Physics 417G : Solutions for Problem set 3

Transcription

1 Poblem Physics 47G : Solutions fo Poblem set 3 Due : Febuay 5 6 Please show all the details of you computations including intemediate steps A thick spheical shell of inne adius a and oute adius b caies a polaization P = k whee k is a constant and is the vecto fom the cente a Calculate the suface bound chages σ a at = a and σ b at = b and ρ inside the thick spheical shell b What is the total chage? c Find the electic fields in the thee egions < a a < < b and > b Sol: a The bound chages ae given by σ = P ˆn with the unit vecto ˆn outwad nomal to the suface and ρ = P Thus σ a = P ˆ = ka σ b = P ˆ = kb ρ = P = k = 3k b The total chage should be This can be checked by adding all the chages 4πb Q total = 4πa σ a + 4πb 3 σ b + 4πa3 ρ = 3 3 c Using the Gauss law E ds = Qencl /ɛ and the chages we can get < a k ɛ a < < b b < whee inside the shell the total chage is Q encl = 4πka 3 3k 4π 3 [3 a 3 ] = 4πk 3 which is the combination of the total suface chage at = a and the volume chage between the adius a and Fo the egion > b the total chage enclosed is thus the electic field vanishes Poblem Let us also conside the same thick spheical shell of inne adius a and oute adius b which now caies a diffeent polaization vecto P = k ˆ whee k is a constant and is the vecto fom the cente We add a metal sphee with adius a inside the inne shell with the same oigin of the spheical shell The inne sphee caies total chage Q a Use a Gauss law to compute the displacement cuent D fo the thee egions < a a < < b and > b b Using the esults of a compute the electic field fo the coesponding egions c Calculate the suface bound chages σ a at = a and σ b at = b and ρ inside the thick spheical shell d Find the electic fields in the thee egions < a a < < b and > b using the bound chages along with the chage Q Compae the esults with b Sol: a Gauss law fo the D goes D ds = Qfee The chage in the conducto is a fee chage and is located in the suface Thus we have D = < a D = D = Q ˆ 4π a < < b D = Q ˆ 4π b < Note : Fo the est of the poblem we need to think about the situation caefully You can think about the poblem as the thick spheical shell a dielectic mateial esponds to the electic field due

2 to the chage Q of the metal sphee and poduce P b Fom the elation between the electic field E and D = ɛ E + P we get Q ɛ 4π k ˆ Q ˆ < a a < < b c The bound chages ae ρ b = P = k = k and σ b = P k/b = b ˆn = k/a = a b < d The electic fields ae the same as Q ɛ 4π k ˆ Q ˆ < a a < < b b < Fo instance in the middle egion a < < b the chage enclosed is Q encl = Q + k/aπa + 4π This gives the electic field fo a < < b a d k/ = Q 4πk 3 Poblem 3 In this poblem we ae going to find the electic field poduced by a unifomly polaized sphee of adius R Fo this you can stat with φ = ds σ b S x x + d 3 x ρ b 4πɛ V x x σ b = P ˆn ρ b = P whee V is a localized volume and S is the bounday with its unit nomal vecto ˆn Pat I: a Show that the electic field E = φ can be ewitten as E = P E E = d V 3 x x x x x 3 b Evaluate E inside and outside of the unifom sphee c Now compute the electic field E Pat II: Fo any chage configuation with Azimuthal symmety potential is independent of the angle φ we can expess the potential as φ θ = A l l + B l l+ P l cos θ in the spheical pola coodinate whee P l epesent Legende Polynomial d Fo unifomly polaized sphee compute the bound chages Ae they familia? e Find the esulting potential inside and outside the sphee Hint: It is impotant to have finite potentials both inside and outside This will detemine possible expansions inside and outside This allows us to detemine B l in tems of A l The suface chage povides a discontinuity on the adial deivative of the potential at = R This detemines A l

3 f Compute electic field inside and outside of the sphee Ae they the same as c? Sol: a One can easily check the ρ b = fo unifom polaization Thus φ = ds S Fo this you can stat with φ = [ ] ds 4πɛ P S x x = ds 4πɛ P x x S x x 3 = d 3 x 4πɛ k x x Pk V x x 3 = P d 4πɛ V 3 x x x x x 3 σ b x x whee we used the divegence theoem the constant natue of P k f x x = f x x in the second line b E is the electic field due to a sphee unifomly chaged ρ = Using the Gauss law we can get < R ˆ > R c Then the electic fields ae E = P E P 3 3 P ˆˆ P < R > R d The bound chages ae ρ b = σ b = P cos θ e Inside and outside the sphee we need to have finite potentials Thus we have φ in θ = φ out θ = A l l P l cos θ R B l l+ P lcos θ R Fo example φ out is finite eveywhee fo R especially at These two functions should be continuously connected at = R Thus we have A l R l B l P l cos θ = R l+ P lcos θ which should be valid fo each l because the Legende polynomial P l cos θ ae othogonal each othe To pove this we multiply P l cos θ and integate to use the othogonality popeties given by Thus we have dcos θp l cos θp l cos θ = l + δ ll B l = A l R L+ To detemine A l we use the fact that the suface chage poduces the discontinuity in the electic field only the nomal component of the electic field with espect to the suface This is adial diection We can expess this as ˆn E = ˆ E = φ Thus we have φ out φ in = σ b =R ɛ which gives l + B l R l+ P lcos θ la l R l P l cos θ = σ b ɛ l + A l R l P l cos θ = σ b ɛ

4 whee we use B l = A l R L+ By multiplying P l cos θ and using the othogonality popeties we get A l = R l ɛ σ b θp l cos θdcos θ Until now we use a geneal fomula that will hold with the Azimuthal symmety in spheical pola coodinate Fo σ b θ = P cos θ = P P cos θ we know that only A will be non-zeo due to the othogonality popety of Legende polynomial Thus A l = P R l ɛ Thus we detemine the potential as P cos θp l cos θdcos θ = P δ l φ in θ = P P cos θ = P cos θ R φ out θ = P R3 P cos θ = P R3 cos θ R f Then the electic fields ae E = φ P 3 3 P ˆˆ P < R > R 4 Poblem 4 a A shot cylinde of adius a and length L caies a fozen unifom polaization P paallel to its axis Find the bound chages Now sketch the electic fields i fo L a ii fo L a and iii fo L a b A vey long cylinde of adius a caies a unifom polaization P pependicula to its axis not adial diection Find the electic field inside and outside the cylinde using one of the two methods below Hint : Method I Think of it as two cylindes of opposite unifom chage density ±ρ P we eseve ρ fo the cylideical coodinate whose centes ae sepaated by a vecto d with d < R see the figue which is a view of the coss section of the cylinde fo method I One can use Gauss law to compute the electic fields of the two cylindes and add them to get the esult we want Now it emains to connect P and d The polaization vecto P is elated to d because the total electic dipole moment is nothing but volume times the polaization vecto Method II You can find the bound chages and solve the scala potential as the second pat of the poblem 3 Sol: a Simple computations give ρ b = σ b = P ˆn = ±P espectively at each end i Fo L a the ends look like point chages and the whole thing is like a physical dipole of length L See Fig a ii Fo L a it is like a cicula paallel-plate capacito Field is nealy unifom inside; non-unifom finging field at the edges See Fig b

5 iii Fo L a See Fig c b Think of it as two cylindes of opposite unifom chage density ±ρ P Inside the field at a distance ρ fom the axis of a unifomly chage cylinde is given by Gauss s law ds E ρ = Q encl πρleρ = ρ P πρ l E ρ S ɛ ɛ = ρ P ρ ɛ whee l is the length of Gaussian suface with adius ρ Fo two such cylindes one plus and one minus the net field inside is E ρ = E + E = ρ P ɛ ρ + ρ = ρ P ɛ d whee we use ρ + ρ = d which is a vecto fom negative to positive axis Now it emains to connect P and d The polaization vecto P is elated to d because the total electic dipole moment is nothing but volume times the polaization vecto P πa L = ρ P πa L d P = ρ d whee L is the total dipole moment of a chunk of length L Thus Fo outside the Gauss s law gives Gauss s law E ρ = P ɛ πρleρ = ρ P πa l E ρ ɛ = ρ P a ˆρ ɛ ρ whee l is the length of Gaussian suface with adius ρ Fo two such cylindes one plus and one minus the net field outside is E ρ = E + E = ρ P a ˆρ + ˆρ = a ρ d ɛ ρ + ρ ɛ ρ ρ ρ d whee we use ρ ± = ρ d and ˆρ + = ρ + d ρ + ρ = ρ + ρ + d 4 ρ d ρ ρ + ρ whee we keep on the tems linea in d and simila elation fo ρ Thus E ρ = a ɛ ρ [ P ˆρ]ˆρ P ρ d ρ d