Today in Physics 217: charged spheres

Transcription

1 Today in Physics 7: chaged sphees Finish example fom Wednesday. Field fom a chaged spheical shell, calculated with Coulomb s Law. z de dq = σ da ˆ da = R sinθ dθ dφ Same, calculated with Gauss Law. θ Analogy with gavity: Gauss Law fo gavity. x R φ y 0 Septembe 00 Physics 7, Fall 00

2 Using Coulomb s Law: de Beak the plane into annuli with adius and width d, and beak the annuli into segments of width dφ. The chage of each segment is α z dq = σddφ φ Hoizontal components dq of field fom segments at φ and φ+π cancel, and thei vetical components add, so above the plane, we have: π dq σ de = cosαzˆ = z ddφzˆ z + z u ( ) E = σ zzˆ dφ + z d = πσ zzˆ u du = πσ zzˆ = πσ zˆ 0 0 z 0 Septembe 00 Physics 7, Fall 00 z

3 Using Gauss Law Fist note that E must point pependicula to, and away fom, the plane, since the plane is infinite and thee s no diffeence between the view to the ight and the view to the left. Then daw a cylinde, bisected by the plane. By symmety, E is pependicula to the aea element vectos on the cylinde walls, paallel to those on the cicula faces, and constant on those faces, so E d a = Eπ = 4πQenclosed = 4 π σ, o E =± πσ zˆ. Hade setup (finding and exploiting symmety), easie math. z E z 0 Septembe 00 Physics 7, Fall 00 3

4 Electic fields fom spheically-symmetical chage distibutions Today we will pove two impotant, though pehaps intuitively obvious, facts about spheical chage distibutions: The field outside a unifomly-chaged spheical shell is the same as that fom a point chage of the same magnitude, the same distance away as the sphee s cente. The field inside a unifomly-chaged spheical shell is zeo. The poof will seve also as anothe useful example of the application of Coulomb s and Gauss laws to the detemination of electic fields fom specified chage distibutions. 0 Septembe 00 Physics 7, Fall 00 4

5 Coulomb s Law example: field fom a unifomlychaged spheical shell Giffiths, poblem.7: What is the electic field a distance z away fom the cente of a spheical shell with adius R and unifom suface chage density σ? z de θ dq = σ da ˆ da = R sinθ dθ dφ R φ y x 0 Septembe 00 Physics 7, Fall 00 5

6 Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) In the plane at azimuth φ, it can be seen moe easily that = R sin θ + zrcosθ ( ) = R sin θ + z + R cos θ Rzcosθ = R + z Rzcosθ Conside two aea elements at azimuth φ and φ + π: as befoe, the hoizontal components of thei contibution to E cancel, and the vetical components add. dq at φ + π d R cosθ α R sinθ θ de E ( φ ) de( φ + π) R dq at φ 0 Septembe 00 Physics 7, Fall 00 6

7 Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) So de = zˆ dq cosα σr sinθ dθ dφ z Rcosθ = zˆ R + z Rzcosθ R + z Rzcosθ sinθ ( z Rcosθ ) = zˆ σr dθ dφ ( R + z Rzcosθ ) 3 E = π π 0 0 ( z Rcosθ ) sinθ zˆ σr dφ dθ ( R + z Rzcosθ ) 3 The fist integal is tivial: it just comes out to π. 0 Septembe 00 Physics 7, Fall 00 7

8 Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) Fo the second, substitute w = cos θ, dw= sin θ dθ, w= : E = zˆ πσ R ( z Rw) ( R + z Rzw) Beak this integal in two. Fo the fist one, substitute 3 dw u = R + z Rzw, du = Rzdw, u= R + z + Rz R + z Rz R + z + Rz 3 z dw = 3 u du R R + z Rz ( R + z Rzw) 0 Septembe 00 Physics 7, Fall 00 8

9 Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) R + z + Rz 3 u du= u R + z + Rz R R + z Rz R R + z Rz = R R + z Rz R + z + Rz The second half of the integal needs to be done by pats. Take Rdw u= w dv = R + z Rzw ( ) 3 du = dw v = z R z Rzw + (as we just saw) 0 Septembe 00 Physics 7, Fall 00 9

10 C Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) udv = uv vdu C C Rw 3 + ( R + z Rzw) w dw = z R z Rzw In the last tem, use (again) dw z R + z Rzw u = R + z Rzw, du = Rzdw, u= R + z + Rz R + z Rz 0 Septembe 00 Physics 7, Fall 00 0

11 Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) and it becomes dw = z R + z Rzw Rz R + z + Rz R + z Rz R + z + Rz = u Rz R + z Rz = Rz u du ( ) R z Rz R z Rz So, putting all these tems togethe (and factoing out as we do), we get z 0 Septembe 00 Physics 7, Fall 00

12 Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) πσ R z E = zˆ z R R z Rz R z Rz z + R z Rz R z Rz ( ) R + z + Rz R + z Rz R This looks like a mess until you notice that ( ) R + z + Rz = z+ R = z+ R ( ) R + z Rz = z R = zr Positive, since they epesent the length of, which is always positive. 0 Septembe 00 Physics 7, Fall 00

13 Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) E πσ R z = zˆ z R z R z R + z ( z R z R ) z R z+ R R πσ R z = zˆ z R z R z+ R z R z R + z + z R z+ R R z R z+ R 0 Septembe 00 Physics 7, Fall 00 3

14 Coulomb s Law example: field fom a unifomlychaged spheical shell (continued) which gives us, finally, πσ R z R z+ R E = zˆ + z z R z+ R Two cases: z lage than, o smalle than, R. (P outside, inside) Lage (outside): z R 4 z + R πσ = = E = zˆ R = zˆ Q z R z+ R z z Smalle (inside): means z R = Rz, so z R z+ R z R + R z + = = 0 E = 0 R z z+ R R z Behaves like a point chage at the sphee s cente. 0 Septembe 00 Physics 7, Fall 00 4

15 Gauss Law example: field fom a unifomlychaged spheical shell (continued) Fist note that the field must be spheically symmetic as well, and point adially outwad o inwad that is, E is pependicula to all sphee s centeed at the same point as the chaged sphee. So daw two Gaussian sphees, one inside and one outside: E d a = 4π Qenclosed > R: ( )( ) ( ) Q E π = π πr σ = 4πQ E = ˆ < R: ( )( E π ) 4 = 0 E = 0 0 Septembe 00 Physics 7, Fall 00 5

16 Gauss Law fo gavity Newton was the fist to ealize these esults, in the context of the othe / foce, gavity. He convinced himself by use of a poof simila to ou Coulomb s law demonstation, Gauss still not having been bon by then. We could have saved Newton a lot of touble by pointing out the following. The foce of gavity on a mass M fom a mass m is F = ˆGmM Gavitational foces supepose: the foce on M fom N chages is N Gm M Gm M m = + + = i i= i ( ) ˆ ˆ i M G ˆ Mg( ) F 0 Septembe 00 Physics 7, Fall 00 6

17 Gauss Law fo gavity (continued) Fo a continuous distibution of mass (density ρ ), the gavitational field g is obtained by letting N : ˆ g( ) = G ρ( ) dτ V Take the divegence of both sides, and cay out the esulting integal on the RHS, as we did on Wednesday, and we get g = 4πGρ ( ) Now integate this esult ove volume, and use the divegence theoem, as we also did on Wednesday : g d a = 4π GM enclosed 0 Septembe 00 Physics 7, Fall 00 7 ( )