1. Tuesday(01/05): 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(01/06): 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(01/07): 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(01/08): REVIEW Ch.23 - Coulomb's Law and Electric Fields.

HW: Complete All Review Handouts.

5. Monday(01/11): 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, 

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 Homework in Scrambled Format: