There is a repulsive force

between electrons, - -
or between protons. + +

There is a attractive force

between electrons and protons.
- +

HOW MUCH
FORCE?
 We first need to
define a certain
quantity of charge.
 1 DOZEN eggs
12 eggs

1 REAM of paper
500 sheets

1 BUSHEL of apples
2152.42 cubic inches
( 42-48lbs of apples)
Kahoot.it

Q3
Q4
 COULOMB
ONE COULOMB of charge (+ or -)
Equals

6.24 x 10 charges
18
A whole heck of
6,240,000,000,000,000,000 a
charges
boophoo bunch of
charge
Kahoot.it

Q5
Q6
Q7
Q8
Coulomb’s
LAW
ELECTROSTATICS

F α q1·q2
1 Coulomb Force 1 Coulomb

2 Coulombs 2 x Force 1 Coulomb

4 x Force
2 Coulombs 2 Coulombs

12 x Force
3 Coulombs 4 Coulombs

Radius
LIGHT INTENSITY

Distance = 1 2 3 4
Area = 1 4 9 16
Intensity = 1 1/4 1/9 1/16
Intensity = 1 1/22 1/32 1/42
GRAVITATIONAL FORCE

Distance = 1 R 2R 3R 4R
Force = 1 g 1/4 g 1/9 g 1/16 g
1/16 N 1/9 N 1/4 N FORCE = 1 N
4R 3R 2R Distance = 1 R
- -
- - - -
- -- - -
- - -- - - -
- - - - --
- --- -- -- --
+ -- --- -- -- - -- -
- -
- - - - - --
-- - - -
- -- - - - - -
- - -- - --
- - - -
- - -
- - --
ELECTROSTATIC
GRAVITY FORCE
FORCE is inversely proportional to the square of the distance.

F α 1
R 2

FORCE is directly proportional to the product of the charges.

F α q1 · q2
FORCE is inversely proportional to the square of the distance.
Coulomb’s Law
F α 1
R 2

F α q 1 · q 2
R of the charges.
2
FORCE is directly proportional to the product

q1 · q2
F =k
F αR2 q1· q2
k = 8.9875 x 109 N·m2 / C2
k ~ 9 x 109 N·m2 / C2
 Electrical Force Very STRONG!

q1 · q2
F =k k = 8.9875 x 109
R2
F = Force (Newtons)
q1= charge on one object (coulombs)
q2= charge on the other object (coulombs)
R = Distance between the charges (meters)

Gravitational Force Very WEAK!

m1· m2
F =G G = 6.67 x 10-11
R 2
 Objectives: After finishing this unit,
you should be able to:
• Explain and demonstrate the first law of electro-
statics and discuss charging by contact and by
induction.
• Write and apply Coulomb’s
Law and apply it to problems
involving electric forces.
• Define the electron, the
coulomb, and the
microcoulomb as units of
electric charge.
 Electric Charge
When a rubber rod is rubbed against fur, electrons
are removed from the fur and deposited on the
rod.
Electrons move negative
from fur to the
rubber rod. -- --
positive
++++

The rod is said to be negatively charged because of an

excess of electrons. The fur is said to be positively
charged because of a deficiency of electrons.
 Glass and Silk
When a glass rod is rubbed against silk, electrons are
removed from the glass and deposited on the silk.
glass Electrons positive

move from ++ ++
glass to the negative

silk silk cloth. - - - -

The glass is said to be positively charged because of

a deficiency of electrons. The silk is said to be
negatively charged because of a excess of electrons.
Rubber Rod and Fur Glass Rod and Silk

Fur +

- + - + - +
-
+
Silk
- - + -
+ - + - + + -
+ +
+ -
- + - - - + - + -
+ -
+ + + - + - +
+ - + - + - -
+ +
- -
+ +
- -
+ +
- -
+
-

Rubber Glass
Rubber is a better insulator than fur.
Silk is a better insulator than glass.
 The Electroscope
Laboratory
Laboratorydevices
devicesused
usedtotostudy
studythe
the
existence
existenceof
oftwo
twokinds
kindsof
ofelectric
electriccharge.
charge.

Pith-ball Electroscope Gold-leaf Electroscope

 Two Negative Charges Repel
1. Charge the rubber rod by rubbing against fur.
2. Transfer electrons from rod to each pith ball.

