1

Welcome to Electromagnetism
Ed Copeland
Arundel 205
e.j.copeland@sussex.ac.uk
http://www.pact.cpes.susx.ac.uk/users/edmundjc/emg_course.htm
I am very grateful to Professor Dam Waddill for making his
slides available to me for this course.
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Some details:
Lectures: Tue 11.30 and Thur 9.15 : Pev1-1A6.
Workshop: Tue 18.00 : Pev1-1A1.
Problem sheets: Hand in Monday following
relevant workshop.
Essay: Handed in first weds after Easter break.
Office hour: Thur 15.00: Arundel 205
Course Book: Tipler, Physics for Scientists and
Engineers, 4
th
edition.
All details to be found at:
http://www.pact.cpes.susx.ac.uk/users/edmundjc/emg_course.htm
3
Electric Charge and Electric Field
Todays menu
Properties of Electric Charges
Insulators and Conductors
Coulombs Law
The Electric Field
The Electric Dipole
4
Electric Charges
Two kinds of charges: Positive and Negative
Like charges repel - unlike charges attract
Charge is conserved and quantized
1909 Robert Millikan : electric charge always occurs in integral
multiples of the fundamental unit of charge, e.
Q is the standard symbol for charge (units-Coulombs)
Q = Ne ; e = 1.602 x 10
-19
C, N is an integer
Proton charge: + e : Electron charge: - e : Neutron charge: 0
Quarks charge : 1/3 e or 2/3 e How come?
Never find isolated individual quarks
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Insulators and Conductors
Classify materials according to their ability to
conduct electrical charge.
Conductors: free moving charge (metal)
Insulators: charge not readily transported (wood)
Semiconductors: electrical properties between
conductor and insulator (silicon)
6
Coulombs Law
1785 Charles Coulomb : fundamental law of electric
force between two stationary charged particles. It is:
inversely proportional to square of separation between
particles
directed along the line joining the particles
proportional to the product of the two charges
attractive if particles have charges of opposite sign
and repulsive if charges have same sign
1 2
12
2

e
QQ
F k r
r
=
r
7
1 2
12
2

e
QQ
F k r
r
=
r
F
12
is the force on charge Q
2
due to Q
1
and
is the unit vector pointing from Q
1
to Q
2
r is the distance between Q
1
and Q
2
k
e
is Coulombs constant and has a value of
8.988 x 10
9
N
.
m
2
/C
2
r

1
Q
2
Q
r
r

r
r
r

r
r
=
8
Coulombs constant k
e
in terms of the permittivity of free space
0
.
-12 2 2
0
1
8.85 10 C /N m
4
e
k

= =
1 2
12
2
0
1

4
QQ
F r
r
=
r
1 2
12
2

e
QQ
F k r
r
=
r
Coulombs law can be written as :
or
9
1 2
12
2

M M
F G r
r
=
r
1 2
12
2

e
QQ
F k r
r
=
r
Coulomb Force
Gravitational Force
Lets compare
Attractive or repulsive Only attractive
36
g
c
27
p
2 2 11
19 2 2 9
2
proton
2
g
c
10 24 . 1
F
F
kg 10 67 . 1 m ; kg Nm 10 67 . 6 G
C 10 6 . 1 e ; C Nm 10 99 . 8 k
protons 2 for
Gm
ke
F
F
=
= =
= =
=


Why dont we worry about electric forces for macroscopic bodies?
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Reminder of direction of Coulomb Force
+
+

1
1
1
+

2
2
2
F
21
F
21
F
12
F
12
F
12
F
21
Recall F
12
is force on charge 2 due to charge 1
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Superposition Principle vital!
When more than two charges are present:
resultant force on any one of them is equal to the
vector sum of the forces exerted by each of the
individual charges.
1 21 31 41
F F F F = + + +
r r r r
L
12
+
+
+
0.5 m
0.5 m
Q
2
Q
1
0.3 m
0.3 m
y
0.4 m
F
23
Q
3
F
13
x

Example 1
3 point charges Q
1
= Q
2
= 2 C
and Q
3
= 4 C are arranged as
shown. Find the resultant force on
Q
3
.

( )
1 3
13
2
0.5 m
e
Q Q
F k =
( )
6 6
9
13
2
(2.0 10 )(4.0 10 )
(9.0 10 ) 0.29 N
0.5
F


= =

( )
2 3
23 13
2
0.5m
e
Q Q
F k F = =
13
+
+
+
0.5 m
0.5 m
Q
2
Q
1
0.3 m
0.3 m
y
0.4 m
F
23
Q
3
F
13
x

Continuation of Example 1
3 point charges Q
1
= Q
2
= 2 C
and Q
3
= 4 C are arranged as
shown. Find the resultant force on
Q
3
.
13
0.29N F =
23 13
F F =
13 13
13 13
( ) cos
( ) sin
x
y
F F
F F

=
=

0.4
cos 0.8
0.5
0.3
sin 0.6
0.5

= =
= =

13 23 13 23
13
( ) ( ) cos cos
2 cos 2(0.29)(0.8) 0.46 N
x x x
x
F F F F F
F F

= + = +
= = =

13 23
13 23
( ) ( )
sin sin 0 N
y y y
y
F F F
F F F
= +
= + =

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The Electric Field
Useful when describing a force that acts at a distance.
Electric field at some point in space is defined as the
electric force acting on a positive test charge, q
0
, placed
at that point divided by the magnitude of the test charge.
It is a vector quantity with units of N/C.
0
F
E
q
=
r
r
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For point charges:
+
Q
q
0
F
r

r
-
Q
q
0
F
r

r
0
2
2
0 0

e
e
Qq
k r
F Q
r
E k r
q q r
= = =
r
r
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Superposition principle for Electric Fields
If the field is due to more than one charge then the
individual fields are added vectorially (superposition
principle).
1 2 3 4
... E E E E E = + + + +
r r r r r
For a series of point charges the electric field is:
2

i
e i
i
i
Q
E k r
r
=

r
r
i
is the distance from the i
th
charge to the point of
evaluation
is a unit vector from the i
th
charge to the point of
evaluation, and Q
i
is the i
th
charge.
i
r

17
+
x
Q
1
Q
2
0.6 m
0.5 m 0.5 m
P
y

E
2
E
1

Example 2
Charges Q
1
and Q
2
are placed 0.6 m apart. Q
1
= +5 C
and Q
2
= -5 C. Find the electric field at point P.
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1 9 5
1 2
2 2
1
5.0 10
(9.0 10 ) 1.8 10 N/C
(0.5)
e
Q
E E k
r

= = = =

1 2
1 2
1
cos cos
2 cos
x x x
x
x
E E E
E E E
E E

= +
= +
=

0.3
cos 0.6
0.5
= =
5 5
2(1.8 10 )(0.6) 2.2 10 N/C
x
E = =

1 2
1 2
sin sin 0 N/C
y y y
y
E E E
E E E
= +
= =

( )
5

2.2 10 N/C
x y
E E i E j
E i
= +
=
r
r

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Electric dipoles
Electric dipole is a system of two
equal and opposite charges Q a
small distance L apart
Electric dipole moment, p, is
vector pointing from negative to
positive charge with magnitude
given by p=QL.
If L is displacement vector of
positive charge from negative
charge, dipole moment is
+
-Q
Q
L
p=QL
L Q p
r
r
=
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Example 3
y
x
-Q
+Q
-a
a
Given two opposite charges at a, and a, find
electric field and dipole moment on the x-axis
at a field point P which is a large distance
away compared to 2a.
P
1. Point P is a dist (x-a) from positive and dist (x+a) from neg charges.
2. Electric field at point P due to the two charges is:
i
) a x (
axkQ 4
i
) a x (
) Q ( k
i
) a x (
kQ
E
2 2 2 2 2
r r r r

=
+

=
3. For x>>a, can neglect a
2
compared to x
2
in the denominator.
Electric field at P is:
i
x
kaQ 4
E
3
r r

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3. For x>>a, can neglect a
2
compared to x
2
in the denominator.
Electric field at P is:
i
x
kaQ 4
E
3
r r

4. Electric Dipole Moment Displacement is: