1. 1/18/07184 Lecture 71 PHY 184 Spring 2007 Lecture 7 Title: Using Gauss’ law

2. 1/18/07184 Lecture 72AnnouncementsAnnouncements  Homework set 3 opened this morning.  Starting Monday, we will have clicker quizzes/questions giving extra credit (up to 5%).  Helproom Hours for the Honors Students start next week.

3. 1/18/07184 Lecture 73 Review - Gauss’ Law  Gauss’ Law  Gauss’ Law says that the electric flux through a closed surface is proportional to the net charge enclosed by this surface.  If we add the definition of the electric flux we get another expression for Gauss’ Law

4. 1/18/07184 Lecture 74 Gauss’ Law for Various Charge Distributions  We have applied Gauss’ Law to a point charge and showed that we get Coulomb’s Law.  Now let’s look at more complicated distributions of charge and calculate the resulting electric field. “charge density”  We will use a “charge density” to describe the distribution of charge.  This charge density will be different depending on the geometry.

5. 1/18/07184 Lecture 75 Cylindrical Symmetry conducting wire with charge per unit length  Let’s calculate the electric field from a conducting wire with charge per unit length using Gauss’ Law  We start by assuming a Gaussian surface in the form of a right cylinder with radius r and length L placed around the wire such that the wire is along the axis of the cylinder

6. 1/18/07184 Lecture 76 Cylindrical Symmetry (2)  From symmetry we can see that the electric field will extend radially from the wire.  How? If we rotate the wire along its axis, the electric field must look the same Cylindrical symmetry If we imagine a very long wire, the electric field cannot be different anywhere along the length of the wire Translational symmetry  Thus our assumption of a right cylinder as a Gaussian surface is perfectly suited for the calculation of the electric field using Gauss’ Law.

7. 1/18/07184 Lecture 77 Cylindrical Symmetry (3)  The electric flux through the ends of the cylinder is zero because the electric field is always parallel to the ends.  The electric field is always perpendicular to the wall of the cylinder so  … and now solve for the electric field

8. 1/18/07184 Lecture 78 Planar Symmetry  Assume that we have a thin, infinite non-conducting sheet of positive charge  The charge density in this case is the charge per unit area,   From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet

9. 1/18/07184 Lecture 79 Planar Symmetry (2)  To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2r, chosen to cut through the plane perpendicularly.  Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder.  Using Gauss’ Law we get  … so the electric field from an infinite non-conducting sheet with charge density 

10. 1/18/07184 Lecture 710  Assume that we have a thin, infinite conductor (metal plate) with positive charge  The “charge density” in this case is also the charge per unit area, , on either surface; there is equal surface charge on both sides.  From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet Planar Symmetry (3)

11. 1/18/07184 Lecture 711 Planar Symmetry (4)  To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height r, chosen to cut through one side of the plane perpendicularly.  The field inside the conductor is zero so the end inside the conductor does not contribute to the integral.  Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the end of the cylinder outside the conductor.  Using Gauss’ Law we get  … so the electric field from an infinite conducting sheet with surface charge density  is

12. 1/18/07184 Lecture 712 Spherical Symmetry charge distributed as a spherical shell  Now let’s calculate the electric field from charge distributed as a spherical shell.  Assume that we have a spherical shell of charge Q with radius R (gray).  We will assume two different spherical Gaussian surfaces r > R (blue) i.e. outside r < R (red) i.e. inside

13. 1/18/07184 Lecture 713 Spherical Symmetry (2)  Let’s start with the Gaussian surface outside the sphere of charge, r > R (blue)  We know from symmetry arguments that the electric field will be radial outside the charged sphere If we rotate the sphere, the electric field cannot change Spherical symmetry  Thus we can apply Gauss’ Law and get  … so the electric field is

14. 1/18/07184 Lecture 714 Spherical Symmetry (3)  Let’s let’s take the Gaussian surface inside the sphere of charge, r < R (red)  We know that the enclosed charge is zero so  We find that the electric field is zero everywhere inside spherical shell of charge  Thus we obtain two results The electric field outside a spherical shell of charge is the same as that of a point charge. The electric field inside a spherical shell of charge is zero.

15. 1/18/07184 Lecture 715 Spherical Symmetry (4) charge distributed uniformly throughout a sphere.  Next, let’s calculate the electric field from charge distributed uniformly throughout a sphere.  Assume that we have a solid sphere of charge Q with radius R with constant charge density per unit volume .  We will assume two different spherical Gaussian surfaces r2 > R (outside) r1 < R (inside)

16. 1/18/07184 Lecture 716 Spherical Symmetry (5)  Let’s start with a Gaussian surface with r 1 < R.  From spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface.  Gauss’ Law gives us  Solving for E we find volumearea inside

17. 1/18/07184 Lecture 717 Spherical Symmetry (6) inside In terms of the total charge Q …

18. 1/18/07184 Lecture 718 Spherical Symmetry (7)  Now consider a Gaussian surface with radius r 2 > R.  Again by spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface.  Gauss’ Law gives us  Solving for E we find area outside full charge same as a point charge!

19. 1/18/07184 Lecture 719 Electric Field from a Ring of Charge We can’t solve it by Gauss’ Law! Use the method of integration …

20. 1/18/07184 Lecture 720 Pointed Surfaces Now consider a sharp point. The field lines are much closer together, i.e., the field is much stronger and looks much more like the field of a point charge. + + + + + + + + + + … … E field is always perpendicular to the surface of a charged conductor.

21. 1/18/07184 Lecture 721 Demo - Electrostatic Wind The strong electric field at the sharp tip polarizes atmospheric molecules. If a molecule is ionized, the ion will be repelled from the point. The recoil force causes the whirl to spin.