# F10HW04 - Problem 2.13 wo hlves of  long hollow...

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Unformatted text preview: Problem 2.13 wo hlves of  long hollow onduting ylinder of inner rdius b re seprted y smll lengthwise gps on eh sideD nd re kept t dierent potentils V nd V 1 b ' V2 2 V1 pigure IX ystem for prolem PFIQ 2.13.a how tht the potentil inside is given yX   Pb os ' V CV V V ¨@; 'A a P C  tn b  where ' is mesured from  plne perpendiulr to the plne through the gpF sing the result from prolem PFIPX b  I   ¨@ a b; 'A d'H ¨@; 'A a P b C  Pb os@'H 'A  = b  b  I  = V HC I a P d' V d'H b C  Pb os@'H 'A P = b C  Pb os@'H 'A =  =  = b  b  I HC I H a P V d' V b C  Pb os@'H 'A P = b C  C Pb os@'H 'A d' = !  = I @b  A @b C  C Pb os@'H 'AA C V @b  A @b C  Pb os@'H 'AA d'H a P V @b C  A @Pb os@'H 'AA @b C  A @Pb os@'H 'AA =  =  = @V C V A @b  A @b C  A C I @V V A Pb os @'H 'A d'H a PI  = @b C  A Rb  os @'H 'A P = @b C  A Rb  os @'H 'A I  @V C V A C I  = @V V A @b  A Pb os @'H 'A H a P P = b C  C Pb  C @Pb  Pb  A Rb  os @'H 'A d' 1 1 2 2 1 2 2 2 2 0 2 1 2 2 2 1 2 2 1 2 22 2 2 2 2 2 2 2 2 2 1 4 4 42 I 2 2 1 2 22 2 2 22 2 2 22 2 2 2 2 2 2 2 2 1 2 2 2 22 2 2 22 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 3 2 2 2 2 2 2 2 2 2 22 2 2 22 22 2   Ab b a PI  @V C V A C PI P @V bV @  AA tn b P  os '     Pb os ' @V C V A C @V V A tn aP  b  1 1 1 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2.13.b. Calculate the surface-charge density on each half of the cylinder  a "0 @¨   a "0 V1 V2   @ =b  IC   PA os '     =b @b  A Pb Pb @ b @b  A b2 2 os '   Rb V V a "  os '  @b  A C @Pb os 'A b 1 0 I 2 2 22 2 = V a " b osV' 1 22 2 2 2 2 0 2 2 2 Problem 2.14 e vrint of the preeding twoEdimensionl prolem is  long hollow onduting ylinder of rdius b tht is divided into equl qurtersD lternte segments eing held t potentil CV nd V F 2.14.a. olve y mens of series solution @PFUIA nd show tht the potentil inside the ylinder isX I    4n+2 V ¨@; 'A a R ¨@; 'A a a C b ln  C 0 0 n=0 b sin [email protected] C PA' Pn C I I ¢ n=1 £ an n sin @n' C n A C bn  n sin @n' C n A ine we9re nding the solution inside the ylinder bn a HX ¨@; 'A a a C 0 I  n=1 I  aA C P 0 an n sin @n' C n A n=1 Ann os @n'A C Bnn sin @n'A P Bn bn I   ¨@ a b; 'A [email protected]'A a P 4 5 P  = @V A [email protected]'A C   @ V A [email protected]'A C  = @V A [email protected]'A C   @ V A [email protected]'A a P =  = ! =   = I @ V A I [email protected]'A I [email protected]'A C I [email protected]'A I [email protected]'A   a 2 0 3 2 0   2 2 n h  3 2 0 n 3  =2 n I a  V os P I @ IAn os P n  I V hP C P @ IAn C R os  n i a n P @ a V n 8 H 2  n  i h  n i   2 2 2 n    =2 3 n a P; T; IH; IR; IV; PP ¡ ¡ ¡ a Rm P otherwise ell the An terms will integrte to zero euse ¨@ a bm'A is n odd funtionF I   VV  I  m sin [email protected] PA' @Rm PA b m m   I   m RV a b @Pm IA sin [email protected] PA' m ¨@; 'A a 4 4 2 2 =1 4 =1 V a R +2 I    4m+2 m=0 b sin [email protected] C PA' Pm C I 2.14.b. um the series nd show thtX  sin V ¨@; 'A a P tn Pbb  P' 1 22 4 Q !  C os Qn @ IAn I os Qn P P 4  !! I    4n+2 V ¨@; 'A a R b n=0 V a R smg a RV smg  4 5 I    4n+2 ei(4n+2)' n=0 4 I  n=0 P b  Pn C I I  i' Pn C I b e 2 2 2n+1 5 2 Q  IU  TH T I C 22 ei2' U b smg Tln d  2  eU T U T I b2 ei2' U R S | {z } R V a R  PH sin [email protected] C PA' Pn C I I  H  IQ 2 i ' 2 i ' I Rd I C  b2 e  e C d I C  b2 e  eS elR a P I 22 ei ' I 22 e i ' b b 2 2 2 I aP  I 4 b4 C Pi 22 b 2   [email protected]'A C I I C 44 P 22 [email protected]'A b b 4 b4 Pi 22 b  [email protected]'A P P 4 I a P 4 2b I C b4 P b2 [email protected]'A I 44 b a 4 P 2 [email protected]'A I C b4 b2 4  2 I H  IQ 2 ei2'  e i2' I Rd I C  b2  e d I C  b2  eS 2 e i2' Pi I 22 ei2' I b2 b    I 44 C Pi 22 [email protected]'A I 44 Pi 22 [email protected]'A b b b b I 2 4 Pi I C 4 P 2 [email protected]'A b b 2 [email protected]'A Ri b2 I 4 P 2 [email protected]'A Pi I C b4 b2 P 22 [email protected]'A b I C 44 P 22 [email protected]'A b b PH smgR a a a a R   R V ¨@; 'A a R tn smgR el 22 3 RV tn P 2 [email protected]'A b a I 44 b   RV tn P b [email protected]'A a 1 1 1  22 b4 4 2.14.c. Sketch the eld lines and equipotentials. Problem 2.23 e hollow ue hs onduting wlls dened y six plnes x a HD y a HD z a HD nd x a aD y a aD z a aF he wlls z a H nd z a a re held t  onstnt potentil V F he other four sides re t zero potentilF 2.23.a. Find the potential [email protected]; y; z A at any point inside the cube. r [email protected]; y; zA a H 2 @ 2¨ @ 2¨ @ 2¨ C C aH @x2 @y2 @z 2 essume [email protected]; y; z A a X @xAY @yAZ @z AX X HH Y Z C XY HH Z C XY Z HH a H S hividing oth sides y XY Z X X HH Y HH Z HH C Y C Z aH X X HH a 2 X Y HH a 2 Y ¡ Z HH a 2 C 2 X X @xA a A sin@ xA C B os@ xA Y @yA a C sin@ xA C D os@ xA Z @z A a Ee z p 2 + 2 C F ez p 2 + 2 X @HA a B a H n X @aA a A sin@ aA a H aA a a Y @HA a D a H m Y @aA a C sin@ aA a H aA a a  n   m  h i p2 2 p  a a [email protected]; y; z A a sin a x sin a y A1e z  n +m C B1ez  n2+m2 m;n sing [email protected]; y; z a HA a V X P P  a  a @V A sin  n x sin  m y dxdy A CB a aa y x a a R V h a os  n xia h a os  m xia A CB a 1 1 =0 =0 a2 n RV Ra 2 A1 C B1 a 2 a mn2 ITV A1 C B1 a mn2 sing [email protected]; y; z a aA a V X 1 p p A1 e a a n2 +m2 C B1 