F10HW08 - 1 Problem 4.9 1.1 " " q...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 Problem 4.9 1.1 " " q d a Figure 1: Setup for problem 4.9 Using the fact that we have azimuthal symmetry, we have inside the sphere: in ( ~ r ) = 1 X l =0 A l r l P l (cos ) (1) And outside the sphere: out ( ~ r ) = q + 1 X l =0 B l r ( l +1) P l (cos ) where q is the potential due to the charge q . q = q 4 " 1 j ~ r ~ r j = q 4 " 1 X l =0 r l < r l +1 > P l (cos ) Because r only points to the single point charge along the z-axis, = . Therefore: out ( ~ r ) = 1 X l =0 q 4 " r l < r l +1 > + B l r ( l +1) P l (cos ) (2) Now, we need to apply the following boundary conditions: 1 a @ in @ r = a = 1 a @ out @ r = a (3) " @ in @r r = a = " @ out @r r = a (4) 1 Applying eqation (3) yields: 1 a 1 X l =0 A l r l _ P l (cos )( sin ) r = a = 1 a 1 X l =0 q 4 " r l < r l +1 > + B l r ( l +1) _ P l (cos )( sin ) r = a 1 X l =0 A l a l = 1 X l =0 q 4 " a l d l +1 + B l a ( l +1) A l = q 4 " d ( l +1) + B l a (2 l +1) (5) Applying eqation (4) yields: " 1 X l =0 A l lr l 1 P l (cos ) r = a = " 1 X l =0 q 4 " lr l 1 d l +1 + B l ( l 1) r ( l +2) P l (cos ) r = a " 1 X l =0...
View Full Document

This note was uploaded on 12/08/2011 for the course PHYSICS 505 taught by Professor Liu during the Winter '08 term at University of Michigan.

Page1 / 7

F10HW08 - 1 Problem 4.9 1.1 " " q...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online