Gauss’s Law: applications

1. Gauss’s Law: applications Chapter 24

2. Gauss’s Law The total flux within a closed surface … … is proportional to the enclosed charge. Gauss’s Law is always true, but is only useful for certain very simple problems with great symmetry.

3. Problem: Infinite charged plane Consider an infinite plane with a constant surface charge density s (which is some number of Coulombs per square meter). What is E at a point a distance z above the plane? y z s x

4. Problem: Infinite charged plane Consider an infinite plane with a constant surface charge density s (which is some number of Coulombs per square meter). What is E at a point a distance z above the plane? y E z s x Step 1: Use symmetry

5. Problem: Infinite charged plane Consider an infinite plane with a constant surface charge density s (which is some number of Coulombs per square meter). What is E at a point a distance z above the plane? y E z s x The electric field must point straight away from the plane (if s>0). Maybe the magnitude E depends on z, but the direction is fixed. And E is independent of x and y. Step 1: Use symmetry

6. Problem: Infinite charged plane Consider an infinite plane with a constant surface charge density s (which is some number of Coulombs per square meter). What is E at a point a distance z above the plane? E y E z s E x The electric field must point straight away from the plane (if s>0). Maybe the magnitude E depends on z, but the direction is fixed. And E is independent of x and y. Step 1: Use symmetry

7. Problem: Infinite charged plane Step 2: Choose a Gaussian surface Passes through the observation point Takes advantage of the symmetry E z s E

8. Problem: Infinite charged plane Step 2: Choose a Gaussian surface Passes through the observation point Takes advantage of the symmetry E Gaussian surface z s z E So make the Gaussian surface a box which has its top above the plane, and its bottom below the plane, each a distance z from the plane. That way the observation point lies in the top.

9. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E

10. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Total charge enclosed by box =

11. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Total charge enclosed by box = As

12. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Outward flux through the top:

13. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Outward flux through the top: EA

14. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Outward flux through the top: EA Outward flux through the bottom:

15. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Outward flux through the top: EA Outward flux through the bottom: EA

16. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Outward flux through the top: EA Outward flux through the bottom: EA Outward flux through the sides:

17. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Outward flux through the top: EA Outward flux through the bottom: EA Outward flux through the sides: E x (some area) x cos(900) = 0

18. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Outward flux through the top: EA Outward flux through the bottom: EA Outward flux through the sides: E x (some area) x cos(900) = 0 So the total flux is: 2EA

19. Problem: Infinite charged plane Step 3:Apply Gauss’s Law Let the area of the top and bottom be A. E Gaussian surface z s z E Gauss’s law then says that As/e0=2EA so that E=s/2e0, outward. This is constant everywhere in each half-space! Notice that the area A canceled: this is typical!

20. Problem: Infinite charged plane Imagine doing this with an integral over the charge distribution: break the surface into little bits dA … dE s Doing this as a surface integral would be HARD. Gauss’s law is EASY.

21. Find the electric field E at point P Example: An infinite thin line of charge. Charge per unit length l P y

22. Find the electric field E at point P Example: An infinite thin line of charge. E Charge per unit length l P y Step 1: use symmetry. Here E is radially outward - along the line through P perpendicular to the line of charge. End view:

23. Example: An infinite thin line of charge. E P y Step 1: use symmetry. Here E is radially outward - along the line through P perpendicular to the line of charge. Step 2: choose a Gaussian surface - here a cylinder centered on the line of charge. Notice that E has the same magnitude at all points on this surface - and points radially outward.

24. Example: An infinite thin line of charge. E P y Step 3: apply Gauss’s law. Say the cylinder has length L. It has radius y since it goes through P. Charge enclosed =

25. Example: An infinite thin line of charge. E P y Step 3: apply Gauss’s law. Say the cylinder has length L. It has radius y since it goes through P. Charge enclosed = lL

26. Example: An infinite thin line of charge. E P y Step 3: apply Gauss’s law. Say the cylinder has length L. It has radius y since it goes through P. Charge enclosed = lL Flux through curved face =

27. Example: An infinite thin line of charge. E P y Step 3: apply Gauss’s law. Say the cylinder has length L. It has radius y since it goes through P. Charge enclosed = lL Flux through curved face = E x area = E(2py)(L)

28. Example: An infinite thin line of charge. E P y Step 3: apply Gauss’s law. Say the cylinder has length L. It has radius y since it goes through P. Charge enclosed = lL Flux through curved face: E x area = E(2py)(L) Flux through flat faces =

29. Example: An infinite thin line of charge. E P y Step 3: apply Gauss’s law. Say the cylinder has length L. It has radius y since it goes through P. Charge enclosed = lL Flux through curved face: E x area = E(2py)(L) Flux through flat faces = 0

30. Example: An infinite thin line of charge. E P y Step 3: apply Gauss’s law. Say the cylinder has length L. It has radius y since it goes through P. Charge enclosed = lL Flux through curved face: E x area = E(2py)(L) Flux through flat faces = 0 Gauss: lL/e0 = E(2py)(L)

31. Example: An infinite thin line of charge. E P y Step 3: apply Gauss’s law. Say the cylinder has length L. It has radius y since it goes through P. Charge enclosed = lL Flux through curved face: E x area = E(2py)(L) Flux through flat faces = 0 Gauss: lL/e0 = E(2py)(L) radially outward

32. Conductors • A conductor is a material in which some of the electrons (~1 per atom) can move relatively freely. • Usually these are metals (Au, Cu, Ag, Al). • The conduction electrons tend to move together, like water in the ocean. An ordinary chunk of conductor is neutral, and the electrons, repelling one another and attracted by the background ions, spread uniformly. • If you place excess charge on a conductor, the charges move as far from each other as possible. • In a static situation, the electric field is zero everywhere inside a conductor. • In a static situation, all the excess charge (- or +) resides at the surface of a conductor.

33. Conductors Why is E=0 inside a conductor in equilibrium?

34. Conductors Why is E=0 inside a conductor in equilibrium? Conductors are full of free electrons, roughly one per cubic Angstrom. These are free to move. If E is nonzero in some region, then the conduction electrons there feel a force -eE and start to move.

35. Conductors Why is E=0 inside a conductor in equilibrum? Conductors are full of free electrons, roughly one per cubic Angstrom. These are free to move. If E is nonzero in some region, then the conduction electrons there feel a force -eE and start to move. In an electrostatics problem, the electrons adjust their positions until the force on every electron is zero (or else it would move!). That means when equilibrium is reached, E=0 everywhere inside a conductor.

36. Conductors Because E=0 inside, the inside of a conductor is neutral. Suppose there is an extra charge inside. Gauss’s law for the little spherical surface says there would be a nonzero E nearby. But there can’t be, within a metal! Consequently the interior of a metal is neutral. Any excess charge ends up on the surface.

37. Example: Charged Conducting Sphere Suppose you add an extra charge Q to a sphere of radius R. Find E everywhere.

38. Example: Charged Conducting Sphere Suppose you add an extra charge Q to a sphere of radius R. Find E everywhere. The excess charge Q goes to the surface, producing a surface charge s=Q/4pR2 s

39. Example: Charged Conducting Sphere Suppose you add an extra charge Q to a sphere of radius R. Find E everywhere. The excess charge Q goes to the surface, producing a surface charge s=Q/4pR2 s Outside: the spherical distribution acts as though it were all concentrated at the origin.

40. Example: Charged Conducting Sphere Suppose you add an extra charge Q to a sphere of radius R. Find E everywhere. The excess charge Q goes to the surface, producing a surface charge s=Q/4pR2 s Outside: the spherical distribution acts as though it were all concentrated at the origin. So

41. Example: Charged Conducting Sphere Suppose you add an extra charge Q to a sphere of radius R. Find E everywhere. The excess charge Q goes to the surface, producing a surface charge s=Q/4pR2 s Outside: the spherical distribution acts as though it were all concentrated at the origin. So Within a spherical shell the net electric field due to the shell is zero. So E(r)=0, r<R.

42. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. b q a

43. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. In the cavity: b q a

44. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. In the cavity: b q a

45. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. In the cavity: b q a Within the shell:

46. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. In the cavity: b q a Within the shell:

47. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. In the cavity: b q a Within the shell: Outside the shell:

48. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. In the cavity: b q a Within the shell: Outside the shell: the only charge enclosed by a Gaussian sphere is q, so it must be the case that

49. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. In the cavity: b q a Within the shell: Outside the shell: the only charge enclosed by a Gaussian sphere is q, so it must be the case that

50. Conducting Spherical Shell with a Charge Inside Suppose you have a neutral conducting spherical shell of interior radius a and exterior radius b. Suppose there is a point charge q at the center of the cavity. Find E everywhere. In the cavity: b q a Within the shell: Outside the shell: the only charge enclosed by a Gaussian sphere is q, so it must be the case that This result is more interesting if we think about why it happens.