Confirmation of Gauss s law

Transcription

1 Gauss s law Consder charge n a generc surface urround charge wth sphercal surface 1 concentrc to charge Consder cone of sold angle dω from charge to surface through the lttle sphere Electrc flu through lttle sphere: q d Φ ˆ ˆ 1 = E da = ( )( r r Ω d ) = r qdω r Electrc flu through surface : q R d Ω ˆr ˆ n d Φ ˆ ˆ = E da = ( r ) ( n ) = qd Ω = qd Ω R cos θ cos θ dφ =dφ 1 Æ Φ =Φ 1 =4πQ Φ= EdA = 4π Q s vald for ANY shape. encl G. colla MIT 8.0 Lecture 3 NB: Gauss s law only because E~1/r. If E ~ anythng else, the r would not cancel!!! Q r ˆn θ R 1 3 Confrmaton of Gauss s law Electrc feld of sphercal shell of charges: Q ˆ r E = r 0 Can we verfy ths epermentally? o utsde the shell n sd e the shell Charge a sphercal surface wth an de Graaf generator Is t charged? (D7 and D8) Is Electrc Feld radal? Does E~1/r, eg: φ~1/r? Neon tube on only when orented radally (D4) (D9?) G. colla MIT 8.0 Lecture 3 4

2 Confrmaton of Gauss s law () Cylndrcal shell postvely charged Gauus tells us that E = 0 nsde E > 0 outsde E=0 E>0 Can we verfy ths epermentally? Demo D6 Charge conduct ve spheres by nducton outsde the cylnder: one sphere wl l be and the other wll be -: t works because E outsde > 0 Try to do the same nsde nsde cylnder Æ nothng happens because E=0 (eplan nducton on the board) G. colla MIT 8.0 Lecture 3 5 Energy stored n E: queeng charges Consder a sphercal shell of charge of radus r How much work dw to squeee t to a radus r-dr? Guess the pressure necessary to squeee t: F QE Q P = = = E = Eσ A A A Q 1 Q E outsde = ; E 0 nsde = E surface = r r 1 Q σ P = E σ = σ = ( 4 π r σ ) = πσ r r We can now calculate dw: dw = Fdr = ( ) = ( πσ )(4 π r ) dr = πσ d (where d = PA dr 4 π ) r dr Rememberng that E =4 πσ created n dr G. colla MIT 8.0 Lecture 3 E dw = d 8 π y 6 3

3 sualaton of gradents Gven the potental φ (,y)=sn()sn(y), calculate ts gradent. φ(, y ) = cos( ) sn( y) ˆ sn cos yy ˆ φ (,y) grad φ The gradent always ponts uphll Æ E=-gradφ ponts downhll G. colla MIT 8.0 Lecture 3 15 sualaton of gradents: equpotental surfaces ame potental φ (,y)=sn()sn(y) φ (,y) NB: snce equpotental lnes are perpendcular to the gradent Æ equpotental lnes are always perpendcular to E G. colla MIT 8.0 Lecture

4 Dvergence n E&M (1) Consder flu of E through surface : 0 da da 1 0 new new 1new 1 1 Cut nto surfaces: 1 and wth new the lttle surface n between Φ= EdA = EdA 1 1 ew EdA n new = E da 1 EdA 1 ew EdA EdA n new = EdA EdA = Φ Φ 1 1 G. colla MIT 8.0 Lecture 3 17 Dvergence Theorem Let s contnue splttng nto smaller volumes If we defne the dvergence of E as Æ Φ= = = largen = largen = largen Φ = EdA = 1 = 1 = 1 largen ( Φ= ) Æ = 1 E da E da E lm 0 E Ed EdA = E d Dvergence Theorem (Gauss s Theorem) G. colla MIT 8.0 Lecture

5 Gauss s law n dfferental form mple applcaton of the dvergence theorem: EdA = Ed ( E 4 πρ ) d = 0 EdA = 4 π Q = 4 d π ρ Ths s vald for any surface : Comments: E = 4πρ Frst Mawell s equatons Gven E, allows to easly etract charge dstrbuton ρ G. colla MIT 8.0 Lecture 3 19 What s a dvergence? Consder nfntesmal cube centered at P=(,y,) Flu of F through the cube n drecton: Φ = nce Æ0 FdA ~ yf [ ( y,, ) F(, y, ) top bottom 1 F Φ = ( y ) lm [ F ( y,, ) F (, y, )] = y 0 P y y mlarly for Φ and Φ y F Φ = y and Fy Φ y = y y G. colla MIT 8.0 Lecture

6 Dvergence n cartesan coordnates FdA We defned dvergence as F lm 0 But what does ths really mean? F lm 0 y 0 0 F F y F y ( ) y = lm y 0 y 0 0 y = FdA F F F y Ths s the usable epresson for the dvergence: easy to calculate! G. colla MIT 8.0 Lecture 3 1 Applcaton of Gauss s law n dfferental form Problem: gven the electrc feld E(r), calculate the charge dstrbuton that created t 4 4π K 3 Er ( ) = π Krr ˆ for r<r and Er ( )= Rr ˆ for r>r 3 3 r Hnt: what connects E and ρ? Gauss s law. EdA = 4 π Q encl (ntegral form) E = 4 πρ ( dfferental form) In cartesan coordnates: E E y 4 K E π when r<r Æ E =... = y 0 when r>r phere of radus R wth constant charge densty K G. colla MIT 8.0 Lecture 3 11