The
Thetwo
twonegative
negativecharges
chargesrepel
repeleach
eachother.
other.
 Two Positive Charges Repel
1. Charge the glass rod by rubbing against silk.
2. Touch balls with rod. Free electrons on the balls
move to fill vacancies on the cloth, leaving each of
the balls with a deficiency. (Positively charged.)

The
Thetwo
twopositive
positivecharges
chargesrepel
repeleach
eachother.
other.
 The Two Types of Charge

Rubber glass
Attraction

fur silk

Note that the negatively charged (green) ball is

attracted to the positively charged (red) ball.

Opposite
OppositeCharges
ChargesAttract!
Attract!
The First Law of Electrostatics

Like
Likecharges
chargesrepel;
repel;unlike
unlikecharges
chargesattract.
attract.

Neg
Pos Neg
Pos Neg Pos
 Electrical
Induction
- - - - - - - - - - - -
- - - - - - - - - - - -
- - - - - - - - - - - -

+
- -
-
+
- +
+
-
+
-
+
-
+
-+
+ -
- +
+ -
- +
+ -
 + -
- +
+ -
- +
+ -
-+
+
-
+
-
+
-
+
- +
+
-
- -
+
+ + + + + + + + + + + +
+ + + + + + + + + + + +
+ + + + + + + + + + + +
Induction
Electrical
 + -
- +
+ -
- +
+ -
-+
+
-
+
-
+
-
+
- +
+
-
- -
+
+ + + + + + + + + + + +
+ + + + + + + + + + + +
+ + + + + + + + + + + +
Induction
Electrical
 +
+
+
+
+
+
+
+
+
-
-
+-
- +
+- -
--
- --
-+
+ + + + + + + + + + + +
+ + + + + + + + + + + +
++ + + + + + + + + + +
Induction
Electrical
 + -
- +
+ -
- +
+ -
-+
+
-
+
-
+
-
+
- +
+
-
- -
+
+ + + + + + + + + + + +
+ + + + + + + + + + + +
Conduction + + + + + + + + + + + +
Electrical
 Electrical
Conduction
+
-
+ +
+

+
-
+

+
+ -
+
+
+
+
 Coulomb’s Law
The
Theforce
forceofofattraction
attractionororrepulsion
repulsionbetween
betweentwo
two
point
pointcharges
chargesisisdirectly
directlyproportional
proportionalto
tothe
theproduct
product
of
ofthe
thetwo
twocharges
chargesand
andinversely
inverselyproportional
proportionalto
tothe
the
square
squareof ofthe
thedistance
distancebetween
betweenthem.
them.

F
- q q’ +
r qq '
F 2
F F r
q q’
- -
 Calculating Electric Force
The proportionality constant k for
Coulomb’s law depends on the choice of
units for charge.
2
kqq ' Fr
F  2 where k 
r qq '
When the charge q is in coulombs, the distance r is
in meters and the force F is in newtons, we have:
2 2
Fr 9 N m
k 9 x 10 2
qq ' C
Example 2. A –5 mC charge is placed 2 mm from a +3
mC charge. Find the force between the two charges.

-5 mC F +3 mC
Draw and label
givens on figure:
q - r + q’

2 mm
9 Nm 2 -6 -6
kqq ' (9 x 10 C2
)( 5 x 10 C)(3 x 10 C
F 2 
r (2 x 10-3 m)2
FF =
= 3.38
3.38 xx 10
104 N;
4
N;
Attraction
Attraction
Note:
Note:Signs
Signsare
areused
usedONLY
ONLYtotodetermine
determineforce
forcedirection.
direction.
 Problem-Solving Strategies
1. Read, draw, and label a sketch showing all
given information in appropriate SI units.
2. Do not confuse sign of charge with sign of
forces. Attraction/Repulsion determines the
direction (or sign) of the force.
3. Resultant force is found by considering force
due to each charge independently. Review
module on vectors, if necessary.
4. For forces in equilibrium: SFx = 0 = SFy = 0.
Example 3. A –6 mC charge is placed 4 cm from a +9 mC charge. What is
the resultant force on a –5 mC charge located midway between the first
charges?

