Chapter 6 Electrostatic Boundary Value Problems. Dr. Talal Skaik

Size: px

Start display at page:

Download "Chapter 6 Electrostatic Boundary Value Problems. Dr. Talal Skaik"

Scott Watkins
5 years ago
Views:

1 Chpter 6 Electrosttic Boundry lue Problems Dr. Tll Skik 1 1

2 Introduction In previous chpters, E ws determined by coulombs lw or Guss lw when chrge distribution is known, or potentil is known throughout the region. when In most prcticl pplictions, however, neither the chrge distribution nor the potentil distribution is known. In this chpter, prcticl electrosttic problems re considered where only electrosttic conditions (chrge & potentil) t some boundries re known, nd E nd re to be found. Such problems re tckled using Poisson's or Lplce s eqution or the method of imges. E

3 6. Poisson s nd Lplce s Equtions They re esily derived from Guss s lw. D E E for n inhomogeneous medium for homogeneous medium ( poisson's eqution) when, (for chrge free region): ( Lplce's eqution) 3

4 Lplce s Eqution Crtesin x y z Cylindricl 1 1 z Sphericl r sin r r r r sin r sin 4

5 Poisson s Eqution Crtesin x y z Cylindricl 1 1 z Sphericl r sin r r r r sin r sin 5

6 Uniqueness Theorem There re severl methods of solving given problem (nlyticl, grphicl, numericl, experimentl, etc). The question is: If solution of Lplce s eqution stisfies given set of boundry conditions, is this the only possible solution? The nswer is Yes, there is only one solution, nd it is unique. This is uniqueness theorem: If solution of Lplce s eqution cn be found tht stisfies the boundry conditions, then the solution is unique. 6

7 Generl Procedure for Solving Poisson's or Lplce's Eqution 1. Solve Lplce's (if ) or Poisson's ( if ) eqution using: () direct integrtion when is function of one vrible, or (b) seprtion of vribles if is function of more thn one vrible. The solution t this point is not unique but expressed in terms of unknown integrtion constnts to be determined.. Apply the boundry conditions to determine unique solution for. Imposing the given boundry conditions mkes the solution unique. 3. Hving obtined, find E using E nd D using D E. 7

8 Exmple 6.1 Current-crrying components in high-voltge power equipment must be cooled to crry wy the het cused by ohmic losses. A mens of pumping is bsed on the force trnsmitted to the cooling fluid by chrges in n electric field. The electrohydrodynmic (EHD) pumping is modeled in the Figure. The region between the electrodes contins uniform chrge ρ which is generted t the left electrode nd collected t the right electrode. Clculte the pressure of the pump if ρ =5 mc/m 3 C nd = k. 8

EHD pumps The technology promises to mke it esier nd more efficient to remove het from smll spces, vstly expnding the cpbilities of dvnced instruments nd microprocessors.

9 EHD pumps The technology promises to mke it esier nd more efficient to remove het from smll spces, vstly expnding the cpbilities of dvnced instruments nd microprocessors. Unlike trditionl therml-control technologies tht rely on mechnicl pumps nd other moving prts, EHD cooling uses electric fields to pump coolnt through tube. Electrodes pply the voltge tht pushes the coolnt through tube. EHD-bsed systems re lightweight nd consume little power. 9

10 Since, we pply poisson's eqution the boundry conditions (z=)= nd (z=d)= show tht depends only on z, 1 1 o Hence becomes: z z d oz Integrting once gives A dz oz Integrting gin gives = Az B where A nd B re integrtion constnts to be determined by pplying the boundry conditions. when z, = B B when z d, Exmple Solution = Ad or A= od od d 1

11 The electric field is given by 1 E= = z E Exmple Continued z d A z z d o o z z z The net force is F= Q E, Q dv d v z d z o S z dz z d T T v v F= Edv ds Edz z F= S F Force per unit re (Pressure) is N/m S z oz = Az B B 3 3 S 11 od A= d

12 Exmple 6. The xerogrphic copying mchine is n importnt ppliction of electrosttics. The surfce of the photoconductor is initilly chrged uniformly s in Figure 6.(). When light from the document to be copied is focused on the photoconductor, the chrges on the lower surfce combine with those on the upper surfce to neutrlize ech other. The imge is developed by pouring chrged blck powder over the surfce of the photoconductor. The electric field ttrcts the chrged powder, which is lter trnsferred to pper nd melted to form permnent imge. We wnt to determine the electric field below nd bove the surfce of the photoconductor. 1

13 Exmple 6. 13

14 Exmple 6. 14

15 Since Exmple 6. - Solution, we pply Lplce's eqution. Also potentil depends only on x. x Integrte twice gives =Ax+B Let the potentils bove x= be, nd below x= be. =A x+b x> =A x+b x< The boundry conditions t the grounded electrodes re, (x=d)= =A d+b B = A d (1) (x=)= =+B B = () 15

