Huiswerk Hoofstuk 23 Chapter 23 Homework

Transcription

1 Huiswerk Hoofstuk 23 Chapter 23 Homework 8 th / 8ste HRW 1, 2(-72 N m 2 /C, 24 N m 2 /C, -16 N m 2 /C,, -48 N m 2 /C), 6, 11, 17, 18( 8. µc,12 µc,-5.3 µc), 22 (, N/C), 27, 29, 35, 34, 37, 46(2.2 µc), 47, 51 9 th / 9 de HRW 1, 2 (-72 N m 2 /C, 24 N m 2 /C, -2 N m 2 /C,, -48 N m 2 /C), 4 ( N m 2 /C), 9, 2 ( 8. µc,12 µc,-5.3 µc), 19, 24 (, N/C) 29, 31, 33, 36 ( N/C up,, N/C down), 37, 44 (2.2 µc), 45, 49 1

2 Lading verspreidings L Charge distributions 1. Lineêre ladingverspreiding 1. Linear charge distribution vb. Lading eweredig versprei oor lang dun staaf/draad. λ q tot L ex. Charge distributed evenly over long thin rod/wire. 2. Oppervlak ladingverspreiding 2. Area charge distribution A vb. Lading eweredig versprei oor oppervlak van n plaat q σ tot A ex. Charge distributed evenly over an area of a plate. 3. Volume ladingverspreiding 3. Volume charge distribution vb. Lading eweredig versprei regdeur volume van n voorwerp. ρ q tot V Asphere Vsphere ex. Charge distributed evenly through volume of an object. 4π r π r 3 2

3 Lading verspreidings Beskou groot aantal puntladings tesame, kontinu versprei op verskillende maniere: Verspreiding / Distribution Puntlading Single charge Lineêre ladingdigtheid Linear charge density Oppervlak ladingdigtheid Surface charge density Volume ladingdigtheid Volume charge density Simbool / Symbol q λ σ ρ Charge distributions Consider large numbers of point charges together, distributed continously in various ways: Definisie/ Definition λ σ ρ total charge length total charge area total charge volume SI eenheid / SI unit C C/m C/m 2 C/m 3 3

4 Elektriese Vloed Electric Flux Beskou plat plaat in n univorme elektriese veld: θ A E Consider flat plate in a uniform electric field: Elektriese vloed: SI eenheid: N m 2 /C Φ E i A EAcosθ Electric flux SI unit: N m 2 /C A area of plate (m 2 ) area vector (direction of plane of A) A oppervlak van plaat (m 2 ) A area vektor A A (rigting van A aan vlak van A) to Φ hoeveelheid elektriese veld deur loodregte oppervlak grootte van A Φ amount of electric field perpendicular through surface area A 4

5 Elektriese Vloed E Electric Flux aan plaat Φ to plate Φ E A θ 9 Φ E i A EAcos9 E E aan plaat Φ EA E to plate Φ EA E A θ Φ E i A EAcos EA 5

6 Gauss se wet Gauss se wet is ekwivalent aan Coulomb se wet. Word gebruik om voordeel te trek uit simmetrie situasies gedurende oplos van elektrostatiese probleme. Gauss oppervlak is n hipotetiese geslote oppervlak wat sekere lading insluit. Gauss se wet gee die verhouding tussen die E-veld by punte op n geslote Gauss oppervlak en die ingeslote netto lading deur daardie q 1 oppervlak. q 2 Gauss law Gauss law is equivalent to Coulomb s law. Is used to take advantage of symmetry situations to solve electrostatic problems. Gauss surface is a hypothetical closed surface enclosing some charge. q 3 Gauss law gives a relationship between the E- field at points on a closed Gaussian surface and the net charge enclosed by that surface ε Φ tot qtot 6

7 Vloed van n E-veld Flux of an E-field Arbitrêre Gauss oppervlak Klein A geen ronding Area vektor A oppervlak, wys na buitekant, uitwaarts E is uniformig oor A, indien A klein genoeg is Φ tot E A Arbitrary Gaussian surface Small squares A no curvature Area vector A surface, away from interior, outwards E is uniform over A, if A is small enough Φ E A EAcosθ 1. θ > 9, E A < 2. θ 9, E A 3. θ < 9, E A > A da Φ E da 7

8 Voorbeeld 1 Example 1 Die figuur wys n Gauss-kubus met syoppervlak A, geplaas in n univorme elektriese veld wat n positiewe rigting langs die z as het. In terme van E en A, wat is the vloed deur (a) die voorkant (wat in die xy vlak is), (b) die agterkant, (c) die bokant, en (d) die hele kubus? The figure here shows a Gaussian cube of face area A immersed in a uniform electric field that has a direction in the positive z axis. In terms of E and A, what is the flux through (a) the front face (which is in the xy plane), (b) the rear face, (c) the top face, and (d) the whole cube? 8

