Aim: How can we explain Electrostatic Force?

+6e -2e

Do Now:What is the net charge of the system when

they come into contact?

+6e -2e

net charge of +4e

Conservation of Charge

• If 2 or more objects come into contact with one another the net charge is distributed evenly among the objects

What will the charge be after separation?

= net charge

# of objects

= +4e = +2e

2

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

Units of Charge • Coulombs (C)

Charles-Augustin de Coulomb 1736-1806

• 1 elementary charge:

(1e = 1.6 x 10-19 C)• 1 Coulomb =

6.25 x 1018 elementary charges

Example• How many

Coulombs in 5 electrons?

• How many protons make up +10 Coulombs?

= 8 x 10-19 C

or

= 8 x 10-19 C

= 6.25 x 1019 C

or

= 6.25 x 1019 C

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

• 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

Graph of Coulomb’s Law

Distance

Force

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

• 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

• Find F between electron and proton, separated by 1.5 x 10-10 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

• If F is positive, the force is

repulsive/attractive

• If F is negative, the force is

repulsive/attractive

Original Force

Changed Variable

New Force

F double q1 2F

F double q1 & q2

4F

F double r ¼ F

F triple r 1/9 F

F half q1 & q2 ¼ F

F half r 4 F

F third r 9 F