1. Electric Charge and Coulomb’s Law

2. Fundamental Charge: The charge on one electron.
e = 1.6 x C
Unit of charge is a Coulomb (C)

3. Two types of charge: Positive Charge: A shortage of electrons.
Negative Charge: An excess of electrons.
Conservation of charge – The net charge of a closed system remains constant.

4. Number of electrons > Number of protons -2e = -3.2 x 10-19C
Nucleus - - + n - n + + n n + n - + + n - - -
Negative Atom
Number of electrons > Number of protons
-2e = -3.2 x 10-19C
Neutral Atom
Number of electrons = Number of protons
Positive Atom
Number of electrons < Number of protons
+2e = +3.2 x 10-19C

5. Electric Forces + + - Like Charges - Repel F Unlike Charges - Attract

6. Coulomb’s Law – Gives the electric force between two point charges.
Inverse Square
Law
k = Coulomb’s Constant = 9.0x109 Nm2/C2
q1 = charge on mass 1
q2 = charge on mass 2
r = the distance between the two charges
The electric force is much stronger than the gravitational force.

7. Example 1
Two charges are separated by a distance r and have a force F on each other. q1 q2 r F
If r is doubled then F is :
¼ of F
If q1 is doubled then F is : 2F
If q1 and q2 are doubled and r is halved then F is :
16F

8. Example 2
Two 40 gram masses each with a charge of 3μC are placed 50cm apart. Compare the gravitational force between the two masses to the electric force between the two masses. (Ignore the force of the earth on the two masses)
3μC
40g
50cm

9. The electric force is much greater than the gravitational force

10. F1 and F2 must be added together as vectors.
Example 3
Three charged objects are placed as shown. Find the net force on the object with the charge of -4μC.
- 5μC
20cm
45º F1
5μC
- 4μC F2
F1 and F2 must be added together as vectors.

11. + F1 = < - 4.5 , 0.0 > F2 = < 1.6 , - 1.6 >
2.3cos45≈1.6
- 2.9 F2
45º
29º θ
- 1.6
3.31
2.3sin45≈1.6
F1 = < , 0.0 > +
F2 = < 1.6 , >
Fnet = < , >
3.31N at 209º

12. Example 4
Two 8 gram, equally charged balls are suspended on earth as shown in the diagram below. Find the charge on each ball. q
20º
L = 30cm FE
10º
30sin10º
r =2(30sin10º)=10.4cm r

13. Draw a force diagram for one charge and treat as an equilibrium problem.
Tsin80º FE
80º
Fg = .08N
Tcos80º