Jackson 3.3 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

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1 Jackson 3.3 Homewok Pobem Soution D. Chistophe S. Baid Univesity of Massachusetts Lowe POBLEM: A thin, fat, conducting, cicua disc of adius is ocated in the x-y pane with its cente at the oigin, and is maintained at a fixed potentia V. With the infomation that the chage density on a disc at fixed potentia is popotiona to ( 2 ρ 2 ) -/2, whee ρ is the distance out fom the cente of the disc, (a) show that fo > the potentia is,, = 2V (b) find the potentia fo <. 2 2 P 2 cos (c) What is the capacitance of the disc? SOLUTION: Waning! The soution that Jackson gives is wong. Let us sove the pobem the wong way (the way Jackson expects), then show why this soution is wong. Then et us sove the pobem the ight way and figue out whee Jackson went wong. The Wong Way: (a) The suface chage density was stated to be: = S 2 2 fo < and 0 othewise The thee-dimensiona chage density is then: =S / fo < and 0 othewise To find out what S is in tems of the potentia V, use Couomb's aw which integates ove a the chage density to find the potentia at the oigin and set it equa to V: = 4 0 x' x x' d x' V = 4 0 x ' x' d x'

2 V = V =S 2 π π 4 π ϵ S δ(θ' π/2) 2 ' 2 ' ' '2 sin θ' d ' d θ' d ϕ' 2 ' 2 d ' V =S 2 0[ sin ' ]0 S= 4 0 V Now knowing S, pugging it back in, we have the fina fom of the chage density: ρ= 4ϵ 0 V π δ(θ π /2) 2 2 fo < and 0 othewise Now appy Couomb's Law to find the potentia at any point on the z axis: = 4 0 x' x x' d x' < Use 0 = + (cos θ') to expand this (which is aowed because we ae on the z axis). > = V ' /2 2 ' 2 ' < > cos ' '2 sin ' d ' d ' d ' = 2V 0 < 2 ' 2 P 0 ' d ' > We have to teat the two egions sepaatey. Let us ook at the > ' egion because its intega is easie. Fo > we aso have > ' so that: = 2V ' 2 ' d ' Make a change of vaiabes u= 2 ' 2 and udu= ' d ' = 2V u 2 / 2 du Now (0) is zeo fo odd, so ony even tems contibute. Let us eabe to take account fo this fact:

3 = 2V 2 0 P u 2 du If we do the intega case by case fo = 0,, 2... the integation is tivia and we soon see a patten:,, = 2V 2 2 Now we must emembe that this is ony vaid on the z axis. We can make use of the handy theoem that fo pobems with azimutha symmety, the genea soution is just the soution on the z-axis, mutipied by P(cos θ): Φ(,θ,ϕ)= 2V π ( ) 2+( 2 ) P 2 fo > This is the soution (the wong one Jackson expects) to the potentia in the exteio egion. (b) To find the potentia in the nea egion ( < ), fist note that the pobem has azimutha symmety and no chage in the nea egion, so the genea soution to the Lapace equation hods: Φ(,θ, ϕ)= ( +B ) We need a finite soution at the oigin, so the soution must have the fom: Φ(,θ,ϕ)= A fo < This soution fo the potentia in the oute egion must match the soution fo the potentia in the inne egion at the inteface whee they touch, = : 2V π,even ( ) / 2 + P = A The Legende poynomias ae othogona, so the coefficients must match up sepaatey, eading to: = 2V π ( ) /2 + and = 0 fo odd The soution fo the nea egion is theefoe: Φ(,θ,ϕ)= 2V π ( ) 2+( ) 2 P 2 Now, the pate is hed at V, so the soution to the potentia shoud educe down to the constant V fo

4 θ=π/2 and <, independent of. It shoud be obvious that the soution above does not educe to V on the disc. The facto (0) is zeo ony fo odd, but this soution ony has even. The coect soution wi have ony odd vaues. The Coect Way: Note that thee ae eay fou egions that we need to teat, sepaatey, as indicted in the diagam. In each egion, thee is no chage, thee is azimutha symmety, and the poes ae incuded, so the soution to the potentia has the fom: Φ out,up z Φ(,θ,ϕ)= ( +B ) V x The inne egions incude the oigin, so they must have a B zeo to have a finite soution at the oigin. Simiay, the oute egions incude infinity, which we can assume to have zeo potentia, eading to a being zeo in these egions. Ou soutions in a egions theefoe become:,down Φ out,down Φ out,up (,θ,ϕ)= B (cos θ) fo > and θ < π/2 Φ out,down (,θ, ϕ)= B,down fo > and θ > π/2 (,θ,ϕ)= A fo < and θ < π/2,down (,θ,ϕ)=, down fo < and θ > π/2 Fist, due to symmety, the potentia at any point in an uppe egion must equa the potentia at the mio point acoss the x-y pane: Φ( z)=φ( z) =,down ( cos θ) and Φ out,up (cos θ)=φ out,down ( cosθ) =, down ( cosθ) and B = B,down ( cosθ), down = ( ) and B, down =B ( ) With these findings, ou soutions now become:

