Field and Wave Electromagnetic. Chapter.4

Transcription

1 Fel an Wave Electromagnetc Chapter.4 Soluton of electrostatc Problems

2 Posson s s an Laplace s Equatons D = ρ E = E = V D = ε E : Two funamental equatons for electrostatc problem Where, V s scalar electrc potental V ( ε V ) = ρ Rao Technology Laboratory

3 Posson s s an Laplace s Equatons Assumng ε constant for smple an homogeneous mea V = : ρ : Posson's equaton. ε Laplacan operator V V V V = + +, for cartesan coornate x y z 1 V 1 V ( r ) V = + +, for cylnrcal coornate r r r r φ z = ( V ) (sn V R + θ ) + V, for sphercal coornate R R R R snθ θ θ R snθ φ ρ = ε note : No free charge V = : Laplace's equaton Rao Technology Laboratory 3

4 Posson s s an Laplace s Equatons ex) Sphercal clou problem ρ = (a) R b, ρ= ρ ρ ρ V = = ε ε 1 V 1 V 1 V ρ + + = V= ( R ) (sn θ ) R R R R snθ θ θ R snθ φ ε By symmetry,, = = θ φ 1 V ( ρ ) R = R R R ε V ρ ( R ) = R R R ε V R 1 R b ρ = ρ ρ C = R + R = s sngular pont unless C 1 = 3ε Rao Technology Laboratory 4

5 Posson s s an Laplace s Equatons V E = V = R( ) R C1 =, snce E can not be nfnte at R= ρ E = R R, R b 3ε V R C1 6 ε ρ = + ( b) R b, ρ= V = 1 Vo Vo C ( R ) = R R R R = R Vo C Eo = Vo = R = R at R= b, E = E (homogeneous meum) R R C ρ ρ 3 = b C = b b 3ε 3ε ρ 3 E = R b, R b 3ε R Rao Technology Laboratory 5

6 Posson s s an Laplace s Equatons Total charge n the clou 4π 3 = Q Q= ρ b Eo R 3 4 πε R Vo b b ' ' V C, As R Vo C= R = ρ ρ 3ε R = + 3 R = Bounary conton at R = b 3 3 o ε ρ ρ b = b + C 3ε 6 ' 1 ε ρ ρ ρ 1 1 ρ C = b b = b ( ) = b ' 1 3ε 6ε ε 3 6 ε ρ 3b R ( ) 3ε V = Rao Technology Laboratory 6

7 Unqueness of Electrostatc Soluton Unqueness theorem A soluton of Posson's equaton that satsfes the gven bounary contons s a unque soluton proof of unqueness theorem charge conuctng boes wth surface S, S,..., S at specfe potental 1 S n Assume two solutons V an V of Posson's equaton n τ 1 ρ ρ V1 =, V = ε ε Assume that both V an V satsfy the same bounary conton on S, S,..., S 1 n 1 Rao Technology Laboratory 7

8 Unqueness of Electrostatc Soluton Defne V = V V V =, 1 then from, On conuctng bounares, the potentals t are specfe an = ( fa) = f A+ A f = + ( V V) V V ( V) ( V) ( V V ) v= V τ τ v = ( V V ) ns, s when n enote the unt normal outwar from τ. ( S, S1, S,..., Sn ) V over the conuctng bounares, V = VV 1,, V V, V but surface area S 3 R. R R R = τ lm ( V V ) ns V v= R s Rao Technology Laboratory 8

9 Unqueness of Electrostatc Soluton V for everywhere V v= V = only for τ n other wors V has the same value at all ponts n τ as well as on the conuctng surface V = throughout the volume τ. V = V 1 an there s only one possble soluton. Rao Technology Laboratory 9

10 Metho of mages V V V V = + + = for y> except at the pont charge. x y z The soluton V( x, y, z) shoul satsfy the followng contons V( x,, z) = Q V as R where Rs the stance to Q 4πε R At p ont P ( x ±, y ±, or z ± ), the potental V The potental functon s even wth respect to the x an z coornates V( x, y, z) = V( x, y, z), V( x, y, z) = V( x, y, z) very ffcult to satsfy all these conton Rao Technology Laboratory 1

11 Metho of mages From fgure (b), Q 1 1 V( x, y, z) = ( ) 4 R R πε + for y> = [ + ( ) + ] = [ + ( + ) + ] where, R x y z, R x y z satsfy all the contons mentone above. Ths s a soluton of ths problems by unquness theorem. E = V for y> Rao Technology Laboratory 11

12 Metho of mages F = F1+ F + F3 Q F1 = y 4 πε ( ) Q F = x 4 πε ( 1) Q F 3 = ( x1+ y ) 3 4 πε [( ) + ( ) ] 1 Q F = x + y πε 1 ( 1 + ) ( 1 + ) Note : Explan whether the force s n the recton to the conuctor or not? Rao Technology Laboratory 1

13 Lne Charge an Parallel Conuctng Cylner The mage must be a parallel lne charge nse the cylner n orer to make the cylnrcal l surface at r = a an equpotental t surface. Because of symmetry w.r.t. the lne OP, the mage lne charge must le somewhere along OP. At a pont P wth a stance from the axs. Two unknowns; ρ, Rao Technology Laboratory 13

