1. TUESDAY(01/06): Intro. to Ch.23 - Electrostatic Force. HW: Read and

Study pages 705-715, then solve problems 2, 3, 7, 9, and 11 on pages 730-31.

2. WEDNESDAY(01/07): Electric Fields and Force. HW: Read and Study pages

715-725, then solve problems 13, 15, 17, 21, and 27 on pages 731-32.

3. THURSDAY(01/08): Motion of Charged Particles in Electric Fields.

HW: Read and Study pages 725-729, then solve problems 42, 43, and 47

on page 734.

4. FRIDAY(01/09): REVIEW Ch.23 - Coulomb's Law and Electric

Fields. HW: Complete All Review Handouts.

5. MONDAY(01/12): TEST on Ch.23. HW: Go to website and

study notes for Ch.24 - Gauss' Law.

INTRODUCTION: In ancient Greece, amber became widely valued around

1600 BC. Greeks were fascinated by it. The ancient Greek word for amber

is "elektron", meaning - originating from the Sun. The Greeks were also

the first to describe the electrostatic properties of amber. Ancient Romans

loved amber as well. From the writings of Thales of Miletus it appears that

Westerners knew as long ago as 600 B.C. that amber becomes charged by

rubbing. There was little real progress until the English scientist William

Gilbert in 1600 described the electrification of many substances and coined

the term electricity from the Greek word for amber. As a result, Gilbert is

called the father of modern electricity.

One of nature's most spectacular display of electricity is the lightning

observed during a thunder storm. Benjamin Franklin (1706-1790) determined

that electricity originates from charges, positive or negative. We know now

that all material bodies possess electric charges. Electrons carry negative

charges while protons carry positive charges in the nucleus of an atom.

1. The electric force that stationary objects exert on each other is called the

electrostatic force. This force depends upon the distance between the two

point charges and the amount of charge on each. Experiments have

demonstrated that the greater the charge and the closer they are to each

other, the greater the force.

2. If charges have unlike signs, each charge is attracted to one other,

whereas like charges repel each other. These attractive forces and repulsive

forces act along the line between the charges, and are equal in magnitude

but opposite in direction (in accordance with Newton's 3rd law).

3. The French physicist Charles-Augustin Coulomb (1736-1806) experimented

with electric force between two point charges (the unit of charge is the

Coulomb, C). His work resulted in a law. Coulombs Law is defined: The

magnitude of the electrostatic force (F), exerted by one point charge on

another point charge is directly proportional to the magnitudes of the two point

charges, and inversely proportional to the square of the distance (r) between

the charges.

4. For a pair of charges q₁ and q₂, separated by a distance r, Coulomb's Law

may be stated as follows: F = k(q₁q₂/r²) .....................Inverse-Square Law

5. The constant of proportionality, k = 8.99x10⁹ Nm²/C². Such a force is

transmitted by the presence of an electric field. The electric field E due to a

point charge q is, E = k(q/r²) ..................................Inverse-Square Law

6. Electric force and electric field are vectors. Hence, they have magnitudes

and directions. The electric force F and electric field E are related as follows:

F=qE .........................................much like F = ma

7. Where the force is on charge q due to the presence of an electric field at

the position of q.

8. The Principle of Superposition also applies to the electric fields produced

by multiple charges. That is, the net electric field at a point due to several

charges is the vector sum of the electric fields due to individual charges.

9. For example, when more than two charges are present, the net force on

any one charge is equal to the vector sum of each of the forces produced by

other charges.

10. In other words, the force on charge q₁ due to the presence of charges q₂

and q₃, is the superposition of the forces exerted by q₂ and q₃. That is, the

net force F on charge q₁ is, F_net = F₁₂ + F₁₃ .................Vector Equation

where, F₁₂ is the force on q₁ due to the presence of charge q₂ and F₁₃ is the

force on q₁ due to charge q₃.

11. While solving a problem, it is useful to rewrite the vector equation in its

component form as follows:

F_x = (F₁₂)_x + (F₁₃)_x

F_y = (F₁₂)_y + (F₁₃)_y

F_net = [(F_x)² + (F_y)²]^1/2

θ = tan^-1 [F_y/F_x]

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

due to various charge distributions. For example, consider the E-field on the

axis of a charged ring:

13. 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.

14. 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.

15. 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.

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

the ring divided by its circumference 2πR.

17. 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,    _.

18. 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,

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

where we are calculating the E-field.

20. 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.

21. 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 vector formula to use

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

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

Answers to review packet: 31. c, 32. d, 33. b, 34. c, 35. a, 36. b,

37. c, 38. b, 39. c, 40. 56 N/C, 41. b, 42. b, 43. b, 44. a.

THE AP PHYSICS C SEMESTER 1 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.