1. ELECTRICITY PHY1013S GAUSS’S LAW Gregor Leigh gregor.leigh@uct.ac.za

2. ELECTRICITY GAUSS’S LAWPHY1013S 2 GAUSS’S LAW Learning outcomes: At the end of this chapter you should be able to… Calculate the electric flux through a surface. Use Gauss’s Law to calculate the electric field due to symmetric charge distributions. Use Gauss’s Law to determine the charge distribution on hollow conductors in electric fields.

3. ELECTRICITY GAUSS’S LAWPHY1013S 3 FLUX is the unit vector normal to a surface. is the surface’s area vector. The amount of flow, , through the surface depends on: the magnitude of the velocity, v ; the area of the surface, A ; the angle between and. Replacing the velocity vector with the electric field vector, we get: (  is a maximum when  = 0°; a minimum for  = 90°.) Hence:  = vAcos  or just: 

4. ELECTRICITY GAUSS’S LAWPHY1013S 4 FLUX For a surface made up of small elements, each  A, the total flux is The electric flux through a Gaussian surface is proportional to the net number of electric field lines passing through that surface. For a closed (Gaussian) surface, in the limit  A  0, [N m 2 /C]

5. ELECTRICITY GAUSS’S LAWPHY1013S 5 1 Gaussian surface 2 3

6. ELECTRICITY GAUSS’S LAWPHY1013S 6  < 0 1 11 1 2 3 Gaussian surface

7. ELECTRICITY GAUSS’S LAWPHY1013S 7 1 2 3 Gaussian surface  = 0 2 22

8. ELECTRICITY GAUSS’S LAWPHY1013S 8 1 2 3 Gaussian surface  > 0 3 33

9. ELECTRICITY GAUSS’S LAWPHY1013S 9 A cylindrical Gaussian surface of radius R is placed in a uniform electric field, with the cylinder axis parallel to the field. What is the (net) flux of the field through this surface?  = 0°  = 90°  = 180° a b c ……

10. ELECTRICITY GAUSS’S LAWPHY1013S 10 – + When will the net flux ever NOT be zero?

11. ELECTRICITY GAUSS’S LAWPHY1013S 11 GAUSS’ LAW The net electric flux through a Gaussian surface is proportional to the net charge enclosed: Substituting for , Q in is the net, enclosed charge. is the total field through the surface. These equations are valid only in vacuum (or air). Gauss’s Law is both easier to use and more universal/fundamental than Coulomb’s Law. Notes:  0  = Q in   Q in

12. ELECTRICITY GAUSS’S LAWPHY1013S 12 S1S1 S2S2 S4S4 GAUSS’S LAW: MULTIPLE CHARGES – + S3S3

13. ELECTRICITY GAUSS’S LAWPHY1013S 13 1.Draw the situation. 2.Choose a Gaussian surface appropriate to the symmetry. 3.Apply Gauss’s law: USING GAUSS’S LAW TO DETERMINE ELECTRIC FIELDS

14. ELECTRICITY GAUSS’S LAWPHY1013S 14 q + (For a point charge, choose a concentric sphere as a Gaussian surface.) The electric field is constant over the surface and directed radially outwards, so GAUSS  COULOMB Apply Gauss’s law… Draw the situation… Choose a Gaussian surface appropriate to the symmetry… r

15. ELECTRICITYGAUSS’S LAWPHY1013S ZERO FIELD If there is no electric field at all, quite obviously the net flux through any Gaussian surface is also zero. (Duh!)

16. ELECTRICITY GAUSS’S LAWPHY1013S 16 CHARGE ON AN ISOLATED CONDUCTOR Any excess charge added to an isolated conductor moves entirely to the external surface of that conductor. The field inside the metal must be zero (otherwise there would be currents inside the conductor)… Therefore there is no flux through the Gaussian surfaces…  there can be no charge within the Gaussian surfaces…  all the charge must lie outside the Gaussian surfaces… i.e.... All the charge lies on the external surface of the conductor. Gaussian surfaces cavity metal

17. ELECTRICITY GAUSS’S LAWPHY1013S 17 For a non-spherical conductor, the surface charge density, , varies over the surface, and the field established around the conductor is very complex. However, for a point just outside the surface, the adjacent section of surface is small enough to be considered as flat, and the charge density as uniform… EXTERNAL FIELD DUE TO A CHARGED CONDUCTOR 

18. ELECTRICITY GAUSS’S LAWPHY1013S 18 1.Draw the situation. 2.Choose a Gaussian surface appropriate to the symmetry. 3.Apply Gauss’s law: USING GAUSS’S LAW TO DETERMINE ELECTRIC FIELDS

19. ELECTRICITY GAUSS’S LAWPHY1013S 19 EXTERNAL FIELD DUE TO A CHARGED CONDUCTOR A cylindrical Gaussian surface with an end cap area of A is embedded in the surface of the conductor as shown. The flux through the outer cap is EA. The total charge enclosed by the cylinder is given by  A. Therefore, according to Gauss’s law:  0 EA =  A and hence:  A

20. ELECTRICITY GAUSS’S LAWPHY1013S 20 The external field lies perpendicular to the surface and is given by. On an irregularly shaped conductor the charge collects around sharp points, but still holds true. EXTERNAL FIELD DUE TO A CHARGED CONDUCTOR The electric field is zero everywhere inside the conducting material. Any excess charge is all on the surface. Summary: 

21. ELECTRICITY GAUSS’S LAWPHY1013S 21 1.Draw the situation. 2.Choose a Gaussian surface appropriate to the symmetry. 3.Apply Gauss’s law: USING GAUSS’S LAW TO DETERMINE ELECTRIC FIELDS

22. ELECTRICITY The area of the curved surface is 2  r L. By symmetry, the total flux through the surface is E 2  r L. The total charge enclosed by the cylinder is L. Therefore, according to Gauss’s law:  0 E 2  r L = L GAUSS’S LAWPHY1013S 22 we choose a cylindrical Gaussian surface with radius r and height L : CYLINDRICAL SYMMETRY For a long, thin, cylindrical insulator with a uniform linear charge density of … and hence: r L 2r2r or:

23. ELECTRICITY GAUSS’S LAWPHY1013S 23 PLANAR SYMMETRY For a large, flat, thin insulating sheet with a uniform surface charge density of  … …we choose a cylindrical Gaussian surface with end cap area A, which pierces the sheet perpendicularly. According to Gauss’s law:  0 (E A + E A) =  A and hence:  A

24. ELECTRICITY GAUSS’S LAWPHY1013S 24 PLANAR SYMMETRY For a charged large, flat, thin conducting plate: surface charge density =  1 When two oppositely charged plates are brought close together, all the charge moves to the inner faces: surface charge density  = 2  1

25. ELECTRICITY GAUSS’S LAWPHY1013S 25 SPHERICAL SYMMETRY A spherical shell of radius R and charge q is surrounded by a concentric spherical Gaussian surface ( S 1 ) with radius r (where r  R ). According to Gauss’s law:  0 E 4  r 2 = q and hence, for S 1 : I.e.A uniform spherical shell of charge acts, on all charges outside it, as if all its charge were concentrated at its centre. [Shell theorem 1] R r S1S1 q

26. ELECTRICITY GAUSS’S LAWPHY1013S 26 According to Gauss, for S 2 : E = 0 SPHERICAL SYMMETRY S 2 is a concentric spherical Gaussian surface with radius r lying within the spherical shell of charge q (i.e. r  R ). R r S2S2 q I.e. A uniform spherical shell of charge exerts no electrostatic force on a charged particle located inside it. [Shell theorem 2]