Post on 31-Dec-2015
First review the gravitational force…
Any two masses are attracted by equal and opposite gravitational forces:
m1 m2
r
F -F
Newton’s Universal Law of Gravitationwhere…… F G
m mr1 2
2
G=Universal Gravitation Constant = 6.67x10-11 Nm2/kg2
This is an Inverse-Square force Gravity is a very weak force
221
r
QQkF
r1Q 2Q
Charge (Q)Coulombs (C)
1 C = 6.2421 x 1018 e
e = 1.602 x 10-19 C
Distance (m)
Coulomb’s constant (k)
k = 8.988 x 109 Nm2/C2
Force (N)
12F 21F
2112 FFF
Notes on Coulomb’s Law
1) It has the same form as the Law of Gravitation: Inverse-Square Force
2) But… (can you spot the most basic difference between these two laws?)
3) The electrostatic constant (k) in this law is derived from a more fundamental constant:
k1
4
0
0= permittivity of free space = 8.85 x 10-12 C2/Nm2
4) Coulomb’s Law obeys the principle of superposition
1. Compare the electric force holding the electron in orbit (r = 0.53 × 10 -l0 m) around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton.What is the ratio of these two forces?
21
21
221
221
G
EmGm
QkQ
r
mGmr
QkQ
F
F
kg10 x .671kg10 x .119/kgmN 10 x .676
C 10 x 602.1 /CmN 10 x .988827312211
219229
3910 x 3.2
Coulomb’s law strictly applies only to point charges.
Superposition: for multiple point charges, the forces on each charge from every other charge can be calculated and then added as vectors.
1Q2Q
3Q
32F
31F3Q
32F
31F
netF
2. At each corner of a square of side L there are four point charges. Determine the force on the charge 2Q.
2
2
2
2o
22x L
kQ 8284.4222
L
kQ45cos
L2
Q4Q2k
L
QQ2kF
2
2
2
2o
22y L
kQ 8284.8226
L
kQ45sin
L2
Q4Q2k
L
Q3Q2kF
2
22y
2xQ2 L
kQ 0625.10 FFF
Q2Q
3Q4QL
3. A +4.75 mC and a -3.55 mC charge are placed 18.5 cm apart. Where can a third charge be placed so that it experiences no net force?
Q1 Q2
d x
Q
21 FF
22
21
x
QQk
xd
QQk
21
2
Qdx
C 10 x 5.3C 10 x .74
C 10 x 5.3cm 5.18x
66
6
cm 116
[Extra] Two 6.7 kg bowling balls are placed in a vertical cylinder. Charge is added to each ball until they repel with enough force to separate the two balls by a distance of 43.7 cm. Assuming the two bowling balls are charged positively and equally, determine the amount of charge (to four significant digits) added to each individual ball.