Jckson.7 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: Consider potentil problem in the hlf-spce defined by, with Dirichlet boundry conditions on the plne = (nd t infinity). () Write down the pproprite Green function G(x, x') (b) If the potentil on the plne = is specified to be Φ = V inside circle of rdius centered on the origin, nd Φ = outside tht circle, find n integrl expression for the potentil t the point P specified in terms of cylindricl coordintes,,. (c) Show tht, long the xis of the circle (ρ = ), the potentil is given by =V (d) Show tht t lrge distnces (ρ + >> ) the potentil cn be expnded in power series in (ρ + ) -, nd tht the leding terms re = V [ 3 3 / 4 53 4 8...] Verify tht the results of prts c nd d re consistent with ech other in their common rnge of vlidity. SOLUTION: () The Green function solution to the potentil problem with this boundry for Dirichlet boundry conditions requires tht G D = on the surfce nd tht G D obeys =G x,x'= F x,x' x x' everywhere else. The problem of finding the Green function mounts to finding the proper function F bove so tht the Green function G disppers on the boundry. This problem is exctly equivlent to the sitution where we hve unit point chrge t x', which cretes the potentil / x-x', in the presence of flt conductor running long the = plne. The solution for the potentil of the equivlent problem will then be identicl to the Green function for the originl problem. To solve the problem of the chrge q t x' in the presence of conducting sheet t the = plne, we use the method of imges. We plce n imge chrge q' t (x', y', -') so tht the potentil is just the sum of the two point chrges:
= q 4 x x' y y' ' q' 4 x x' y y ' ' Apply the boundry condition of conductor t the = plne: x, y, == = 4 q '= q q x x' y y ' ' 4 q' x x' y y ' ' The imge chrge is just the perfect mirror imge in loction nd in chrge, which mkes sense becuse perfectly conducting surfce tht is perfectly flt is just perfect mirror. The potentil is then: = q 4 [ x x ' y y ' ' x x' y y ' ' ] Now go bck to the originl problem without the conductor nd point, but some potentilly complex chrge distribution in the presence of some potentilly complex boundry condition long the = plne. The solution to this originl problem is, ccording to the Green function method: x= 4 x ' G D d 3 x' 4 d G D d n' d ' where we now know the Green function used by this eqution. It is just the solution to the equivlent problem with unit chrge (q = 4πε ): G D x,x'= x x' y y ' ' x x' y y' ' (b) If the potentil on the plne = is specified to be Φ = V inside circle of rdius centered on the origin, nd Φ = outside tht circle, find n integrl expression for the potentil t the point P specified in terms of cylindricl coordintes,,. Although not stted, it cn be ssumed tht there is no chrge present nywhere in this problem. when tht is the cse, then the Green function solution for Dirichlet boundry conditions becomes: x= 4 S d G D d n' d' The enclosing surfce S in this cse is box with one side on the plne = nd the other sides t infinity. The potentil dies off to ero t infinity, so the sides t infinity mke no contribution to the integrl. The potentil is lso ero everywhere on the plne = outside the circle, so tht those loctions mke no contribution to the integrl. The only piece of the integrl left is inside the circle, so tht:
x= 4 d G D d n' ' d ' d ' x= V 4 d G D ' d ' d ' d n' The norml n' is defined s pointing out of the volume enclosed so tht for this problem n' = -' x= V 4 d G D ' d ' d ' d ' Plugging in the Green function we found bove: x= V 4 d d '[ x x ' y y' ' ' d ' d ' x x ' y y ' ' ] Putting everything in cylindricl coordintes: x= V 4 x= V 4 d d ' [ ' ' cos ' ' ] ' d ' d ' ' ' cos ' ' ' [ / ' ' cos ' ' 3/ ' / ] ' d ' d ' ' ' cos ' ' 3/ We now evlute everything t ' = becuse our integrtion surfce (the primed system) is completely contined in this plne. x= V ' d ' d ' ' 'cos ' 3/ The problem is imuthlly symmetric, so tht we re free to mke chnge of vribles: ' ' x= V ' d ' d ' ' 'cos ' 3/
(c) Show tht, long the xis of the circle (ρ = ), the potentil is given by =V Tke the generl solution found bove nd plug in (ρ = ) to get the on-xis solution: x= V x= V x= V ' d ' d ' ' 3 / d ' ' d ' ' 3/ ' d ' ' 3/ Mke chnge of vribles u= ', d u= ' d ' x= V du u 3/ x= V [ u ] x=v [ ] (d) Show tht t lrge distnces (ρ + >> ) the potentil cn be expnded in power series in (ρ + ) -, nd tht the leding terms re = V [ 3 3 / 4 53 4 8...] Verify tht the results of prts c nd d re consistent with ech other in their common rnge of vlidity. The generl solution found bove ws: x= V ' d ' d ' ' ' cos ' 3/ Divide top nd bottom by 3 /
x= V 3 / d ' 3/ d ' ' ' 'cos' Expnd the lst fctor using the Binomil series: x n =n x n n x= V 3 / We cn now integrte term by term x= V x= V x= V x= V 3 / [ [ x... d ' d ' ' [ 3 ' ' cos' 5 8 ' 'cos'...] d ' 3 / [ 3 [ 5 8 d ' '] 3 5 8 d ' d ' d ' '[ ' 'cos '] d ' d ' '[' 'cos' ]...] d ' '[' ] d ' d ' ' [' cos']] d ' d ' '[ ' 4 4 ' cos ' 4' 3 cos ']...] [ [ 3/ 3 ] 4 5 8 6 [ 4 6 4 4 ]...] 3/ [ Along the xis this becomes x= V 3 4 4 53 8...] [ 3 4 54 8 ]... 4 x=v [ 3 4 8 4 5 6 6 6... ] [ x=v 34 8 56 4 6... ] 6 Now we recognie the expnsion: x / = x 3 8 x 5 6 x3
[ x=v / ] x=v [ ] This mtches prt c.