For the flux through a surface: Ch.24 Gauss s Law In last chapter, to calculate electric filede at a give location: q For point charges: K i r 2 ˆr

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1 Ch.24 Guss s Lw In lst hpter, to lulte eletri filed t give lotion: q For point hrges: K i e r 2 ˆr i dq For ontinuous hrge distributions: K e r 2 ˆr However, for mny situtions with symmetri hrge distribution, Guss s Lw provides onvenient nd powerful tool to lulte eletri field letri Flux : The number of eletri field lines penetrting surfe. Are=A Norml For the flux through surfe: = lim A i 0 i A i = da i surfe For losed surfe (lso lled Gussin Surfe): A lwys points from inside to outside., if n eletri field line goes in., if n eletri field line goes out. The eletri flux through Gussin surfe: = da = n da = Aos = A Define: A = A ˆn Slr Produt: A B = ABos () (b) () = 0 > 0 < 0 net q = 0 net q > 0 net q < 0 Here we see the ler onnetion between nd hrges inside Gussin surfe.

2 xmple: The net eletri flux going through ubi Gussin surfe in uniform eletri field, with no hrge distribution inside the surfe. = top Notie tht: top = bottom beuse bottom = front = bk = 0 A, os = 0. front bk right left C = 0 C = q 1 / 0 q 1 q2 q C = ( q2 + q ) / 0 right = A right = Aos0 =A left = Aleft = Aos180 =A = right + left = A A = Guss s Lw The net flux through ny losed surfe is C = da= q in 0 q in is the net hrge inside the losed surfe. 0 = 1 4k e = C 2 / N m 2, permittivity of free spe. Q: Will C hnge, if (1) The hrges inside the surfe re moved? (2) The surfe hnges shpe? () The surfe hnges size? (4) The hrges re doubled? (5) An eletri dipole is dded inside the surfe?

3 Generl Proedures For Applying Guss s Lw: (1) Find the DIRCTION of the eletri field bsed on symmetry. (2) Choose suitble Gussin surfe with prt of its surfe either prllel or perpendiulr to the eletri field. () Clulte the eletri flux pssing through the Gussin surfe C = da (4) Clulte the totl net hrge, q in, inside the Gussin surfe. (5) Clulte the mgnitude of eletri field using Guss s 24. Applition of Guss s Lw to Chrged Insultors xmple: An insulting sphere of rdius hs uniform hrge density nd totl positive hrge Q. (1) Clulte outside the sphere (R>) Use spheril Gussin surfe is pointed rdil outwrd. is prllel to = da= d A everywhere da = Q / 0 da= (4R 2 ) = Q / 0 r R lw: C = da= q in 0 = Q / 4 0 R 2 Q = k e R 2 (for R>) (Similr to point hrge, but we obtined it from Guss s Lw) (2) Clulte inside the sphere (r<) 4 r Totl hrges inside sphere of rdius r: = ( ) q in Guss s Lw: da = da = ( 4r ) = qin / 0 2

4 q r in = = = r r 4 r xmple: Find the eletri field due to nononduting, infinite plne with uniform hrge per unit re. By definition: 4 = Q / keq = r (for r<) A All fields must be norml to the plne A good hoie of Gussin surfe: ylindril surfe. No eletri flux from the side of ylinder. Guss s Lw: q in = 2A = = 0 = 2 0 A 0 r Notie tht: does not depend on distne. Therefore, n infinite lrge hrged plne retes uniform eletri field.

5 24.4 Condutors in letrostti quilibrium letri field just outside hrged ondutor = 0 = Condutors: in ondutor, hrges (eletrons) n move =0 A freely. letrostti equilibrium: there is no net motion of hrge within the ondutor. Properties of Condutor in letrostti quilibrium (1) = 0 everywhere inside ondutor. is norml to the surfe. (2)Any hrge n only distribute on ondutor surfe. Inside the ondutor, = 0 () just outside hrged ondutor is perpendiulr to the Apply Guss s Lw: surfe nd hs mgnitude / 0. (4)Chrge tends to umulte t shrp points. = q in = A n da = n A = q in 0 n = 0 Notie tht, for uniform hrged insultor plne, n = 2 0 Why they re different by ftor of 2?

6 xmple. A ondutor sphere with net hrge 2Q inside ondutor spheril shell with net hrge -Q -Q Q b () Determine hrge distribution. (b) letri Fields t 4 lotions: 1. 1 = 0 2. Apply Guss s Lw: 2 A = 2 (4r 2 ) = 2Q 0 2 =. = 0 2Q 4 0 r 2 = k 2Q e r 2 (for < r < b) 4. 4 = k e Q r 2 (for r > )