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 ƒystem for pro˜lem PFIQ 2.13.a ƒhow 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   PA ™os '     =b @b  A Pb Pb @ b @b  A b2 2 ™os '   Rb V V a "  ™os '  @b  A C @Pb ™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. ƒolve ˜y me—ns of series solution @PFUIA —nd show th—t the potenti—l inside the ™ylinder isX I ˆ   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 ˆ¢ n=1 £ an n sin @n' C n A C bn  n sin @n' C n A ƒin™e we9re nding the solution inside the ™ylinder bn a HX ¨@; 'A a a C 0 I ˆ n=1 I ˆ aA C P 0 an n sin @n' C n A n=1 ‘Ann ™os @n'A C Bnn 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 ˆ  VV  I  m sin [email protected] PA'“ @Rm PA b m m   I ˆ  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 ˆ   4m+2 m=0 b sin [email protected] C PA'“ Pm C I 2.14.b. ƒum the series —nd show th—tX  sin V ¨@; 'A a P t—n Pbb  P' 1 22 4 Q !  C ™os Qn @ IAn I ™os Qn P P 4  !! I ˆ   4n+2 V ¨@; 'A a R b n=0 V a R sm—g a RV sm—g  4 5 I ˆ   4n+2 ei(4n+2)' n=0 4 I ˆ 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 dened ˜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 ˆ 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 xia h a ™os  m xia A CB a 1 1 =0 =0 a2 n RV Ra 2 A1 C B1 a 2 a mn2 ITV A1 C B1 a mn2 …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 ƒolving 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 + mn2 ITV Aa nm 1 A1 a 2 ITV e p  e 1 ITV Aa nm 1 VV Aa nm n2 +n2 n2 +m2 p nm2 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 mn2  VV e a nm nm2 ™osh  pn2 +m2 2 2 VV a nm [email protected]; y; z A a 2  n  ˆ  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 nm2 ™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 ˆ 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 ˆ 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 ƒee 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 ˆ 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 ˆ  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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