16 At the surfce of the photoconductor, (x=)= (x=) A +B A (3) D D = E E 1n n S x 1 1n n Exmple 6. - Continued d d d d Since E= = x y z En dx dy dz dx d1 d Therefore 1 S S 1A1 A (4) dx dx Solve equtions (1) to (4) to find the constnts. d1 S x E 1= x A1 x, x d dx d d S 1 x d E = x A x, x dx d S =A x+b x> =A x+b x< B = A d, B =

17 Exmple 6.3 Semi-infinite conducting plnes t Ф= nd Ф=π/6 re seprted by n infinitesiml insulting gp s in Figure6.3. If (Ф=)= nd (Ф=π/6 )=1, Clculte nd E in the region between the plnes. 17

18 Exmple 6.3 Solution Since depends only on, Lplce's eqution in cylindricl coordintes becomes d z d d since = is excluded due to the insulting gp, we cn multiply by to get d Integrting twice gives =A +B. Apply boundry conditions: when, =+B B= when, =A, Hence = A 1 d E d 6 6 Substituting =1 nd / 6 gives: = nd E= check ( )=, ( / 6)=1, 18

19 Exmple 6.4 Two conducting cones (Ѳ=π/1 nd Ѳ=π/6) of infinite extent re seprted by n infinitesiml gp t r =. If (Ѳ=π/1 )= nd (Ѳ=π/6)=5, Find nd E between the cones. 19

20 Exmple 6.4 Solution Since depends only on, Lplce's eqution in sphericl 1 d d r sin d d coordintes becomes: sin since r= nd =, re excluded, we cn multiply by r sin to get d d sin d d d d A Integrting once gives: sin A d d sin d d Integrting this results in = A A sin cos( / )sin( / ) d A A cos( / )sin( / ) cos( / ) tn( / ) cos( / ) 1 sec ( / ) d Let u= tn /, du=1/ sec ( / ) d du = A Aln( u) B Aln(tn / ) B u

21 Apply boundry conditions to determine the integrtion consttnts And B. ( ) Aln(tn / ) B B Aln(tn / ) tn / Hence Aln(tn / ) Aln(tn 1 / ) Aln tn 1 / tn / Also ( ) Aln A tn 1 / tn / ln tn 1 / Thus = Exmple 6.4 Continued ln(tn / ) tn / ln tn 1 / tn / ln tn 1 / A B 1

22 1 d Exmple 6.4 Continued A E r d r sin tn / r sin ln tn 1 / tn( / ) Tking 1,, nd 5 gives Aln 1 6 tn( 1 / ) tn / 5ln tn / A tn / = tn( 95.1ln / ) tn / ln ln tn( 1 / ) tn / tn( / ) ln nd tn 1 / = 95.1 tn( / ) E= /m ln r sin tn( 1 / ) Check: ( /1)=, ( / 6)=, d A d sin

23 6.5 Resistnce nd Cpcitnce For conductor with uniform cross section, the resistnce is R l S If the cross section of the conductor is not uniform, the resistnce is obtined from: R E. dl I E. ds 3

24 Resistnce The resistnce R ( or conductnce G=1/R) of conducting mteril cn be found by the following steps: 1. Choose suitble coordinte system.. Assume o s the potentil difference between conductor terminls. 3. Solve Lplce's eqution to obtin then determine E from E nd find I from I E. ds. 4. Finlly, obtin R s o /I. 4

25 Cpcitnce To hve cpcitnce, we must hve two (or more) conductors crrying equl but opposite chrges. The conductors my be seprted by free spce or dielectric. The conductors re mintined t potentil difference: 1 E. dl (note E is lwys norml to conducting surfces) 5

26 Cpcitnce The cpcitnce C of the cpcitor is defined s the rtio of the mgnitude of the chrge on one of the pltes to the potentil difference between them: Q E. ds C E. dl Obtin C for ny two conductor cpcitors by either: 1. Assume Q nd determine in terms of Q (involves Guss Lw).. Assume nd determine Q in terms of (involves Lplce s eqution) Method (1) involves the following: 1. Choose suitble coordinte system.. Let the two conducting pltes crry chrges +Q nd -Q. 3. Determine E using Coulomb or Guss lw nd find from 4. Finlly, obtin C from C=Q/. 6 E. dl

Cpcitors Storing Chrges- Cpcitors: A cpcitor

these conductors with n insulting mteril clled

27 Cpcitors Storing Chrges- Cpcitors: A cpcitor consists of conductors of ny shpe plced ner one nother without touching. It is common; to fill up the region between these conductors with n insulting mteril clled dielectric. We chrge these pltes with opposing chrges to set up n electric field. 7

