Lecture 5 Solving Problems using Green s Theorem. 1. Show how Green s theorem can be used to solve general electrostatic problems 2.

Size: px

Start display at page:

Download "Lecture 5 Solving Problems using Green s Theorem. 1. Show how Green s theorem can be used to solve general electrostatic problems 2."

Ashlyn Garrett
6 years ago
Views:

1 Lectue 5 Solving Poblems using Geen s Theoem Today s topics. Show how Geen s theoem can be used to solve geneal electostatic poblems. Dielectics A well known application of Geen s theoem. Last time we deived Geen s theoem.. We also deived the fee space Geen s function in a sphee and cylinde. 3. These functions wee then used to deive the integal fom of electostatics fom which the potential is deived by an integal involving the chage density. 4. This was eassuing but we aleady knew these esults fom pio wok. 5. Today we focus on the moe inteesting and geneal poblem of solving multidimensional electostatic poblems in complex geometies, including the pesence of conductos (and dielectics) 6. Let s set up a typical poblem. We want to solve φ ρ/ ε φ( S ) φ known S 7. Although we could solve this poblem numeically it becomes inconvenient and computationally time consuming to do so fo a lage vaiety of bounday conditions φ S. Often this is what we need to do. 8. It is time consuming because each new bounday condition equies a whole new numeical calculation. 9. Geen s theoem helps if we now change the bounday condition on G fom the fee space condition at infinity to a diffeent one specified on S. ( ) G( ) Old G δ ( ) G( S) New G δ

2 . Geen s theoem fo an abitay inteio point becomes ( S ) G φ φ() Gρ( ) d φ( S ) G ds ε + n n V S G Gρ( ) d + φ( S ) ds ε n V S. Fo any φ S we need to evaluate a volume integal and a suface integal to detemine φ, a simple numeical task.. If we want to edo the poblem fo a diffeent φ S we only need to e-evaluate the lowe dimensional suface integal. 3. This seems too good to be tue! What is the catch? 4. In geneal it is of compaable difficulty to detemine the Geen s function satisfying G( S ) as it is to solve the oiginal poblem. This is a majo stumbling block. A less well known but moe impotant application. We show below how to use Geen s theoem to solve the geneal poblem without having to deal with the complicated poblem of detemining G such that G( S ).. Let s etun to the oiginal poblem φ ρ/ ε. φ o φ ( S )/ n is specified. 3. Fo geneality assume that eithe ( S ) 4. The fist step is to convet fom Poisson s equation to Laplace s equation. We define φ() φ () + φ () whee p h φ () p 4πε V ( ) fo 3-D ρ ( ) ln ds fo -D ε S ρ d

3 5. Fo simplicity call φ () ψ(). The homogeneous solution satisfies h ψ( S) φ( S) φp ( S) ψ ψ( S) φ( S) φp ( S) o n n n 6. Geen s theoem becomes αψ () G ψ ψ G d S n n S Inteio point α Exteio point / Suface point 7. Now, choose the obsevation point to lie on the suface so that α /. Then ψ ( ) ψ( ) S ( S ) G ψ S S G ds n n 8. Choose the Geen s function to coespond to the fee space Geen s function. This is easy to do. We aleady know this function. No complicated bounday conditions on G ae equied. 9. Since we know G it is an easy task to calculate G/ n.. If we know ψ ( S )/ n then Geen s theoem yields an integal equation fo ψ ( S ).. Similaly if we know ( S ) ψ we have an integal equation fo ψ ( S )/ n.. This is the desied fomulation. If we assume that we can solve the integal ψ and ψ ( S )/ n. Thus ψ and hence φ equation then we will know both ( S ) can be easily found by using Geen s theoem fo an intenal point and simply evaluating known integals. 3. The next step is to show how to solve the integal equation, guaanteeing that we will always be able to avoid the poblem of choosing impope expansion functions as was the case using sepaation of vaiables. 3

4 Solving the integal equation. The integal equation is a linea equation. Theefoe, expansion techniques ae a good appoach.. Note that in a 3-D poblem we need to solve the integal equation on a closed - D suface bounding the volume of inteest. Fo a -D poblem we need to solve the integal equation on a closed -D cuve bounding the suface of inteest. 3. Hee is the absolutely citical point!!! On a closed suface o cuve the solutions must be peiodic. Theefoe, we ae guaanteed that a Fouie seies must exist that can epesent any abitay bounday data. 4. Fo example fo a -D poblem whee l is the ac length along the bounday, the potential ψ ψ( l) can always be witten as ψ() l ψ im me θ 5. The existence of the Fouie seies guaantees that the poblem of impope expansion functions is eliminated. 6. Futhemoe, one does not have to use the angle θ as the independent vaiable. We could choose any othe angle v v( θ) that might be moe convenient (i.e. could put moe esolution in cetain sections along the cuve) Details of the pocedue. Thee ae a fai numbe of details to obtain the solution to the integal equation. We demonstate the steps fo a geneal -D poblem using an elliptical suface as a special example.. Assume the bounday cuve is paameteized in tems of an abitay angle-like vaiable v (i.e. v ) as follows x x( v) acosv y y( v) bsin v 4

