Gauss’s Law

1. Gauss’s Law Alan Murray

2. a a q b b a a b Revision : Vector Dot (Scalar) Product a.b = ab cos(90) = 0 a.b = ab cos q a.a = aa cos(0) = a² a.b = ab cos(0) = ab In Cartesian co-ordinates, a.b = (ax,ay,az).(bx.by,bz) = axbx + ayby + azbz Alan Murray – University of Edinburgh

3. 2a -2a a Revision : Vector x Scalar In Cartesian co-ordinates, for example, 2a = 2(ax,ay,az) = (2ax,2ay,2az) Alan Murray – University of Edinburgh

4. Gauss’s Law : Crude Analogy • Try to “measure” the rain on a rainy day • Method 1 : count the raindrops as they fall, and add them up • cf Coulomb’s Law • Method 2 : Hold up an umbrella (a “surface”) and see how wet it gets. • cf Gauss’s Law • Method 1 is a “divide –and-conquer” or “microscopic” approach • Method 2 is a more “gross” or “macroscopic” approach • They must give the same answer. Alan Murray – University of Edinburgh

5. 1C 1C 1C Electric Field Lines These are all “correct” as E-field lines are simply cartoons For now, adopt a drawing scheme such that 1C = 1 E-line. Alan Murray – University of Edinburgh

6. 8C 16C 32C Lines of Electric Field How many field lines cross out of the circle? 8C Þ 8 lines 16C Þ 16 lines 32C Þ 32 lines Alan Murray – University of Edinburgh

7. 8C 16C 32C Lines of Electric Field How many field lines cross out of the surface? 8C Þ 8 lines 16C Þ 16 lines 32C Þ 32 lines Alan Murray – University of Edinburgh

8. Gauss’s Law : Cartoon Version • The number of electric field lines leaving a closed surface is equal to the charge enclosed by that surface • S(E-field-lines) a Charge Enclosed N Coulombs ÞaN lines Alan Murray – University of Edinburgh

9. 8C 16C 32C Lines of Electric Field How many field lines cross out of the surface? 8C Þ 0 lines 16C Þ 0 lines 32C Þ 0 lines i.e. charge enclosed = 0 Alan Murray – University of Edinburgh

10. Gauss’s Law Proper (L) • S(E-lines) proportional to (Charge Enclosed) • D = εE • òòD.ds = òòòr(r)dv = òòòr(r)dxdydz • òòD.ds = charge enclosed • ε= ε0 = 8.85 x 10-12 in a vacuum Alan Murray – University of Edinburgh

11. Digression/RevisionArea Integrals This area gets wetter! Alan Murray – University of Edinburgh

12. Rainfall Rainfall ds ds Area Integrals – what’s happening? This area gets wetter! Alan Murray – University of Edinburgh

13. Rainfall Rainfall ds ds Area Integrals – what’s happening? Clearly, as the areas are the same, the angle between thearea and the rainfall matters … Alan Murray – University of Edinburgh

14. Rainfall, R Rainfall, R Area Integrals – what’s happening? ds ds Extreme casesat 180° - maximum rainfallat 90°, no rainfall Alan Murray – University of Edinburgh

15. Flux of rain (rainfall) through an area ds • Fluxrain = R.ds • |R|´|ds|´cos(q) • Rds cos(q) • Fluxrain = 0 for 90° … cos(q) = 0 • Fluxrain = -Rds for 180° … cos(q) = -1 • Generally, Fluxrain = Rds cos(q) • -1 < cos(q) < +1 Alan Murray – University of Edinburgh

16. Area Integrals : Take-home message • Area is a vector, perpendicular to the surface • Calculating flux of rain, E-field or anything else thus involves a scalar or “dot” product a.b = abcos(q) • This is what appears in a surface integral of the form òòD.ds, or òòR.ds, which would yield the total rainfall on whatever surface is being used for integration (here, the hills!) Alan Murray – University of Edinburgh

17. ds ds r L ds Gauss’s law – Worked ExampleLong straight “rod” of charge Construct a “Gaussian Surface” that reflects the symmetryof the charge - cylindrical in this case, then evaluate òòD.ds E, D E, D rlCoulombs/m Alan Murray – University of Edinburgh

18. E, D ds r Evaluate òòD.ds • òòD.ds = òò D.dscurved surface +òò D.dsflat end faces • End faces, D & ds are perpendicular • D.ds on end faces = 0 • òò D.dsflat end faces = 0 • Flat end faces do not contribute! Alan Murray – University of Edinburgh

19. D & ds parallel, D.ds= |D|´|ds| = Dds E, D ds rlCoulombs/m L ds Evaluate òòD.ds • òòD.ds = òò D.dscurved surface only Alan Murray – University of Edinburgh

20. D has the same strengthD(r) everywhere on thissurface. E, D rlCoulombs/m L Evaluate òòD.ds • òò D.dscurved surface only = òò Dds Alan Murray – University of Edinburgh

21. 2pr rlCoulombs/m r L Evaluate òòD.ds D • òò D.dscurved surface only = òò ds • = Dòò ds = D ´ area of curved surface • = D ´ 2 p rL • So 2Dp rL = charge enclosed • Charge enclosed? • Charge/length ´ length L = rl ´L Alan Murray – University of Edinburgh

22. x x Evaluate òòD.ds • òò D.ds = charge enclosed • 2pDr´L = rl ´L • D(r) = rl2pr • D(r) =rl âr2pr Alan Murray – University of Edinburgh

23. Discussion • |D| is proportional to 1/r • Gets weaker with distance • Intuitively correct • D points radially outwards (âr) • |D| is proportional to rl • More charge density = more field • Intuitively correct Alan Murray – University of Edinburgh

24. Spherical charge distribution ra r-2, r-3, e –r … Choose a spherical surface for integration Then D and ds will once again be parallel on the surface Check it out! r Other forms of charge distribution? Alan Murray – University of Edinburgh

25. Solid sphere of 4C, uniformly distributed r>R R Spherical charge distribution- worked example Alan Murray – University of Edinburgh

26. Sheet of charge Mirror symmetry Choose a surface that is symmetric about the sheet Then D and ds will once again be parallel or perpendicular on the surfaces Check it out! D, ds ` ds D Other forms of charge distribution? ` Alan Murray – University of Edinburgh

27. Gauss’s Law This is Maxwell’s first equation And we can have Maxwell’s second equation for free! As there is not such thing as an isolated “magnetic charge”, no Gaussian surface can ever contain a net “magnetic charge” – they come in pairs (North and South poles). Alan Murray – University of Edinburgh