1 nC = 1 x 10-9 C
1. Draw and label. F1
-6 mC F2 +9 mC
2. Draw forces. q3
3. Find resultant;
q1 - r1 - r2 + q2

2 cm 2 cm
right is positive.

kq1q3 (9 x 109 )(6 x 10-6 )(5 x 10 -6 )

F1  2  2
; F1 = 675 N
r1 (0.02 m)
kq2 q3 (9 x 109 )(9 x 10 -6 )(5 x 10 -6 )
F2  2  2
; F2 = 1013 N
r1 (0.02 m)
Example 3. (Cont.) Note that direction (sign) of forces are
found from attraction- repulsion, not from + or – of charge.

+ F1
-6 mC F2 +9 mC
F1 = 675 N q3
q1 - r1 - r2 + q2
F2 = 1013 N 2 cm 2 cm

The resultant force is sum of each independent

force:
FR = F1 + F2 = 675 N + 1013 FFRR==+1690 +1690NN
N;
Example 4. Three charges, q1 = +8 mC, q2 = +6 mC and q3 = -4 mC are
arranged as shown below. Find the resultant force on the –4 mC charge
due to the others.

+6 mC 3 cm q Draw free-body diagram.

q2 + - 3
-4 mC
F2
q3
4 cm 5 cm - -4 mC
53.1 o 53.1o
+ F1
q1 +8 mC

Note the directions of forces F1and F2 on q3

based on attraction/repulsion from q1 and q2.
Example 4 (Cont.) Next we find the forces F1 and F2 from
Coulomb’s law. Take data from the figure and use SI units.

kq1q3 kq2 q3 +6 mC 3 cm q
F1  2 ; F2  2
q2 + F 2 - -4 mC
3
r1 r2
(9 x 109 )(8 x 10-6 )(4 x 10 -6 ) 4 cm
F1
5 cm
F1 
(0.05 m) 2
53.1o
(9 x 109 )(6 x 10-6 )(4 x 10 -6 ) +
F2  q1 +8 mC
(0.03 m) 2
Thus, we need to find resultant of two forces:

FF11==115
115 N, 53.1 SSof
N, 53.1 oo
ofW
W FF22==240
240N,
N,West
West
Example 4 (Cont.) We find components of each force
F1 and F2 (review vectors).
F2
F1x = -(115 N) Cos 53.1 o
240 N F1x
q3
= - 69.2 N - -4 mC
F1y 53.1o
F1y = -(115 N) Sin 53.1o
= - 92.1 N F1= 115 N
Now look at force F2:

F2x = -240 N; F2y = 0 Rx = SFx ; Ry = SFy

Rx = – 69.2 N – 240 N = -309 N RRxx==-92.1

-92.1NN

Ry = -69.2 N – 0 = -69.2 N RRyy==-240

-240NN
Example 4 (Cont.) Next find resultant R from
components Fx and Fy. (review vectors).

RRxx==-309
-309NN RRyy==-69.2
-69.2NN Rx = -309 N
q3
We now find resultant R,q: f - -4 mC
R
Ry
R  R  R ; tan  =
2
x
2
y
Rx Ry = -69.2 N

R  (309 N) 2  (69.2 N) 2 317 N

Thus, the magnitude RR =

= 317
317 N
N
of the electric force is:
Example 4 (Cont.) The resultant force is 317 N. We now
need to determine the angle or direction of this force.

R  Rx2  Ry2 317 N -309 N q

f -
Ry  309 N R -69.2 N
tan   -62.9 N
R x -69.2 N
The reference angle is: f = 77.40S
of W
Or, the polar angle q is: q = 1800 + 77.40 =
257.40
Resultant
Resultant Force:
Force: RR =
= 317
317 N,
N, qq =
= 257.4
0
257.40
 Summary of Formulas:
Like
LikeCharges
ChargesRepel;
Repel;Unlike
UnlikeCharges
ChargesAttract.
Attract.

kqq ' N m 2
F 2 k 9 x 10 9

r C 2

11mC
mC = 1 x 10 CC
= 1 x 10 -6-6
11nC
nC = 1 x 10 CC
= 1 x 10 -9-9

11pC
pC = 1 x 10-12CC
= 1 x 10 -12
11electron:
electron: e = -1.6 x 10-19CC
e --
= -1.6 x 10 -19