1. Electrostatics Electric Force

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

Vector addition review:

Two forces acting on an object

tail-to-tip method

Components method

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

QQ

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.

More informationhttp://ap-physics.david-s.org