28 Cpcitnce or Cpcitnce Resistnce 8 1 E l E Guss or Coulombs lw (in terms of Q) Q C d D S D= E E= = or = / Q C Q d = J S J= E E= = or = / R I I d

Prllel Plte Cpcitor Suppose ech plte hs n re S nd

29 Prllel Plte Cpcitor Suppose ech plte hs n re S nd the pltes re seprted by distnce d. Assume pltes 1 nd crry uniformly distributed chrges +Q nd Q Q S Q, D or E ( ) S S S S x x x 1 Q Qd Hence E dl x dx x S S nd thus for prllel-plte cpcitor d Q S C d 1 1 The energy stored in cpcitor is WE C Q Q C 9

30 Coxil Cpcitor Assume Conductors 1 nd crry uniformly distributed chrges +Q nd Q. Apply guss lw to rbitrry Gussin cylindricl surfce of rdius ρ such tht < ρ< b, Q Q E ds= E L E= L 1 Q Q b E dl d ln L L b Thus the cpcitnce of coxil cylinder is given by Q C= L b ln 3

31 Sphericl Cpcitor Assume chrges +Q nd Q re on inner nd outer spheres. Apply guss lw to rbitrry Gussin sphericl surfce of rdius r such tht < r< b, Q Q E. ds= Er4 r E= 4 r r 1 Q Q 1 1 E dl r dr r 4 r 4 b Thus the cpcitnce of sphericl cpcitor is given by C= Q b b 31

32 Series nd Prllel Cpcitors If two cpcitors with cpcitnce C 1 nd C re in series ( hve sme chrge on them), the totl cpcitnce is: CC or C= 1 C C C C C 1 1 If the cpcitors re in Prllel (hve sme voltge cross them), the totl cpcitnce is: C C C 1 3

33 It hs been shown tht: Which is the relxtion time T r of the medium seprting the conductors. 33 E l E S E S E l The product of these expressions yields d R I d d Q C d RC

34 Assuming homogeneous medi, the resistnce of vrious cpcitors mentioned erlier cn be obtined using RC = ε/σ : S d For prllel-plte cpcitor C, R d S For cylindricl cpcitor b ln L C, R b ln L For sphericl cpcitor C, R b b Note: R in these equtions is NOT the resistnce of the cpcitor plte, but the lekge resistnce between the pltes. Therefore, σ is the conductivity of the dielectric medium seprting the pltes. 34

35 Exmple 6.8 A metl br of conductivity σ is bent to form flt 9 o sector of inner rdius, outer rdius b, nd thickness t s shown in the Figure. Show tht :- () The resistnce of the br between the verticl curved surfces t r = nd r = b is: ln b R t (b) The resistnce between the two horizontl surfces t z= nd z=t is R' 4t ( b ) 35

36 () Between the verticl curved ends locted t = nd =b, the br hs nonuniform cross section. Let potentil difference be mintined between the curved surfces t = nd =b so tht ( = )= nd ( = b)=. Solve using Lplce's eqution in cylindricl coordintes, 1 d d d d As is excluded, multiply by nd since =( ), d d A integrte once, A, integrting gin yields, d d =A ln +B, where A nd B re constnts of integrtion. Use the boundry conditions: Exmple 6.8 Solution ( = )= =A ln +B or B Aln b ( = b)= =A ln b+b=a ln b Aln Aln or A b ln 36

37 Hence, Exmple 6.8 Continued A A A ln ln = ln ln d d E J E, ds ddz / t A b ln I J ds dz d b z ln b ln Thus R= s required I t b ln t b ln =A ln +B A b ln B Aln 37

38 (b) Let d Exmple 6.8 Continued be the ptentil difference between the two horizontl surfces so tht ( z )= nd ( z t)=. since ( z),lplce's eqution becomes, integrting twice gives dz Az B, pply boundry conditions to get A nd B, ( z )= B or B ( z t)= At or A t Hence, z t d E z= z 38 dz t

39 Exmple 6.8 Continued E ds dd t J z, b / b b ( ) I J ds dd. t t 4 t 4t Thus R'= I b Alterntively, for this cse, the cross section of the br is uniform between the horizontl surfces t z= nd z=t, then l t 4t R'= S ( b ) b 4 z 39

40 Exmple 6.9 A coxil cble contins n insulting mteril of conductivity σ. If the rdius of the centrl wire is nd tht of the sheth is b, show tht the conductnce of the cble per unit length is: G= ln b 4

41 Exmple 6.9 G= ln b Trnsmission Line Equivlent circuit: (See Chpter 11) G=Conductnce per unit length for dielectric. G=1/(Lekge Resistnce of Dielectric) 41