5 3. To use Geen s theoem note that ds dl dz L dl. Since the facto cancels eveywhee, we heeafte suppess it fo convenience. 4. The following geometic elation fo vecto ac length is needed fo the solution (whee ove dot denotes d/ dv) x y ( x y) ( asinvex bcosvey) dv dl dxe + dye x e + y e dv + z L z 5. Fom this we find that x e + y e asinve + bcosve t x y x y ( x + y ) ( a sin v + b cos v) / / y e x e bcosve + asinve n t e z x y x y ( x + y ) ( a sin v + b cos v) / / ( ) ( sin cos ) / / unit tangent unit nomal dl x + y dv a v + b v dv ac length 6. Similaly { xv ( ) xv ( ) yv ( ) yv ( ) } + / / a ( cosv cosv) + b ( sinv sinv ) 7. Fom this we can evaluate the Geen s function { } G ln x ( v ) x ( v ) y ( v ) y ( v ) 4π + ln a ( cos v cos v) + b ( sin v sin v 4π ) 5

6 8. We also need the nomal deivative of the Geen s function. 4π x y ( x + y ) n G y x ln ( x x) + ( y y) / ( ) ( ) ( x x) + ( y y) cos( v v) ( ) + ( ) y x x x y y ab a cosv cosv b sinv sin v 7. Obseve that G has a logaithmic singulaity when v v. Howeve, this is an integable singulaity ln xdx xln x x finite 8. Because of this one might think that G/ n would be singula as / when v v. It is actually finite. Using L Hospital s ule twice we find that fo the case of the ellipse v v ( ) / L x + y n G ab 4π a sin v + b cos v 9. In fact it can be shown that in geneal G/ n is finite on any suface as. v v ( ) / L x + y n G xy yx 4π x + y. Let s etun now to the integal equation of inteest which can be witten as ( v ) G ψ ψ() v ψ( v ) G dl n n / / ψ ( x y ) G G( x y ) ψ + n + n dv 6

7 . We shall solve this equation by Fouie analysis leading to a elation between ψ ( v) and ψ ( v)/ n. The expansion is as follows ψ () / ( ) n ψ( ) v a e m imv x + y v b e m imv 3. The goal now is to find a elation between the a and b. One quantity is given by the bounday condition. The othe is obtained by solving the integal equation. 4. Let us assume that ψ ψ( v) is specified on the suface. This means that we know the a m S coefficients. m m an ψ inv () v e dv 5. Now Fouie analyze the Geen s function and its nomal deivative on the bounday cuve G( v, v ) B e mm mn, x + y G v v A e mm / ( ) n (, ) imv im v mn, imv im v 6. The matix elements A, B ae known quantities that can be evaluated mm mm numeically in a staightfowad manne. inv+ inv B (, nn G v v ) e dvdv / A ( x y ) G( v, v nn + ) e n inv+ inv dvdv 7

8 7. These expansions ae substituted into Geen s theoem ae aa e e dv m imv in v ipv ip v m n pp π n, p, p in v ipv ip v bb n pp e e dv n, p, p 8. Cay out Fouie analysis by multiplying by e inv dv 9. The vaious tems ae evaluated as follows im ( nv ) m m π a e dv a in ( p) v ip ( nv ) a A e n pp e dv dv A a nn n n, p, p π n in ( p) v ip ( nv ) b B e n pp e dv dv B b nn n n, p, p π n n. Combine tems an ( A ) nn a B n nn b n n. In compact fom this can be witten as I A a B b. Consequently, if ψ ( v) is specified, then the Fouie coefficients fo the nomal deivative ae given by b B I A a 8

9 3. Convesely, if the nomal deivative ψ ()/ v n is given then the Fouie coefficients of the potential ae given by a I A B b 4. Once both a and b ae known, then ψ () can be found at any inteio point by evaluating the now known Geen s function integals. ψ () G ψ ψ G d S n n S 5. This complicated pocedue has been used extensively used in the NSE fusion pogam. 6. One poblem involved an accuate detemination of the magnetic field in the pesence of lage amounts of ion in the PHENIX detecto on the RHIC facility at Bookhaven National Laboatoy. 7. Anothe application involves detemining the best set of coil cuents in the Alcato C-Mod poloidal field system to achieve a given desied plasma shape. Dielectics. A new topic now dielectics. What is a dielectic? 3. A dielectic is an insulating mateial one with no fee chages and no conduction electons (as in a metal) 4. Dielectics consist of neutal atoms which become polaized when placed in an electic field. 5. We shall see that the diection of polaization is such as to cancel pat of the applied field. 9

