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

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 Force Distance

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

F = (8.99x109 Nm2/C2)(-1.6x10-19C)(1.6x10-19C) 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

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

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