1. • Charles Coulomb
(1736-1806: French)
– Studied relationship between
force and charge
Coulomb’s Law
• 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.
2. • The LAW!
– Force directly related to magnitude of
the charges
– Force inversely related to the square of
the distance between the two charges
Coulomb’s Law
Fα q Q
Fα
1 kq Q
F= 2
r2 r
F α qQ
r2
3. • Coulomb’s Law:
Should look familiar
Large Magnitude
9 x 109 N·m2/C2
Attractive or
Repulsive Product of
kQ q charges
F= 2
r
Small Magnitude Both are
6.67 x 10-11 N·m2/kg2 inverse-square
laws
GMm Product of
F= r2 masses
Only Attractive
4. • 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
Coulomb’s Law
them?
(Remember that μ means 10-6 and n means 10-9)
F= kq Q (9 x (5.3 x 10- (12 x 10-6)
r2 = 109) 6
)
(0.69) 2
F = 1.2 N
5. • 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)
Coulomb’s Law
F= kq Q (9 x (1.6 x 10- (1.6 x 10-
r2 = 109) 19
) 19
)
(3.0 x 10 -15
)
2
F =25.6 N
6. • 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)
Coulomb’s Law
F= kq Q (9 x (1.6 x 10- (1.6 x 10-
r2 = 109) 19
) 19
)
(3.0 x 10 -15
)
2
F =25.6 N
Hinweis der Redaktion
1 coulomb = The amount of electricity transported by a current of one ampere flowing for one second. About as many electrons that pass through 100-W bulb every second
What do we need to replace the ‘ α ’ with?
Electrostatic interactions may be attractive or repulsive, but gravity is ONLY attractive
Because both protons have the same charge we replace q 1 q 2 with q 2 .
Because both protons have the same charge we replace q 1 q 2 with q 2 .