1. TUESDAY(01/12): Intro. to Ch.24 - Gauss's Law. HW: Read and Study

pages 740-46, then solve problems 1, 4, 7, 13, and 15 on pages 755-56.

2. WEDNESDAY(01/13): Applications of Gauss's Law. HW: Read and Study

pages 746-50, then solve problems 19, 23, 25, and 29 on pages 756-57.

3. THURSDAY(01/14): Conductors in Electrostatic Equilibrium. HW: Read

and Study pages 750-54, then solve problems 35, 39, and 41 on pages 757-58.

4. FRIDAY(01/15): REVIEW I Ch.24 - Gauss's Law. HW: Complete All

Review Handouts.

5. TUESDAY(01/19): REVIEW II Ch.24 - Gauss's Law. HW: Complete All

Review Handouts.

6. WEDNESDAY(01/20): TEST on Ch.24. HW: Go to website and study notes for

Ch.25 - Electric Potential.

WEBSITE NOTES: Ch.24 - Gauss's Law.

This chapter (24) is devoted to the work of one of the greatest mathematicians

of all time. Carl Friedrich Gauss (1777-1855) influenced the work in many areas

of mathematics from number theory and geometry through differential equations.

He also made equally significant contributions to theoretical physics.

1. When you finish this chapter, you should be able to, (a) define and compute

electric flux through a surface, (b) use Gauss' Law and the concept of Gaussian

surfaces to find the flux, electric field, or total charge enclosed by the surface,

and (c) describe and explain the properties of a conductor in electrostatic

equilibrium.

2. Electric flux, Φ (Phi), is proportional to the number of lines of electric field

through an area and calculated by Φ = E·A = E·cos(θ)·A.

3. The number of lines is infinite, because each one represents the force per unit

charge on a test charge placed at the point!

4. Some fields are larger than others, so some infinities are bigger than others!!

Have you heard this before?

5. Well, the mathematicians do indeed talk about different infinities. If we don't

let this little thing worry us, we can assign a number to electric flux in this way:

(a) If the field lines are perpendicular to the area that they intersect (see Figure

24.1, page 740), and the field is constant, the flux is E times A, or EA, (b) If the

field lines are not perpendicular to the area, but the field is constant, the flux is

given by EAcos(θ).

6. This last expression suggests a dot product. And if the field is non-uniform,

we'll need an integral: given by equation 24.4 on p.742, (the circle on the integral

sign says that this is to be done over a closed surface - the entire area).

7. There are hard ways to do this integral - and there are easy ways! The example

- Flux Through a Hemisphere - would be very difficult if we tried to sum (an

integral is a sum) the dot product contributions over the entire curved surface.

8. Enter, Gauss' Law: The net electric flux through a closed surface (called a

Gaussian surface) is given by the enclosed charge q divided by the permittivity

constant, ε_o ,epsilon nought.   ∫E·dA = q/ε_o             (Always on AP exam.)

9. In section 24.2, Serway does this integral for a spherical surface and then

generalizes the resulting expression to any surface that encloses the charge(s)

that created the field.

10. Suppose there's a solid insulating sphere containing a uniform charge density

and you need to find the field inside and outside it.

11. Not your everyday problem, is it? Anyway, here's the solution - it's mostly the

same as Example 24.5, page 747. Serway's other examples are very good also.

12. The method to use when studying them is to try to do them without help. (Of

course I'll be helping you in class, but soon you will be on your own.)

13. You'll have to do this several times before you can finally do all of the

examples without looking at them, but THAT'S THE WAY THIS IS!

14. Properties of a Conductor: (a) The electric field is zero everywhere inside the

conductor, (b) Any charge placed inside the conductor will move immediately to its

surface, (c) The electric field just outside the conductor is perpendicular to the

surface at that point, (d) On an irregularly shaped conductor, charge tends to

accumulate where the radius of curvature is smallest (at sharp points, like a

lightning rod).

15. Thus, the electric field at such points is the largest.

16. Electric flux through a surface is defined as the dot product of the electric field

and the "vector area" of the surface, where the "vector area" points in the direction

of the normal to the surface.

17. Units of electric flux are N·m²/C. Flux going INTO a closed surface is negative;

flux coming OUT OF a closed surface is positive.

18. Remember, Gauss' Law states that the net electric flux through any closed

surface is equal to the electric charge enclosed by that surface, divided by the

permittivity of free space, ε_o.    ∫E·dA = q/ε_o

19. We will still need the charge density equations: (a) Linear charge density is

given by λ = q/L , (b) Area charge density is, σ = Q/A , and (c) volume charge

density is  ρ = Q/V .

20. We will also use Integral Calculus to determine the electric field (E-field)

due to various charge distributions that can't be solved by Gauss's Law.

For example, consider the E-field on the axis of a charged ring:

21. This is a very limited solution to the more general problem of finding the

E-field at any point in space near a ring with a charge Q spread uniformly

around the ring.

22. Along the axis (say the x-axis) the perpendicular components of the E-field

due to charges spread around the ring cancel each other out. There is just as

much charge on one side of the ring and the other.

23. The net E-field (on the axis) is along the axis, outwards from the ring if the

charge Q is positive and towards the ring in the charge Q is negative.

24. The charge density λ = Q/length on the ring is just the total charge Q on

the ring divided by its circumference 2πR.

25. Treating the differential charge dq as a point charge, the differential electric

field at a distance x from the center of the ring (the origin) is given by the

equation,    _.

26. Since the sum of the y-components cancel out, the magnitude of the electric

field is equal to the sum of the x-components of the E-field. We can express

this as an integral over the differential arc length ds,

27. Note that x and r are both constant from any location on the ring to the point

where we are calculating the E-field.

28. There are two interesting limits. At the very center of the ring the E-field is

zero as one would expect. The E-fields of the charges spread around the ring

cancel each other out.

29. When one gets very far from the ring so that x >> R, then R can be neglected

in the denominator compared to x, and the loop looks like a point charge at

large distances.

And still, we need these steps to solve any problem in Physics:

(i) read the problem and identify the given variables

(ii) determine what you are asked to solve for

(iii) find the correct equation to use

(iv) use Algebra, Trigonometry, and/or Calculus to isolate the unknown

(v) substitute-in the given information and simplify.

THE AP PHYSICS C ARCHIVES Ch.1: Physics Intro. Ch.2: Linear Motion. Ch.3: Vectors. Ch.4: 2-Dim Motion. Ch.5&6: Newton's Laws. Ch.7&8: Work&Energy. Ch.9: Momentum. Ch.10&11: Rotary Motion. Ch.12: Elasticity. Ch.13: Gravitation. Ch.15: SHM. Mechanics Review. Ch.23: Electric Fields. Ch.24: Gauss's Law. Ch.25: Electric Potential.