Coulom

b’s Law

• Charles Coulomb (1736-1806: French)–Studied relationship between

force and charge• SI unit of charge = 1 coulomb (C)

–NOT charge of 1 e- (1.6 x 10-19)–Charge of 6.24 x 1018 e-

•We’ll get real definition later

1 Coulomb is about how many electrons flow through a 100 W light bulb every second.

Co

ulo

mb

’s La

w

• The LAW!– Force directly related to magnitude of

the charges– Force inversely related to the square of

the distance between the two charges

F = r2

kqQF α q Q

r2

1F α

r2

q QF α

F= r2k qQ

• Coulomb’s Law: Should look familiar

Only Attractive

Attractive or Repulsive

F= r2G mM Product of

masses

Product of charges

Both areinverse-square

laws

Large Magnitude9 x 109 N·m2/C2

Small Magnitude 6.67 x 10-11 N·m2/kg2

Coulom

b’s Law

F= r2k q Q

• Sample problem 0! YEAH!–A negative charge of -5.3μC and a

positive charge of 12 μC are separated by 0.69 m. What is the force of attraction between the two of them?

(9 x 109)=

(5.3 x 10-

6)(0.69)2

F= 1.2 N

(12 x 10-6)

(Remember that μ means 10-6 and n means 10-9)

Coulom

b’s Law

F= r2k q Q

• Sample problem 1! YEAH!–What is the magnitude of the

electrostatic force between 2 protons in the nucleus? (assume the distance to be 3.0x10-13 cm)

(9 x 109)=

(1.6 x 10-

19)(3.0 x 10-15 )

2

F=25.6 N

(1.6 x 10-

19)

Coulom

b’s Law

F= r2k q Q

• Sample problem 1! YEAH!–What is the magnitude of the

electrostatic force between 2 protons in the nucleus? (assume the distance to be 3.0x10-13 cm)

(9 x 109)=

(1.6 x 10-

19)(3.0 x 10-15 )

2

F=25.6 N

(1.6 x 10-

19)