42 Let be the potentil difference between the inner nd outer conductors, so tht ( ) nd ( b) Solve b ln 1 d d d d ln nd E= J E, ds ddz ln b l Exmple 6.9 Solution ln. From exmple 6.8, we got: L I J ds dzd b b z ln ln b ln 1 The resistnce per unit length is R=. I L 1 nd the conductnce per unit length is G= R b ln b 4

43 Exmple 6.1 Conducting sphericl shells with rdii = 1 cm nd b = 3 cm re mintined t potentil difference of 1 such tht (r=b) = nd (r=) = 1. Determine nd E in the region between the shells. if εr =.5 in the region, determine the totl chrge induced on the shells nd the cpcitnce of the cpcitor. 43

44 1 d d r r dr dr since r in the region of interest, we multiply through by r to obtin d d d d A r, Integrting once gives: r A or dr dr dr dr r A Integrting gin: B (use boundry conditions to find A nd B) r A A 1 1 when r b, = B or B=, Hence A b b b r 1 1 when r, A, or A= b 1 1 b Thus = 1 1 r b 1 1 b Exmple 6.1 Solution 44

45 d A E r r r dr r 1 1 r b To determine cpcitnce, first find Q on either shell Q E ds r sindd 1 1 r b b r r Or ds, Q= S For inner shell, Q= S[4 ], bu t 4 Dn En Q=.4 r r b b b The cpcitnce is esily determined s Q C= b Exmple 6.1 Continued r r r S D n r 1 1 A b r A= 1 1 = b 1 1 r b 1 1 b 45

46 Substituting =.1 m, b=.3 m, 1 1 r = / 3 r E /m r r r 1 1 / 3 r (.5)(1) 36 Q= nc 1 1 / 3 =1 yields, positive chrge is induced on the inner shell, negtive chrge is induced on the outer shell. 9 Q C= pf 1 Exmple 6.1 Continued = 1 1 r b 1 1 b 46

Exmple 6.11 In Section 6.5, it ws mentioned tht the cpcitnce C=Q/ of cpcitor cn be found by either ssuming Q nd finding or by ssuming nd finding Q. The former pproch ws used in Section 6.

47 Exmple 6.11 In Section 6.5, it ws mentioned tht the cpcitnce C=Q/ of cpcitor cn be found by either ssuming Q nd finding or by ssuming nd finding Q. The former pproch ws used in Section 6.5 while we hve used the ltter method in the lst exmple. Using the ltter method, derive cpcitnce for prllel plte cpcitor: Q S C d Assuming the potentil difference between the prlle pltes is so tht (x=)= nd (x=d)=. 47

48 Exmple 6.11 Solution d Solve Lplce's eqution: Integreting twice gives dx Ax b At x, B or B At, or /. Hence = d x d Ad A d x To find the cpcitnce, first find the chrge on either plte Q d But D = E, where E A dx d S On the lower plte n x, So S nd Q d d S On the upper plte n x, So S nd Q d d (Q is equl S n n x x x but opposite on ech plte). C= Q S d ds S 48

49 Exmple 6.1 Determine the cpcitnce of ech of the cpcitors in the Figure. Tke εr1 =4, εr=6, d=5 mm, S=3 Cm. () Since D nd E re norml to the dielectric interfce, the cpcitor in Fig. () cn be treted s two cpcitors C nd C in series. C Exmple 6.1 Solution = S S S, C = d / d d r1 r1 r

50 Exmple 6.1 Solution Continued The totl cpcitor is given by C C C=... C=5.46 pf C C d S ( r1 r) r1 r (b) D nd E re prllel to the dielectric interfce, the cpcitor in Fig. (b) cn be treted s two cpcitors C nd C in prllel (the sme voltge cross C nd C ) 1 1 r1s / r1s rs C1=, C = d d d The totl cpcitnce is 9 4 S C C1 C= ( r1 r) pf 3 d 36 (51 ) 5

51 Q Q E ds= E L. Hence E= L Q Q d Q d d o rl L 1 L 1 b b b Q Q 1 b ln(1 ) ln L L 1 b Exmple 6.13 A cylindricl cpcitor hs rdii = 1 cm nd b =.5 cm. If the spce between the pltes is filled with n inhomogeneous dielectric with ε r =(1+ρ)/ρ, where ρ is in centimeters, find the cpcitnce per meter of the cpcitor. Thus the cpcitnce per meter is (L=1m) 9 Q 1 1 C= pf/m 1 b ln ln

CAPACITORS AND DIELECTRICS

Importnt Definitions nd Units Cpcitnce: CAPACITORS AND DIELECTRICS The property of system of electricl conductors nd insultors which enbles it to store electric chrge when potentil difference exists between