Coulomb = V m. The line integral of the electric field around any closed path is always zero (conservative field)

Transcription

1 Chapter 3 Static Electric Fields Cheng; 3//009; 3-3. Overview Static electric fields are prduced by statinary electric charges N change in time N change in space 3. Fundamental Pstulates f Electrstatics in Free Space The electric field is the frce experienced by a unit charge F E = lim : Newtn q 0 q Culmb = V m () The frce n a charge q F = qe : [N] The tw fundamental pstulates f electrstatics ρv E = ε (3) E = 0 (4) where ρ v : vlume charge density ε : permittivity f free space Vlume integral f (3) EdV = ρ vdv E ds = V ε V S divergence therem ε : Gauss s law Surface integral f (4) with Stkes s therem E dl = 0, C The line integral f the electric field arund any clsed path is always er (cnservative field) Vltage r Wrk dne by E in mving a unit charge The line integral f the electric field depends nly n the end pints. Kirchff s vltage law: The algebraic sum f vltage drps arund any clsed circuit is er. P P

2 Cheng; 3//009; Culmb s Law A single pint charge, q Electric field with radial symmetry Apply Gauss s law n a hypthetical surface (Gaussian surface) q E ds = S ε E a a b g ds E 4π S Therefre q E = Ea = a A pint charge at an arbitrary lcatin E p = q a, where a qp qp = E p e q = j 3 A charge q is in the electric field made by q The frce n q qq F = qe = a 4 : Culmb s law πε A. Electric Field Due t a System f Discrete Charges n pint charges at,,... n The ttal E at q q q E = n ( ) ( ) n ( n ) n qk ( k ), Vectr sum 3 k = k

3 Example Electric diple Frm Culmb s law Cheng; 3//009; 3-3 The first term The secnd term Therefre We define the electric diple mment ewrite the electric field ewrite p in plar crdinates The electric field in plar crdinates

4 Cheng; 3//009; 3-4 B. Electric Field Due t a Cntinuus Distributin f Charge A vlume charge density ρ v ( x, y, ). b g The electric field at P(x,y,) due t a charge element ρ v x, y, dv ρvdv de = a The ttal electric field ρv x, y, E = a bx, y, g b g dv (6) V b g b g b g where = x x + y y + (Primed crds. fr surce pint, unprimed crds. fr field pint) A surface charge density ρ s ρs E = a ds (7) πε S 4 A line charge density ρ l ρl E = a dl (8) πε l 4 Example Find the electric field f an infinitely lng, line charge f a line charge density ρ l C / m. E is independent f φ. (cylindrical symmetry) The electric field at P E = ρ l d 3 = r a a The electric field due t ρ l d ρl d r a a de = r + Therefre L ρl r E = M πε NM r L O rd l d L ρ ar r + r NM ρ + P e j e j + NM e j r l 3 / 3 / 3 / d a r er + j / P 4 3 O P E O P a Canceled during integratin frm = t = ρl = r a r πε

5 3.4 Gauss s Law and Applicatins Gauss s law : E ds = S ε The ttal charge enclsed in the Gaussian surface The ttal utward flux f E thrugh a clsed surface Cheng; 3//009; 3-5 Gauss s law is very useful t find E in the presence f a symmetry If n symmetry, use (6), (7), r (8) Example Use Gauss s law t find E f a line charge, ρ l. Infinitely lng line charge E = E a : radial symmetry r r Chsse cylindrical Gaussian surface L π E ds Errd d rler S φ π 0 0 N cntributin frm the tp and bttm faces ds = a rdrdφ ds = a rdrdφ ρ L l ρl πrler = E = Erar = ar ε πε r dr rdφ d rdφ L Nte: The cylindrical Gaussian surface des nt wrk if the line charge is finite Example Find E f an infinite planar charge, ρ s. E is perpendicular t the sheet Chse rectangular Gaussian surface On the tp face E ds E a a ds E ds b g b g On the bttm face E ds E a a ds E ds Frm Gauss s law s A E ds = S ρε ρs E = Ea = a ε E s = ρ ε b g b g a, < 0, > 0 E ds EA, : N cntributin frm the side faces A Nte: Impssible t chse a Gaussian surface if the charge sheet is finite