9 Gauss se wet Gauss se wet verbind die netto vloed Φ van n E-veld deur n geslote oppervlak met die netto lading q enc wat deur die oppervlak ingesluit word. ε Φ q ε E i da tot enc Gauss law Gauss law relates the net flux Φ of an E-field through a closed surface to the net charge q enc that is enclosed by that surface. q enc S1 : q enc > Φ outwards / na buite S2 : q enc < Φ inwards / na binne S3 : q enc Φ S4 : q enc Φ 9

10 Voorbeeld 2 Example 2 n Elektriese veld wat beskryf word deur E 4î -3(y 2 + 2) ĵ sny deur die Gauss-kubus in die figuur (E is in N/C en x is in m). Wat is die elektriese vloed deur (a) die boonste oppervlak, (b) die onderste oppervlak, (c) die linkerkant en (d) die agterkant? (e) Wat is die netto elektriese vloed deur die hele kubus? An electric field given by E 4î -3(y 2 + 2) ĵ pierces the Gaussian cube in the figure (E is in N/C and x is in m). What is the electric flux through (a) the top surface, (b) the bottom surface, (c) the left face and (d) the back face? (e) What is the net electric flux through the whole cube? 1

11 Voorbeeld 2B Watter netto lading word omsluit deur die Gauss-kubus in die figuur? Example 2B What net charge is enclosed by the Gaussian cube in the figure? Voorbeeld 3 Example 3 q 1 q nc, q 2 q nc, q nc Total flux through S? 11

12 Gauss se wet Coulomb se wet Lei Coulomb se wet af m.b.v. Gauss se wet: Konstrueer konsentriese Gauss-oppervlak om positiewe punt lading q Verdeel oppervlak in differensiaal area segmente da da by enige punt aan oppervlak en na buite gerig E ook aan oppervlak en na buite gerig θ tussen E en da is nul Gauss law Coulomb s law Derive Coulomb s law from Gauss law: Construct concentric Gaussian surface around point charge q Divide surface into differential area segments da da at any point to surface and directed outward E also to surface and directed outward θ between E and da is zero 12

13 Gauss Coulomb se wet Gauss se wet: Maar vir elke area segment op Gauss oppervlakte Ingeslote lading bestaan slegs uit q ε E da q enc E da E dacos E da ε E da q enc qenc + q Gauss law Coulomb s Gauss law: But, For each area segment over Gaussian surface Enclosed charge consist of only q E-veld is konstant oor buite oppervlak Integraal van da oor geslote oppervlak gee oppervlak van sfeer: ε E da q ε E π r q 2 (4 ) E da 4π r 1 4πε 2 q r 2 E-field is constant over outer surface Integral of da over closed surface results in area of sphere 13

14 Gauss se wet: silindriese simmetrie E op afstand r van lang reguit gelaaide staaf? E radiaal na buite Φ vir bo- & onderkant λ linieêre ladingdigtheid Gauss law: cylindrical symmetry E at distance r from a long straight charged rod? E radially outward Φ for top & bottom λ linear charge density Φ deur silindriese oppervlak: Φ EAcosθ q enc ε Φ EA cylinder E( 2π rh) Φ through cylindrical surface: Lyn van lading λ h ε E(2π rh) λ E 2πε r Line of charge 14

15 Gauss se wet: vlak simmetrie Gauss law: planar symmetry Dun nie-geleidende plaat met uniforme ladingverspreiding σ Gelaaide isolerende plaat ε E da qenc ε EA + EA) σa ( σ E q tot A σ 2ε Thin non-conducting sheet with uniform charge distribution σ Insulated sheet of charge 15

16 Voorbeeld 4 Example 4 Figuur wys dwarsdeursnitte deur twee groot, parallel, nie-geleidende plate met identiese verspreidings van positiewe ladings met oppervlak ladings digtheid σ. Wat is E by punte (a) bokant die plate, (b) tussen (c) en onder die plate? Figure shows cross-sections through two large, parallel, non-conducting sheets with identical distributions of positive charge with surface charge density σ. What is E at points (a) above the sheets, (b) between them, and (c) below them? 16

17 Gelaaide geleiers 1. In n toestand van statiese ewewig is die elektriese veld binne n geleier null. 2. Die elektriese veld van n geleier in statiese ewewig is loodreg op die oppervlak van die geleier. 3. Indien n geleier in statiese ewewig verkeer, is al die netto lading op die oppervlak van die geleier. 4. Indien n netto lading binne die ruimte van n hol geleier gebring word, verskyn n ekwivalente lading op die buitekant van die geleier. Charged conductors 1. The electric field inside a conductor is zero in a state of static equilibrium. 2. The electric field is perpendicular to the surface of a conductor when in static equilibrium. 3. When a conductor is in static equilibrium, all the nett charge will be on the surface of the conductor. 4. When a nett charge is put inside the space of a hollow conductor, an equivalent charge appears on the outside surface of the conductor. 17