5 Φ out,up (,θ,ϕ)= B (cos θ) fo > and θ < π/2 Φ out,down (,θ, ϕ)= B ( ) fo > and θ > π/2 (,θ,ϕ)= fo < and θ < π/2,down (,θ,ϕ)= ( ) fo < and θ > π/2 Note that by focing the uppe and owe egion potentias to be mio images, we automaticay made them match up at thei inteface, and have aeady taken cae of this bounday condition. Next, the potentia in the inne egions must become V on the disc. V = eading to: A 0 =V, = (cos(π/2)) and V = ( ) (cos(π/2)) (0)=0, and = ( ) (0)=0 Note that (0) = 0 fo a odd, in which case the ast two equations ae automaticay satisfied. Fo even, (0) is not zeo, so: =0 fo even,>0 Ou soution so fa is: Φ out,up (,θ,ϕ)= B (cos θ) fo > and θ < π/2 Φ out,down (,θ, ϕ)= (,θ,ϕ)=v + B ( ) fo > and θ > π/2 =,3,5..,down (,θ,ϕ)=v + =,3,5... fo < and θ < π/2 ( ) (cos θ) fo < and θ > π/2 Note that now that is odd, (-) is aways just -. The potentia in the inne-down egion theefoe becomes,down (,θ,ϕ)=v =,3,5... combined into (,θ,ϕ)=v +sgn =,3,5.... The soutions in the two inne egions can now be (cos θ) whee sgn(cos θ) is + fo θ < π/2 and - fo θ > π/2. Aso note that because the potentia in the inne egions and oute egions must match at =, and due to othogonaity, ony the = odd tems wi contibute in the oute egions as we. The oute egion soutions can theefoe be combined in the same way. Ou soution so fa is thus:

6 Φ out (,θ, ϕ)= B 0 +sgn =,3,5... (,θ,ϕ)=v +sgn =,3,5... B (cos θ) Next, the potentias of the inne and oute egions shoud match at = : (,θ, ϕ)=φ out (, θ, ϕ) V +sgn(cos θ) = B 0 =,3,5... +sgn B (cos θ) =,3,5.. The Legende poynomias ae othogona, so we match up coefficients. Matching up a coefficients, we find: B 0 =V and B = 2 + The soution so fa becomes: Φ out (,θ,ϕ)=v +sgn(cos θ) =,3,5... (,θ,ϕ)=v +sgn =,3, fo > (cos θ) fo < A the at this point ae abitay, so et us edefine as / to make these equations symmetic, eading to: Φ out (,θ, ϕ)=v +sgn(cos θ) =,3,5... (,θ,ϕ)=v +sgn =,3,5... ( + ) (cos θ) fo > ( ) fo < The ast set of coefficients can be found by eating the eectic fied acoss the pate: (E 2 E ) n 2 = σ ϵ 0 (E in,down E in,up ) θ= σ ϵ 0 whee σ= S 2 2 fo a conducting pate To find out what S is in tems of the potentia V, use Couomb's aw to integate ove a the chage density and find the potentia at the oigin and set it equa to V: V = σ (x ') 4π ϵ 0 x ' d a

7 V = V =S 2π 4 π ϵ S 'sin θ' d ' d ϕ' 2 ' 2 ' 2 ' 2 d ' V =S 2 0[ sin ' ]0 S= 4ϵ 0 V π Using this vaue, the bounday condition on the eectic fied acoss the pate now becomes: (,down + ) θ= 4V π 2 2,down + θ θ = 4V π 2 2,down + θ θ = 4V π ( / ) 2 Pefom a binomia expansion on the ight side, using,down + θ,down θ,down θ θ = 4V π [ + 2( ) 3 + θ = 4V π [ =,3, ( ) ( 2)!! ( )!!( ) ] x =+ 2 2 x x x ( ) θ = 4V [ π ( ) 2 (0) =,3,5... ( ) ] +...] A Legende poynomia identity was used in the ast step to get the ight side in a fom that we anticipate wi be on the eft side. Now evauate the deivatives: 2 A =,3,5... ( ) 2 =,3,5... [ ( ) [ θ ]θ=π/ 2 x ( x) ]x=0 = 4V π [ = 4V π [ =,3,5... =,3,5... ( ) 2 (0) ( ) ] ( ) 2 (0) ( ) ]

8 2 2 =,3,5... =,3,5... ( ) = 2 V + π ( ) 2 [ x ( x) ( x) x 2 ]x=0 = 4V π [ =,3,5... ( ) [ P (0) ]= 4V [ π ( ) 2 (0) =,3,5... ( ) ] Ou fina soution is theefoe: Φ out (,θ, ϕ)=v +sgn(cos θ) 2V π (,θ,ϕ)=v +sgn 2V π =,3,5... =,3,5... ( ) + 2 ( ) ( ) + 2 ( ) + ( ) 2 (0) ( ) ] (cos θ) fo > fo < Note that this soution obeys a the bounday conditions that it shoud. On the pate, (cos θ) becomes (0), which is zeo fo odd, eaving just the constant V as it shoud. So whee did Jackson go wong? The symmety of the pobem equies Legende poynomias with odd, but Jackson's soution had even, indicating that he got the symmety wong. (c) The tota chage is: 2 Q= ', ', ' ' 2 sin ' d ' d ' d ' Q=8 0 V 0 2 ' 2 ' d ' Q=8 0 V [ 2 ' 2 ] 0 Q=8 0 V Q V =8 0 This is the capacitance: C=8 0

Jackson 4.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jackson 4.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell Jackson 4.7 Homewok obem Soution D. Chistophe S. Baid Univesity of Massachusetts Lowe ROBLEM: A ocaized distibution of chage has a chage density ρ()= 6 e sin θ (a) Make a mutipoe expansion of the potentia