6 Example Determine E f a spherical electrn clud with a vlume charge density ρ = ρ fr 0 b ρ v v = 0 fr > b Cheng; 3//009; 3-6 Slutin: E must be spherically symmetric, E Spherical Gaussian surface = E a Slve the prblem in the tw regins Inside and utside the electrn clud S i S (a) 0 b On Gaussian surface S i E = E a ds = a ds Frm Gauss s law E ds = ρ vdv S ε V ρ 4π 3 dv ρ ε V 3ε E ds E 4π S i E = ρ 3ε a, 0 b (b) b S is the Gaussian surface in this case 3 ρb 4π 3 E ds = ρ vdv E = aˆ ˆ ρ b a S ε V 3ε 3, b 4π 3 ρ b 3ε E π Cncentrated charge n a pint 4

7 3.5 Electric Ptential We define a scalar electric ptential, V E = V Cheng; 3//009; 3-7 Wrk dne in mving a charge q frm P t P in E P W = q E dl : independent f the path(energy cnservatin) P - sign means the wrk is dne by us against the frce. The electric ptential, V, is the wrk dne with a unit charge P V V E dl : Vltage () P Zer-ptential pint is at infinity if nt specified L P O L P O V V E dl E dl NM P NM P Nte (a) The minus sign in () The directin f increasing V is ppsite t the directin E (b) E = V The directin f V The surfaces f cnstant V

8 Electric Ptential Due t a Charge Distributin The electric ptential due t a pint charge F q I V = a a d HG KJ b g V The ptential difference between tw pints P q F I V = V V = HG A system f n charges, q, q,..., q n at V = n k= l KJ and P q =,,..., n qk : Scalar sum, easier t handle k q Cheng; 3//009; 3-8 A cntinuus distributin f charge, ρ v, ρ s, ρ l ρv V = dv ρs, V = V ds, V S ρl = dl L Example An electric diple: Oppsite pint charges +q and -q separated by d The ptential at P V = q 4 πε F HG + When d << d + cs θ, d + cs θ I KJ Then q F dcs θ I qd cs θ V = G J d cs θ H 4 p a V = p = qd, electric diple mment K The electric field V V E = V = aˆ ˆ aθ θ p cs θ aˆ sin ˆ 3 + θ a ( ) θ

9 3.6 Material Media in Static Electric Field Three types f media accrding t electrical prperties = Aggregatin f atms(nucleus + rbiting electrns) Cheng; 3//009; 3-9 Cnductrs Semicnductrs Insulatrs (r Dielectrics) : Lse uter electrns migrate easily : Small number f free charges : Tightly held electrns A. Cnductrs in Static Electric Field Charges in a gd cnductr epelling frces amng them. All charges n the surface N electric field inside. N tangential electric field n the surface E is nrmal t the cnductr surface The cnductr surface is an equiptential surface Inside a cnductr ρ v = 0 Gauss s law E = 0 Interface between a cnductr and free space Free space E E n a n Δw Δs Δh b c a d Δh ρ s () Apply E = 0 r E dl = 0 t the cntur abcda C E dl abbetg + bcb E cd E da E abcd f s ng left + b tg + b cnd ng.. right b g b g b g b g Δw E + E + Δh E + E b g b g t f. s. t cnd n left n right = 0 Ignre since Δh 0 () Apply Gauss s law t the bx E ds En S E S S fs Δ + n cnd Δ + E... t x[ side surfaces ] = ρ s ΔS ε = 0 Negligible as Δh 0 The bundary cnditins at the cnductr surface ρs Et = 0, En = ε Cnductr = 0 E t = 0 s E n = ρ ε

