2 where the denominator in this notation is written 1 ß 1 +
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: K 2 Ý J 2 ? 2 _J cos S Þà 3/2 or, after expanding, d Ý x 3 Þ = z 2 ^ V r 3 X a J d J X 2 ^ 1 ? 3 2 K 2 J 2 + 3 K 2 _J cos S + 15 8 K 4 J 4 ? 15 2 K 4 J 3 _ cos S + 15 2 K 4 _ 2 J 2 cos 2 S d S Integrating over S gives d Ý x 3 Þ = z 2 ^ V r 3 X a J 2 ^ + 15 4 K 4 J 4 ^ ? 3 K 2 J 2 ^ + 15 2 K 4 _ 2 J 2 ^ d J Integrating over J yields d Ý x 3 Þ = z 2 ^ V r 3 5 8 K 4 ^ a 6 ? 3 4 a 4 K 2 ^ + 15 8 a 4 K 4 _ 2 ^ + ^ a 2 or d Ý x 3 Þ = Va 2 2 Ý _ 2 + z 2 Þ 3/2 1 ? 3 4 a 2 Ý _ 2 + z 2 Þ + 5 8 a 4 + 3 _ 2 a 2 Ý _ 2 + z 2 Þ 2...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

2_7 - K 2 Ý J 2 2_J cos S Þà 3/2 or after expanding d Ý...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online