1. Aim: How can we explain Electrostatic Force?
Do Now:
What is the net charge of the system when they come into contact?
+6e
+6e
-2e
-2e
net charge of +4e

2. Conservation of Charge
If 2 or more objects come into contact with one another the net charge is distributed evenly among the objects

3. What will the charge be after separation?
= net charge
# of objects
= +4e = +2e 2

4. Example
What is the net charge when the spheres come into contact?
-6e
What is the charge on each sphere after separation?
- 2e
+4e
-8e
-2e

5. Units of Charge
Coulombs (C)
Charles-Augustin de Coulomb

6. 1 elementary charge:
(1e = 1.6 x C)
1 Coulomb =
6.25 x 1018 elementary charges

7. Example How many Coulombs in 5 electrons?
How many protons make up +10 Coulombs?
= 8 x C or
= 8 x C
= 6.25 x 1019 C or
= 6.25 x 1019 C

8. Coulomb’s Law
The electrical force of attraction or repulsions between 2 charged objects

9. k = electrostatic constant 8.99 x 109 N•m2/C2
q1 = charge on 1st object (C)
q2 = charge on 2nd object (C)
r = distance between objects

10. Graph of Coulomb’s Law
Force
Distance

11. Examples q1 = +2.0C, q2= +2.0C, r = 5m. Find F F = kq1q2 r2
F = (8.99x109 Nm2/C2)(+2.0C)(+2.0C)
(5.0 m)2
F = 1.4 x 109 N

12. q1 = -10.0C, q2= -10.0C, r = 2.0m.
Find F
F = kq1q2 r2
F = (8.99x109 Nm2/C2)(-10.0C)(-10.0C)
(2.0 m)2
F = 2.2 x 1011 N

13. F = (8.99x109 Nm2/C2)(-1.6x10-19C)(1.6x10-19C)
Find F between electron and proton, separated by 1.5 x m
F = kq1q2 r2
F = (8.99x109 Nm2/C2)(-1.6x10-19C)(1.6x10-19C)
(1.5x10-10 m)2
F = -1.0 x 10-8 N

14. repulsive/attractive
If F is positive, the force is
repulsive/attractive
If F is negative, the force is

15. Original Force
Changed Variable
New Force F
double q1 2F
double q1 & q2 4F
double r
¼ F
triple r
1/9 F
half q1 & q2
half r
4 F
third r
9 F