14 Lne Charge an Parallel Conuctng Cylner Assume ρ = ρ ntellgent guess. l procee to check f ths fals to satsfy the B.C. or not. If ths satsfy all B.C.'s, then t s the only soluton by the unquness theorem. em Electrc fel ntensty an potental at a stance r from a lne charge ρl ρl E = r πε r r ρ r l 1 ρl r V = Err = r ln r πε = r πε r r note) reference pont for zer o potental r can not be, snce potental s fnte. Rao Technology Laboratory 14

15 Lne Charge an Parallel Conuctng Cylner The potental at a pont on or outse the cylnrcal surface: superposton of contrbutons by ρl an ρ, at a pont M ρl r ρl r ρl r VM = ln ln = ln πε r πε r πε r where reference pont for zero potental s at the same stance from ρl an ρ equpotental surface are specfe by r r = constant f an equpotental surface s to conse wth the cylnrcal surface (OM= a) Rao Technology Laboratory 15

16 Lne Charge an Parallel Conuctng Cylner a r cf) If an are constant an MOP common, then =constant. a r Snce ΔOMP an ΔOMP s smlar, r a = = = constant ( a,, fxe) r a r a can be constant over the cylnrcal surface then, P = Inverse pont r cf) As the pont M moves along the cylnrcal surface, r an r changes but ther rato s constant cf) By symmetry, parallel cylnrcal surface surrounng the orgnal lne charge ρ wth raus a an ts axs at a stance to the rght of P s also l equpotental surface. Rao Technology Laboratory 16

17 Ex 4-4) 4) Cross secton of two-wre wre transmsson lne Determne the capactance per unt length between two long gp parallel, crcular conuctng wres of raus a. the axes of the wre are seperate by a stance D. Rao Technology Laboratory 17

18 Ex 4-4) 4) Cross secton of two-wre wre transmsson lne Conuctng surface equpotental surfaces equpotental surfaces can be generate by two parallel lne charge + ρ, ρ seperate by a stance ( D ) =. Potental on 1 Potental on l l V V 1 = ρl a ln πε postve quantty. ρl a = ln πε negatve quantty. because (a<) ρl r r a (cf V = ln ), = = const. πε r r ρ f r s at the same stance from both ρl an l Rao Technology Laboratory 18

19 Ex 4-4) 4) Cross secton of two-wre wre transmsson lne ρ l V1 = ln πε a Q per unt length = ρl C ( per unt length) = Q πε = V 1 ln( ) a [ F / m] a = D = D, 1 ( = D± D a ) choose only +sgn ( D, a) πε πε C = = 1 ln ( D ) + ( D ) 1 cosh ( D / a ) a a cf) ln + 1 = cosh 1 x x x Rao Technology Laboratory 19

20 Pont Charge an Conuctng Sphere Conserng symmetry, assume the mage charge Q (negatve pont charge) nse the sphere an on the lne jonng O an Q. stance from orgn O Q Q to make the sphercal surface R= a a zero-potental surface. both an Q : unknowns Rao Technology Laboratory

21 Pont Charge an Conuctng Sphere At an arbtrary pont M on the equpotental surface V M 1 Q Q ( = + ) = 4πε r r r Q Q a = = const, = r Q Q a a Q = Q, = a E, an V can be calculate from two pont charge Q, Q Rao Technology Laboratory 1

22 Bounary value problem Develop a metho for solvng three-mensonal problems where the bounares, over whch the potental or ts normal ervatve s specfe, conce wth the coornate surfaces of an orthog onal, curvlnear system The metho of seperaton of varables Bounary value problems Drchlet problems : the value of potental s specfe everywhere on the bounares Neumann problems : the normal ervatve of the potental s specfe everywhere on the bounares Mxe bounary-value problems : the potental s specfe over the remanng ones. Rao Technology Laboratory

23 Cartesan coornate Laplace's equaton for scalar electrc potental V n Cartesan coornates V V V x y z + + =, Assume V( x, y, z) = X( x) Y( y) Z( z), where X( x), Y( y), Z( z) are functons of only xy,, an z, respectvely. Then, X ( x ) Y ( y) Zz ( ) Y ( y) Z( z) + X ( x) Z( z) + X ( x) Y ( y) = x y z ve by V( x, y, z), 1 X ( x ) 1 Y ( y ) 1 Z ( z ) + + = X ( x) x Y ( y) y Z( z) z 1 X( x) 1 Y( y) 1 Z( z) = k x x X ( x ) x Y ( y ) y + Z ( z ) z s nepenent of X( x) Y( y) Z( z) + k ( ), ( ), ( ) xx x = + k yy y = + k zz z = x y z kx, ky, kz :separaton constant Rao Technology Laboratory 3

24 Cartesan coornate Remns Possble soluton of X x k X x ''( ) + x ( ) = A, B, A1, B1, A, B can be etermne by the gven bounary conton Rao Technology Laboratory 4