1.  like charges repel  unlike charges attract  charges can move but charge is conserved Law of conservation of charge: the net amount of electric charge produced in any process is zero. There are two kinds of charge. + - Electric Charge and The Electric Field Static Electricity (Demo) Properties of charges

2. Although there are two kinds of charge in an atom, electrons are the charges that usually move around. A proton is roughly 2000 times more massive than an electron. The charge on an electron is 1.6x10 -19 coulombs. Conductors and insulators There is no such thing as a perfect conductor. A superconductor has no resistance to the flow of current, but is not the same as a perfect conductor. There is also no such thing as a perfect insulator.

3. Induced Charge: The Electroscope Demonstrate (if electroscope available). Coulomb’s Law attractive for unlike OSE: Coulomb’s law gives the force between charges Q 1 and Q 2 (in coulombs), where r is the distance in meters between the charges, and k=9x10 9 Nm 2 /C 2. We’ve seen attractive and repulsive electrical forces at work. Coulomb’s law quantifies the magnitude of the electrostatic force.

4. To make it into a “really good” OSE I should specify “repulsive for like,” but that makes it too wordy. You’ll just have to remember that! attractive for unlike OSE: Force is a vector quantity, so the equation by itself gives the magnitude of the force. Note how the direction is specified. If the charges are opposite in sign, the force is attractive; if the charges are the same in sign, the force is repulsive. Also, the constant k is equal to 1/4  0, where  0 =8.85x10 -12 C 2 /N m 2. I have to do it this way for my Physics 35 class, the algebra-based second semester of College Physics. Later I’ll show a better way to write it.

5. The equation as written is valid for point charges. If the charged objects are spherical and the charge is uniformly distributed, r is the distance between the centers of the spheres. If more than one charge is involved, the net force is the vector sum of all forces (let’s not make that a required OSE). For objects with complex shapes, you must add up all the forces acting on each separate charge (calculus!). Note that we could have agreed that in this formula, the symbols Q 1 and Q 2 stand for the magnitudes of the charges. In that case, the absolute value signs would be unnecessary. We’ll do that in later OSE’s, but for now I want to emphasize the magnitude part. On your homework diagrams, show both the magnitudes and signs of Q 1 and Q 2.

6. Solving Problems Involving Coulomb’s Law and Vectors You may wish to review vectors. Example: Calculate the net electrostatic force on charge Q 3 due to the charges Q 1 and Q 2. x y Q 2 =+50  C Q 3 =+65  C Q 1 =-86  C 52 cm 60 cm 30 cm  =30º

7. Step 0: Think! This is a Coulomb’s Law problem (all we have to work with, so far). We only want the forces on Q 3. Don’t worry about other forces. Forces are additive, so we can calculate F 32 and F 31 and add the two. If we do our vector addition using components, we must resolve our forces into their x- and y-components.

8. Step 1: Diagram Draw a representative sketch—done. Draw and label relevant quantities— done. Draw axes, showing origin and directions— done. Draw and label forces (only those on Q 3 ). Draw components of forces which are not along axes. x y Q 2 =+50  C Q 3 =+65  C Q 1 =-86  C 52 cm 60 cm 30 cm  =30º F 31 F 32

9. Step 2: OSE ”Do I have to put in the absolute value signs?” Yes, for now. x y Q 2 =+50  C Q 3 =+65  C Q 1 =-86  C 52 cm 60 cm 30 cm  =30º F 31 F 32

10. Step 3: Replace Generic Quantities by Specifics (from diagram) Can you put numbers in at this point? OK for this problem. You would get F 32,y = 330 N and F 32,x = 0 N. x y Q 2 =+50  C Q 3 =+65  C Q 1 =-86  C 52 cm 60 cm 30 cm  =30º F 31 F 32

11. Step 3 (continued) Can you put numbers in at this point? OK for this problem. You would get F 31,x = +120 N and F 31,y = -70 N. (- sign comes from diagram) (+ sign comes from diagram) x y Q 2 =+50  C Q 3 =+65  C Q 1 =-86  C 52 cm 60 cm 30 cm  =30º F 31 F 32

12. Step 4: Complete the Math F 3x = F 31,x + F 32,x = 120 N + 0 N = 120 N F 3y = F 31,y + F 32,y = -70 N + 330 N = 260 N x y Q 2 =+50  C Q 3 =+65  C Q 1 =-86  C 52 cm 60 cm 30 cm  =30º F 31 F 32 You know how to calculate the magnitude F 3 and the angle between F 3 and the x-axis. (If not, holler!) F3F3 The net force is the vector sum of all the forces on Q 3.

13. Remember I said I’d show you a better way to write Coulomb’s law? I have to write like this because my College Physics students are not supposed to have to use unit vectors: attractive for unlike OSE: If you define as the unit vector (a vector of unit length) pointing from Q 1 to Q 2, and F 12 is the force on Q 1 due to Q 2, and you include the correct signs in Q 1 and Q 2, then + - r 12 Q1Q1 Q2Q2 Much more satisfying!

14. I did a sample Coulomb’s law calculation using three point charges. How do you apply Coulomb’s law to objects that contain distributions of charges? We need another tool to do that.