10 B. Dielectrics in Static Electric Field Cheng; 3//009; 3-0 Nnplar mlecules Mlecules in dielectrics Displacement f psitive and negative charges (electrically neutral) (bund charges) External electric field Induced Electric diples Plar mlecules Permanent diple mment withut an external field (Water mlecule, H O ) N external field andm rientatin N net diple mment Applied external field Alignment f the mlecules Plariatin vectr, P (diple mment density) P lim Δv 0 N k= p Δv k The electric ptential due t the diple mment Δp Δp a ΔV = where Δp = P Δv The ttal electric ptential P a V = dv v Interpretatin f P () Equivalent surface charge density, ρ ps : Psitive surface charges at the right bundary Negative surface charges at the left bundary Cnsider P ΔS ( P aˆ n ) ΔS q N ( dcs θ ΔS) surface nrmal, vlume Nqd ΔS θ P a n The surface charge density = P a ρ ps n d cs θ

11 () Equivalent vlume charge density, ρ pv Psitive surface charges Negative charges inside the vlume The net charge inside the vlume = ep a njds P ds = e Pj dv s s v The vlume charge density = P ρ pv Cheng; 3//009; 3- Dielectric materials are electrically neutral Ttal charge = ρ ds + ρ dv P ds Pdv 0 s ps pv v s v The ptential due t ρ ps and ρ pv V ρps = ds + S v ρ pv dv 3.7 Electric Flux Density and Dielectric Cnstant In a dielectric material E = dρv + ρpvi eεe + Pj = ρv ε = P D, Electric flux density, Electric displacement. D = ρ v, free charges D = ε E + P () The integral frm Ddv = ρ vdv D ds =, Gauss s law v v s The ttal utward electric flux thrugh a clsed surface is equal t the ttal enclsed free charge. In a linear, istrpic and hmgeneus dielectric material P =ε χ E χ e : Electric susceptibility () e Frm () and () D = ε + χ E = εe b eg permittivity ε r, relative permittivity, dielectric cnstant

12 Dielectric Strength Very strng electric field in a dielectric material Detachment f electrns frm nuclei Acceleratin f electrns Cllisins f the electrns with ther atms (Large amunt f iniatin, and large current) Permanent damage in the material Dielectric breakdwn Cheng; 3//009; 3- Dielectric strength: The maximum electric field withut dielectric breakdwn Dielectric strength f air: 3 kv / mm Example Charge tends t cncentrate at sharp pints Cnductr surface is equiptential The ttal charge is = + Therefre The electric fields at the surfaces Therefre

13 3.8 Bundary Cnditins fr Electrstatic Fields Cheng; 3//009; 3-3 An interface between tw media Medium D a n Δh b c E Δw E a d Δh a n D Δs Medium Apply E = 0 r E dl = 0 t the cntur abcda C L O E dl E ab+ E cd+ E bc+ da E Δw E Δ w E abcda = E t t NM P = side t t 0 Δh 0 The tangential cmpnent f E is cntinuus acrss the dielectric interface. Apply Gauss s law, D ds =, t the hexahedrn s D ds D a ΔS + D a ΔS + D a side areas a D D ΔS =ρ s Δ S s a D D = n b g e j n n side t n e j ρ s, Dn Dn = ρ s Δh 0 The nrmal cmpnent f D is discntinuus acrss the dielectric interface by the surface charge density. If medium is a cnductr : D n =, D = εe = ρ If n surface charge : D = D 0 n n n n s

14 Example Tw dielectric media with n free charge Find E when E makes an angle α at the surface Cheng; 3//009; 3-4 The tangential cmpnents E shuld be cntinuus E sin α = E sin α The nrmal cmpnents f D shuld be cntinuus ε E cs α = ε E cs α Cmbining the tw eqs. tan α ε = tan α ε The magnitude f E L NM F / = + Mb g + H G I cs α K J P E E E ε t n Esin α E ε Frm the figure ne can tell that ε < ε O P 3.9 Capacitances and Capacitrs Charges shuld be n the surface f the cnductr The electric ptential ρs V = s ds An increase f the charge by a factr f k ρ s increases by the same factr k (N change in the charge distributin) The ptential increases by the same factr k V is linear t (V = a ) ewrite this = CV C : capacitance

15 Cheng; 3//009; 3-5 The capacitr Tw separate cnductrs A d-c vltage surce Charge transfer resulting in + and. Electric field lines frm + cnductr t cnductr. Perpendicular t the surfaces (equiptential) C = V [F] Hw t find the capacitance () Chse a crdinate system () Assume charges + and - n the cnductrs (3) Find E frm frm Gauss s law r ther relatins (4) Find V = E dl (5) Find C = / V Example Parallel-plate capacitr Tw parallel cnductrs. The gap is filled with a dielectric with permittivity ε. Find the capacitance. Slutin: Chse Cartesian crds. Put charges + and - n the upper and lwer cnductrs Assume unifrm distributin f the charge ρ s = S Ignre the fringing at the edges ρs E = ay S a y ε ε The ptential difference y= d V = E dl S d y= 0 ε The capacitance S C = = ε V d

16 Example A cylindrical capacitr. The gap is filled with a dielectric with permittivity ε. Chse cylindrical crds. Assume charge + n the inner cnductr, - n the uter cnductr Frm Gauss s law / L E = E a = r a πε r r r The ptential difference r= a b Vab E dl ln r= b πεl a = F H G I K J The capacitance C = L V = πε ln b/ a ab b g Cheng; 3//009; Electrstatic Energy and Frces Mve a charge frm infinity twards The wrk dne is L W = N M O P L 4 N M O 4 P W = b V + V πε πε V V The wrk is stred in the assembly f tw charges as ptential energy A third charge 3 is brught frm infinity t a pint W = L NM F HG I F HG KJ KJ πε πε 3 πε πε 3 πε 3 3 V V V 3 I F HG g IO KJ P N discrete pint charges W = N V e k k k= 3 The electric ptential f a vlume charge density We = vvdv ρ v

17 Example Find the energy required t assemble a unifrm sphere f charge f radius b and vlume charge density ρ v Cheng; 3//009; 3-7 Accumulate spherical layers f thickness d (a) The charge enclsed by the sphere : The ptential due t the charge : V = π ρ = (b) The differential charge in d : d =ρ 4π d (c) The wrk in bringing up d : dw = Vd 4 π 4 v d 3ε ρ (d) The ttal wrk 4π b 4 4 vb 3 W = dw v d 3ε ρ 0 5 πρε 0πε b 5 v 3 Ttal charge πb ρ v v = 4 3 Electrstatic Energy in terms f Field uantities Apply Gauss s law t We = vvdv v ρ We = D Vdv VD dv D V dv v v v e j e j b g : e j = + = E evdj ds 0 s As,,, We = D Edv, r We = ε E dv W w dv v v e = e, where v VD V D D V we ε E, energy density Example Find the stred energy in a parallel-plate capacitr Ignre edge effect: E = V d W e F = H G I K J F H G I K J b g F H G I K J v We = CV V d dv V S ε ε Sd ε d d V C, capacitance

18 Electrstatic Frces Principle f Virtual Displacement Tw separate charges. One charge pushes away the ther charge by dl Cheng; 3//009; 3-8 F The wrk dne by the system dw = F dl eductin f the stred energy dwe = F dl = W dl : frm the gradient therem F b e g = W e dl 3. Slutin f Electrstatic Bundary-Value Prblems Find V and E frm the bundary values withut knwing the charge distributins Bundary-value prblems Pissn s and Laplace s Equatins Tw fundamental Eqs. fr electrstatics D = ρ v () E = 0 () Frm () E = V Using D =ε E, Eq. () becmes ε V = ρ b g v V = ρ v ε : Pissn s equatin (3) In Cartesian crds. x y x y x y V V V V V V V V aˆ ˆ ˆ ˆ ˆ ˆ x + ay + a ax + ay + a + + In cylindrical crds. V V r r r r φ F = H G I V r K J + + In spherical crds. V V V θ sin θ θ θ sin F = H G I K J F + H G I V sin K J + In a medium where ρ v = 0 V θ φ V = 0 : Laplace s equatin