10 6. A simple physical pictue is shown below 7. In a eal mateial not evey atom stays polaized. Othe foces, such as themal foces, ae also pesent which tend to andomize the polaization. Thus the amount of polaization depends upon the detailed atomic stuctue of the mateial unde consideation. 8. Let s see if we can ceate a model to detemine the induced electic field in a simple atom. Keep in mind that this is a qualitative, not quantitative model. 9. In the diagam below assume the nucleus of the atom is infinitely massive (compaed to the electon). An electon cloud encicles the nucleus with a adius detemined by quantum mechanics.. An electic field is applied causing a slight shift d in the location of the electon cloud. Thee is now moe cloud below the nucleus than above it. This geneates a net Coulomb foce on the cloud.. Note that diection of the electic field induced by this chage sepaation is opposite to that of the applied field.. We can appoximate the elationship between d and E by a simple foce balance as follows. 3. Assume the sphee of electon chage has a unifom chage density and a thickness Δ. 4. The net upwad foce on the cloud F F e F cos θ is found by integating Coulomb s foce law ove the volume of the cloud z z ρ

11 +Δ Fz qe q d d d cos θ π ρ φ sin θ θ 4 πεl l + d dcos θ dcos θ 5. This integal can easily be evaluated in the limits d and Δ. We obtain (with q e) qd 6πε qe d E 6πε 3 3 e 6. Hee we have balanced the shift in the obit d caused by the electic field E (equal to the applied field plus the induced field due to all othe atoms) against the attactive Coulomb foce. 7. Next, note that in a dielectic the negative chage due to a downwad shift of one electon is balanced by a deficit of negative chage fom the atom located one laye lowe. Thee is only a net effect at the sufaces of the dielectic whee no futhe compensating chages ae available. 8. As shown, this poduces a suface chage whose value is estimated by assuming that on aveage the numbe density of atoms in the mateial is n paticles pe cubic mete. 9. Each electon caies a chage equal to q e. The total numbe of unbalanced electons is elated to the aveage shift d due to the polaization. Thus nv nad. The total chage in the unbalanced laye is qnv which is equivalent to a suface chage density σ qnv A qnd /. Clealy thee is also a net deficit of electons on the uppe edge of dielectic poducing a suface chage σ σ.. The net macoscopic effect of the polaization is to induce a macoscopic opposing electic field within the dielectic which is calculated as the field between two equal and opposite suface chages.

12 E ind σ edn 3 ( 6π ) ε ε n E. We now intoduce the concept of the elative dielectic constant as follows. Wite down the -D fom of Poisson s equation. εe z ρ fee + ε ρ ind ε 3. Hee, ρ ind epesents the induced suface chage due to the polaization while ρ fee epesents any othe fee chage that may be pesent in the mateial (e.g. such as due to a beam of chaged paticles popagating though the mateial) 4. If we integate Poisson s equation acoss the dielectic we obtain ε E ρ dz + ε E fee ind 5. This can be ewitten as ( p) ε + χ E ρ χ p 6πn 3 fee d z 6. We can now define the elative dielectic constant as ε + χ p and Poisson s equation becomes ( εe ) ρ 7. Fo a simple dielectic we show that afte all this wok we simply eplace with ε ε ε. ε

13 Bounday conditions fo a dielectic. It is customay in E&M theoy to intoduce the displacement vecto D εe so that Poisson s equation becomes D ρ D εe. The bounday conditions acoss a dielectic-vacuum inteface ae conveniently expessed in tems of E and D. They ae found as follows. 3. Conside the aea integal shown below and use the fact that in electostatics E. Stoke s vecto theoem then implies that E ds E dl n E 4. Next, we integate Poisson s equation ove the volume shown below, assuming that no infinitesimally thin fee suface chages exist. Dd D nds n D 5. Acoss a dielectic-vacuum inteface the tangential electic field and nomal displacement vecto ae continuous. 3

14 The dielectic filled capacito. As a simple application of dielectics conside the dielectic filled capacito as shown below. The goal is to calculate the capacitance of the system and the voltage pofile.. In paticula, the electic field in each egion is a constant and we want detemine thei values fom which the othe infomation can then be easily obtained. 3. The solution is obtained as follows. Fist use the voltage elation V Edz V ( Ea + E c) c b a b 4. Second, fom symmety we see that the condition n E acoss the inteface is automatically satisfied. 5. Thid, acoss the inteface the condition on the displacement vecto educes to n D ε E E 4

15 6. We can solve these two simultaneous equations fo E and E E E V a + εc V ε a + ε c 7. The capacitance can easily be calculated fom the enegy definition εe CV d Aε E c + E a ( ε ) AεV εc εa 4 + ( a εc) ( a εc) + + AεV ε 4 a + ε c 8. We see that C ε A ε a εc/ + a 9. The voltage dop acoss each egion can now be easily calculated. V V V Ea + ε c/ a V εc E c + ε c/ a a. Lastly, conside the inteesting limit of a stongly diamagnetic mateial ε 5

16 . Then V V V V a cε. Note that most of the voltage dop occus acoss the vacuum. It had bette be a good vacuum to avoid beakdown. 6

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum 2. Electostatics D. Rakhesh Singh Kshetimayum 1 2.1 Intoduction In this chapte, we will study how to find the electostatic fields fo vaious cases? fo symmetic known chage distibution fo un-symmetic known