Consider a potential problem in the half-space dened by z 0, with Dirichlet boundary conditions on the plane z = 0 (and at innity).

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1 Problem.7 Consier otentil roblem in the hlf-sce ene by z 0, with Dirichlet bounry conitions on the lne z 0 (n t innity)..7.. Write own the rorite Green function G(~x; ~x 0 ). G D (~x; ~x 0 ) (x x 0 ) (x x 0 ) (x x 0 ) (x x 0 ) (x x 0 ) (x x 0 ) where x, x, n x enote the x, y, n z coorintes, resectively..7.b. If the otentil on the lne z 0 is secie to be V insie circle of rius centere t the origin, n 0 outsie tht circle, n n integrl exression for the otentil t the oint P secie in terms of cylinricl coorintes (; '; z). We're going to use eution. from Jckson: I (~x) (~x 0 D 0 0 Note tht (~x 0 ) V insie the circle of rius centere t the origin. Let's convert G D (~x; ~x 0 ) to cylinricl coorintes: G D (~x; ~x 0 ) ( cos ' 0 cos ' 0 ) ( sin ' 0 sin ' 0 ) (z z 0 ) ( cos ' 0 cos ' 0 ) ( sin ' 0 sin ' 0 ) (z z 0 ) s 0 0 (cos ' cos ' 0 sin ' sin ' {z } 0 ) (z z 0 ) cos(' ' 0 ) s 0 0 (cos ' cos ' 0 sin ' sin ' {z } 0 ) (z z 0 ) cos(' ' 0 ) S 0 0 cos(' ' 0 ) (z z 0 ) 0 0 cos(' ' 0 ) (z z 0 )

2 We nee to n the norml erivtive of G D. Note tht the norml oints wy from the region of interest { since we're consiering z 0, let ^n ^z 0 D 0 D (~x; ~x 0 ) ( ^z 0 ) 0 * > D 0 (0 G D D z 0 0 " # (z z 0 )( ) ( 0 0 cos(' ' 0 ) (z z 0 ) ) (z z 0 ) ( 0 0 cos(' ' 0 ) (z z 0 ) ) z z ( 0 0 cos(' ' 0 ) z ) ( 0 0 cos(' ' 0 ) z ) z ( 0 0 cos(' ' 0 ) z ) Note tht the terms which \cncel to zero" o so becuse the erivtives of the two terms of G D sum to zero when evlute t z 0 0. Now, lug this into eution (). Note tht we only nee to integrte over the circle which hs otentil V becuse the integrn is zero elsewhere.! z (~x) ' ' (V ) ( 0 0 cos(' ' 0 ) z ) 0 ( 0 0 cos(' ' 0 ) z ) 0 ' ' 0 z c. Show tht, long the xis of the circle ( 0), the otentil is V z z. Letting 0: (~x) ' ( 0 z ) 0 ' 0

3 Using the substitution u 0 z n u 0 : " z.7.. (~x) V ' 0 0 ' 0 0 ' 0 0 uz ( ) u # u ' 0 u z z z z z z uz ' 0 Show tht t lrge istnces ( z ) the otentil cn be exne in ower series in ( z ), n tht the leing terms re: V z ( z ) ( z ) 5 ( ) : : : 8 ( z ) Verify tht the results of rt c n re consistent with ech other in their common rnge of vliity. z ' 0

4 (~x) ' ' z 0 ( 0 0 cos(' ' 0 ) z ) 0 ' cos(' ' 0 ) z 0 ' 0 Using the Tylor exnsion ( x) n nx n(n )x : : : cos(' ' 0 ) 5! 0 0 cos(' ' 0 ) : : : ( z ) ' z z cos(' ' 0 ) cos(' ' 0 ) : : : 0 ' 0 ( z ) ' z z ( z ) ( z ) ( z ) ' ' ' cos(' ' 0 ) z 05 0 cos (' ' 0 ) 0 cos(' ' 0 ) ( z ) : : : 0 ' cos(' ' 0 ) z cos (' ' 0 ) 5 05 cos(' ' 0 ) : : : ( z ) 0 cos(' : ' 0 ) z 6 6 cos (' ' 0 ) cos(' : ' 0 ) : : : ' 0 ( z ) Note tht the \cncele" terms integrte to zero. ( z ) : : : z ( z ) ( z ) ( z ) 5 ( ) : : : ( z ) 0 0 ' 0 0 ' 0