ea a n2 +m2 A1 e  p n2 +m2 a 1 C B e 1 p n2 + 0 m a @IA PP a aa m2 a ITV mn a a y=0 x=0     @V A sin n x sin m y dxdy a a @PA 2 olving eqution @IA for B nd plugging it into eqution @PAX  p p  n2 m2 C ITV A e n2 m2 a ITV Ae 1 1 + 0 mn 1 2 T + mn2 ITV Aa nm 1 A1 a 2 ITV e p  e 1 ITV Aa nm 1 VV Aa nm n2 +n2 n2 +m2 p nm2 e 2 ITV Aa nm p pI e  n2 +m2 n2 +m2 p +n e n2 2 I p2 2  p e 2 n +m e 2 n2 +m2 p C e 2 p p n2 +m2 p  2 2 2 2  e 2 n +m e 2 n +m p 2 e 2 n2 +n2  2 n2 +m2 e 2 p p 2 2  p 2 2 p 2 2  e 2 n +m 2  n2 +m2  n +m  n +m e2 e 2 n + m C e 2 e 2  pn2 +m2 2 e2  pn2 +m2 e2 p e2 2 C e  p n2 +m2 p n2 +m2 ¡ osh  n C m lugging this into eqution @IAX IT VV Ba 1 p 2 2 1 2 2 mn2  VV e a nm nm2 osh  pn2 +m2 2 2 VV a nm [email protected]; y; z A a 2  n    pn2 +m2 e2  osh  osh p 2 sin a x sin a y m;n  p ¡ 42 2 n2 C m2 2 n2 C m2  m  n2 C m2  p p  n2 +m2 e  n2 +m2 2 2 Ce  pn2 +m2 e 2 p 2 ¡ ¡  pn2 +m2 VV nm2 osh e2  p 2 3 n2 C m2  pn2 +m2 3 e 2 VV  pn2 +m2 e z a ¡ p z  n2 +m2 a p ¡e C nm osh  n C m 3 2p p 1z 1z aC VV  I sin  n x sin  m y e n2 m2@ 2 A p e  n2 ¡m2@ 2 a A a nm a a osh  n C m m;n 2 2 2 2 + + 2 2 2 ¢ 2 p 2 1 z C I sin  n x sin  m y osh  n p m a ¡  m;n nm a a osh  n C m [email protected]; y; z A a ITV  5 2 2 2 2 2 ¡£ 3 2 2 2.23.b. ivlute the potentil t the enter of the ue numerillyD urte to three signint guresF row mny terms in the series is it neessry to keep in order to ttin this uryc gompre your numeril result with the verge vlue of the potentil of the wllsF ee prolem PFPVF U he following wtl ode ws usedX for T = 1:10 Phi = 0; for n=1:T for m=1:T Phi = Phi + sin(n*pi/2)*sin(m*pi/2)/cosh((pi/2)*sqrt(n^2+m^2)); end end Phi*16/(pi^2) end xote tht the cosh term in the numertor is extly I t z a a=PF he upper limit of the summtion of oth m nd n ws inresedF he results re shown in the tle elowX T ¨=V I HFQRUSS P HFQRUSS Q HFQHTSR R HFQHTSR S HFQHVHT T HFQHVHT U HFQHVHH V HFQHVHH W HFQHVHH IH HFQHVHH xote tht three deiml ples of ury is rehed t T a SD where oth summtions in m nd n re evluted from I to T @iFeFD PS terms in totlAF he verge potentil over the six fes of the ue isX V CV V T aQ his is lose to the ¨ a H:QHVV we found t the enter of the ueF V 2.23.c. Find the surface-charge density on the surface z a a.  @¨   a "0  @z z=a  3 2 ¢p z ¡£ nC ITV  I sin  n x sin  m y sinh  2p m2 1 a   pn2 C m2  2 a "0 2   n2 C m2 ¡  nm a a a osh 2 m;n a " ITV a 0 p I sin x sin y tnh nm a a P n Cm m;n   n   m   W 2 2  p z =a n2 C m2 ...
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## 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.

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