18 Gelaaide geleiers Indien n lading op n geisoleerde geleier geplaas word sal hierdie lading as geheel na die oppervlak van die geleier beweeg. ε E da qenc E binne geleier Φ deur Gauss oppervlak q enc binne geleier alle lading op buitekant van geleier Selfde argumente vir geleier met holte alle lading steeds op buitekant van geleier Charged conductors If an charge is placed on an isolated conductor this charge will move entirely to the surface of the conductor. ε E da qenc E inside conductor Φ through Gaussian surface q enc inside conductor all charge on outside of conductor Same arguments for conductor with cavity all charge still on outside of conductor 18

19 Voorbeeld 5 Example 5 Die diagram wys n dwarsdeursnit van n sferiese metaal dop met binne radius R. n Puntlading van -5. µc word dan geplaas n afstand R/2 vanaf die middelpunt. (a) Indien die dop elektries neutraal is, wat is die geïnduseerde ladings op die binne- en buitekant van die dop? (b) Is die lading uniform versprei aan die binnekant? (c) Wat is die elektriese veld patroon binne en buite die dop? The figure shows a cross section of a spherical metal shell of inner radius R. A point charge of -5. µc is located at a distance R/2 from the center of the shell. (a) If the shell is electrically neutral, what are the induced charges on its inner and outer surfaces? (b) Are the charges uniformly distributed on the inside? (c) What is the field pattern inside and outside the shell? 19

20 E-veld buite n geleier Beskou n klein area op n gelaaide geleier met ladingsdigtheid: σ lading per eenheids area Kies silindriese pyp as Gauss Oppervlak. E geleier se oppervlak E aan deksel Vloed deur deksel EA q enc σ A ε E da qenc E-field outside a conductor Consider a small area on a charged conductor with surface charge density: σ charge per unit area Choose cylindrical Gaussian surface. E conductors surface E to cap Flux through cap EA q enc σa Toepaslik slegs naby die oppervlak van enige geleier ε EA σa σ E ε Applicable only close to any conductor s surface 2

21 Enkel geleidende plaat Ladingverspreiding op die oppervlakte van plaat. Geen eksterne elektriese veld Lading eweredig versprei: σ 1 E σ 1 / ε net buite geleier Positiewe lading E weg van plaat. Negatiewe lading E na die plaat. Twee geleidende plate Lading herversprei Lading net op binne oppervlak 2σ 1 σ Tussen die plate E 2σ ε Links & Regs E 1 σ ε Single conducting plate Charge distribution on surface of plate. No external electric field Charge distributed evenly: σ 1 E σ 1 / ε just outside conductor Positive charge E away from plate. Negative charge E towards the plate Two conducting plates Redistribution of charge Charge on inner surface only 2σ 1 σ Between the plates Left & right 21

22 Gauss se wet: sferiese simmetrie OPPERVLAK STELLING 1: Uniforme ladingverspreiding op buite oppervlak van n dop totale lading oor oppervlak gekonsentreer by middelpunt as puntlading. Dun uniform gelaaide sferiese dop met radius R Pas Gauss se wet toe op sferiese Gauss oppervlakte (r R) ε φ ε E da q enc ε E (4πr ) q Gauss law: spherical symmetry SHELL THEOREM 1: Uniform charge distribution on ouside surface of a shell total charge over shell concentrated at center as point charge. 2 E 2 4πε r (r R) R ε E da q enc 1 r q Thin uniformly charged spherical shell of radius R Apply Gauss law on spherical Gausian surface (r R) 22

23 OPPERVLAK STELLING 2: Lading binnekant dop met uniforme ladingverspreiding geen elektrostatiese krag daarop. SHELL THEOREM 2: Charge inside shell of uniform charge no electrostatic force on it. Dun uniform gelaaide sferiese dop met radius R Pas Gauss se wet toe op sferiese Gauss oppervlakte (r < R) R r Thin uniformly charged spherical shell of radius R Apply Gauss law on spherical Gausian surface (r < R) Dieselfde argumente soos voorheen q enc ε E da q E F enc Same arguments as previously (r < R) 23

24 Nie-uniform gelaaide sfeer Ladingdigtheid ρ verander met r Non uniformly charged sphere Charge density ρ, changes with r r > R: r R: E 1 4π ε q tot 2 r E 1 4π ε q r 2 Uniform gelaaide sfeer Ladings digtheid ρ, konstant ρ 4 3 q tot V q tot π R 3 R Uniformly charged sphere Charge density ρ, constant q tot 2 1 E 4π ε r r > R E q tot r 3 4π ε R r R 24

25 Voorbeeld 6 Example 6 Die figuur wys n sferiese gelaaide dop met univorme volume ladingsdigtheid ρ. Plot E a.g.v. die dop vir die afstande r van die middelpunt van die dop vir die afstand nul tot 3 cm. Veronderstel dat ρ C/m 3, a 1 cm en b 2 cm. The figure shows a spherical shell of charge with uniform volume charge density ρ. Plot E due to the shell for distances r from the centre of the shell ranging from zero to 3 cm. Assume that ρ C/m 3, a 1 cm and b 2 cm. 25