19 Bundary-value Prblems in Cartesian Crdinates Cheng; 3//009; 3-9 Example Tw parallel cnducting plates are separated by d and are at ptentials 0 and V. Electrn distributin between the plates is ρv = ρ y / d. Ignre the edge effect and (a) Find the ptential between the plates (b) Find the surface charge densities n the plates y d V (a) Slve Pissn s equatin dv dy ρy = ε d E V bg y = ρ d y 3 6ε + C y + C The bundary cnditins at the plates Vby = 0g = 0 V y = d = V C b g V = ρ d d 6ε, C = 0 Therefre ρ V y d y 3 V d bg F ρ I = + y 6ε d 6ε The electric field frm E = V Ey a d y V d bg L ρ F = ρ ym + ε d 6ε NM HG HG KJ IO KJ P (b) Apply Gauss s law at y=0 V ρd ρ ˆ ˆ s = εea y y ay ε d 6ε Apply Gauss s law at y=d ρ V ρ ˆ ( ˆ s = εea y y ay) + ε 3 d d Different charges at tw plates N meaning in finding the capacitance

20 Bundary-value Prblems in Cylindrical Crdinates Laplace s eq. in cylindrical crds. F I V HG K J + + = V V r r r r r φ 0 Cheng; 3//009; 3-0 A cylindrical symmetry : V 0 =, φ Lnger in -directin than r-directin r F V I r HG r K J = 0 The slutin Vr bg = C ln r+ C : V 0. Example Tw semi-infinite cnductrs are arranged in a wedge-shaped cnfiguratin. Find the ptential fr the regins. (a) 0 < φ < α (b) α < φ < π V + - α φ We nte that V r = 0, and V = 0 Laplace s eq. 0 φ V bg φ = K φ + K Find K and K frm the bundary cnditins (a) At φ=0, 0 = K = 0 At φ = α, Vbg α = K α = V V Vbg φ = α φ fr 0 < φ < α (b) At φ = α, Vbg α = Kα + K = V At φ = π, V π K π K 0 V bg V φ b g b g = + = = π α π φ fr α < φ < π

21 Bundary-value Prblems in Spherical Crdinates Cheng; 3//009; 3- Example Tw cncentric spheres with dielectric material in between. Find the ptential in the sandwiched regin. Spherical symmetry V is independent f θ Laplace s eq. becmes The slutin is V and φ F I HG K J = d d dv d = C + C Apply bundary cnditins At = : V i At = : V Therefre V bg L = b g inm 0 C = + C i C = + C i V V + V V i fr i O P Methd f Images Uniqueness Therem: A slutin f Pissn s eq. satisfying a given bundary cnditin is a unique slutin Methd f Images: Bunding surfaces are replaced by image charges Example A psitive charge at a distance d abve a large grunded cnductr (a) Find the ptential in the regin y>0 (b) Find the induced charge distributin n the surface The nrmal way. Slve Pissn s eq. in y>0. Apply the bundary cnditins, V=0 at y=0 and infinity (Very difficult)

22 Cheng; 3//009; 3- eplace the cnductr by an image charge N change in the bundary cnditin N change in E fr y>0 The ptential due t the tw charges F I L O HG + KJ NM, y > x + by dg + x + by + dg + P b g = 0, y Vbx, y, g = V x, y, The electric field is btained frm E = V as xaˆ + ( y d ) aˆ + aˆ xaˆ + ( y + d ) aˆ + aˆ E = { x + ( y d) + } { x + ( y + d) + } x y x y 3/ 3/ The surface charge density (y=0) d ρs = εey = π x + d + Example Find d i e j / 3 and ρ i f the image charge, y > The cylindrical surface shuld be equiptential The image shuld be a line charge inside the cylinder We try ρi = ρl The electric ptential due t the line charge ρ l r ρ r l V E dr r dr ρl r = r = ln, r : er ptential pint r πε r πε r The electric ptential at M ρl r ρl r ρl ri V = ln ln ln Cnstant n the circle πε r πε r πε r i Enfrce OMP i t be similar t OPM ri di a = = = cnstant r a d d i = a d