5 For 0: (~x) V z V z V V 6 5 z z : : : 5 6 8z 8z : : : z 8z 8z : : : {z 6 7 } 5 z z z Hence, rts c n gree in the limit where z. Problem.9 An insulte, shericl, conucting shell of rius is in uniform electric el E 0. If the shere is cut into two hemisheres by lne ereniculr to the el, n the force reuire to revent the hemisheres from serting.9.. If the shell is unchrge. From Jckson's exmle roblem in section.5, we know tht the surfce-chrge ensity is given by: " 0 E 0 cos Using the eution shown in gure.: jf z j F " 0 9" 0 E 0 cos " 0 F z 9" 0 E 0 cos cos " 0 '0 0 9 " 0E 0 cos sin ' 5

6 Using the substitution u cos, u sin : jf z j 0 '0 u 9 " 0 E 0 9 " 0 E 0 9 " 0E 0 u ( u)' 0 '0 '0 u 0 u u u' ' 9 " 0 e 0.9.b. If the totl chrge on the shell is Q. The surfce chrge ensity on the shere of chrge Q is: Q Q Now, we n the sme metho s bove to n the force between the two hemisheres of eul chrge: jf z j '0 F " 0 Q 6 F z Q 0 Using the substitution u sin, u cos : jf z j " 0 " 0 cos Q cos " 0 sin ' Q " 0 Q " 0 '0 '0 Q " 0 Q " 0 u0 ' uu' 6

7 The totl force on the shere ue to the shere's own chrge Q n the electric el is the sum of the force foun in rt n the force we just foun: jf totl j 9 " 0 e 0 Q " 0 Problem.0 A lrge rllel lte ccitor is me u of two lne conucting sheets with sertion D, one of which hs smll hemishericl boss of rius on its inner surfce D. The conuctor with the boss is ket t zero otentil, n the other conuctor is t otentil such tht fr from the boss the electric el between the ltes is E 0. E E 0 D Figure : Setu for roblem Clculte the surfce-chrge ensities t ny rbitrry oint on the lne n on the boss, n sketch their behvior s function of istnce (or ngle). Assuming the lnes to be innite n very fr from ech other, we see tht this system cn be roximte by groune shere in uniform electric el (we re given tht the urose of the non-groune lte is to cuse the electric el between the ltes to be constnt n uniform). Hence, eution. from Jckson gives the electric otentil between the ltes: E 0 r cos () On the boss, the surfce-chrge ensity is the sme s eution.5 from Jckson: r " 0 E 0 cos 7

8 To n the surfce-chrge ensity on the groune lne (locte t z 0), we rst convert eution () to Crtesin coorintes: E 0 r cos {z } r z E 0 z E 0 z0 r r.0.b. Show tht the totl chrge on the boss hs the mgnitue " 0 E 0. We will integrte the surfce-chrge ensity over the surfce re of the boss to n its net chrge: Q " 0 E 0 cos " 0 E 0 '0 " 0 E 0 '0 Using the substitution u sin, u cos : Q " 0 E 0 cos sin ' '0 " 0 E 0 '0 " 0 E 0 " 0 E 0 cos sin ' u0 uu' u ' u0 8

9 .0.c. If, inste of the other conucting sheet t ierent otentil, oint chrge is lce irectly bove the hemishericl boss t istnce from its center, show tht the chrge inuce on the boss is: 0 This system is shown in gure. z z z z Figure : Setu for roblem.0.c The otentil for this system is: " 0 x x (x ) x x x x x x Converting to shericl coorintes: " 0 cos cos 5 x x (x ) cos 5 cos 9

10 @ " 6 ( cos ) ( cos ) cos cos cos cos ( cos ) 7 5 ( cos ) Q '0 8 '0 " 8 '0 " 8 () sin ' 6 ( cos ) ( cos ) cos cos ( ) cos cos ( cos ) 7 5 sin ' ( cos ) # # ' 0