electromagnetism around the equator and it balance with the rain

physics in Africa

1. PHY3612: ELECTROMAGNETISM
Mr. S. A. Shimboyo
Department of Physics, Chemistry &
Material Science
University of Namibia
Main Campus
© 2024

2. MR. S.A. SHIMBOYO
Department of Physics,
Chemistry & Material Science
OFFICE: W-Block WM16
E-mail:[email protected]

7. COURSE INFORMATION
Course description:
E
LECTROSTATICS
:Static electric charge, Electric interaction; Gauss’s Law,
electric force and field, Electric potential, Capacitors;
DC C
IRCUITS
:Electric current, Ohms law, Kirchhoff’s laws, resistance, Joule
effect and emf;
M
AGNETOSTATICS
: Magnetic interaction, Lorentz force, Electromagnetic field of
a moving charge, Electric flux of a moving charge, Magnetic field and electric
current, Ampere’s law, Time dependent electric field; Maxwell’s equations.
KindlyconsulttheadditionalcourseinformationprovidedontheMOODLE
platform.

8. COURSE INFORMATION
B
RIEF
H
ISTORYOF
E
LECTROMAGNETISM
TheancientGreeksdiscoveredasearlyas600B.C,thataftertheyrubbedamberwithwool,
theambercouldattractotherobjects.Sincethenthisphenomenawasfurtherinvestigatedby
manyfamousscientistssuchasFranklin,Ørsted,Ohm,Coulomb,Ampére,Faraday,Henry,
etc.Theelectromagneticphenomenawassignificantlyenhancedandbeautifullypackaged
byJamesClerkMaxwellin1865–throughasetofwell-knownequationsthatare
customarilyknownastheMaxwell’sequations.Theelectromagneticforce,whichisoneof
thefourfundamentalforcesofnature,holdsatoms,moleculesandmaterialstogetherand
playsavitalroleinhumanunderstandingofalmostallexistingandpotentialtechnological
developments.Asitturnsout,theelectromagneticforceistheoneresponsiblefor
practicallyallphenomenaencounteredindailylife,withtheexceptionofgravity.

9. Chapter 1
Electric Interaction
In nature motions are brought about by interactions;
The fundamental types of interactions that exist in nature are;
(i) Gravitational interaction
(ii) Electromagnetic interaction
(iii) Strong (nuclear) interaction
(iv) Weak interaction –e.gbeta decay

10. The concept of a field is used to describe the earlier mentioned interactions
Definition:
Field:-A physical entity extending over a region of space and is a function of positionand time.
The interaction between two particles depend on their relative positionsand motions.
For each interactiona particle produces a corresponding field around it and this field thus
acts on any other second particle, hence we have interaction.
But the second particle also has its own field that interacts with the field of the first particle,
hence we have mutual interaction.
Now electromagnetic interaction is described by two fields;
(i)Electric field (E-field) and
(ii)Magnetic field (B-field)
The electric and magnetic fields are not independent entities, but intimately related.
A requirement of a field to describe interaction between particles;
(i) The field must possess energyand momentum
(ii) The field must be able to transport energyand momentumfrom one particle to another.

11. Electric Charge
We will introduce a new property of matter known as “electric charge”
(symbol q).
We will explore the charge of atomic constituents.
Moreover, we will describe the following properties of
charge:
•Types of electric charge
•Forces among two charges (Coulomb’s law)
•Charge quantization
•Charge conservation

12. Electric Charge
Around 600 BC ancient Greeks discovered that if you rub amber with wool, thenit acquires
the property of attracting light objects such as feathers.
Today we say that the amber has net charge (or has become charged)
The Greek word for amber is “elektron” and this is where the word “electric” is derived from.
In later years after numerous experiments by Benjamin Franklin (1706 -1790) he discovered
that there exists two types of electric charges and he named one “ positive” and the
other “negative”.
When we rub a glass rod with silk cloth, both objects acquire electric charge.
The sign on the charge on the glass rod is defined as positive. In a similar fashion, when we
rub a plastic rod with furboth objects acquire electric charge.
The sign on the charge on the plastic rod is defined as negative.

13. Q:Do we have enough information so as to be able to
determine the sign of all other charges in nature? To
answer this question we need one more piece of
information.
Charges of the same sign repel each other.
Charges of opposite sign attract each
other.
Further experiments on charged objects showed that:
1.Charges of the same type (either both positive or
both negative) repel each other (fig. a).
2.Charges of opposite type on the other hand attract
each other(fig. b).
The force direction allows us to determine the sign of
an unknown electric charge.

14. repulsive
force
attractive
force
The recipe is as follows:
We charge a glass rod by rubbing it with silk cloth.
Thus we know that the charge on the glass rod is
positive. The rod is suspended in such a way so that it
can keep its charge and also rotate freely under the
influence of a force applied by charge with the
unknown sign. We approach the suspended class rod
with the new charge whose sign we wish to determine.
Two outcomes are possible. These are shown in the
figure to the left:
Fig. a: The two objects repel each other. We then
conclude that the unknown charge has a positive sign.
Fig. b: The two objects attract each other. We then
conclude that the unknown charge has a negativesign.

15. InBenjaminFranklin’sday(18
th
century)it
wasassumedthatelectricchargeissome
typeofweightlesscontinuousfluid.
Investigationsofthestructureofatomsby
ErnestRutherfordatthebeginningofthe
20
th
centuryrevealedhowmatteris
organizedandalsoidentifiedthatchargeof
itsconstituents.
Thuselectricchargeisafundamentalpropertyoftheelementaryparticles(electrons,
protons,neutrons)outofwhichatomsaremade.
Atoms consist of electronsand the nucleus.
Atoms have sizes 510
-10
m.
Nuclei have sizes 510
-15
m.
The nucleus itself consists of two types of
particles: protons and neutrons.
The electrons are negatively charged. The
protons are positively charged. The neutrons
are neutral (zero charge).

16. ThechargeofanelectronwasfirstmeasuredbytheAmericanphysicistRobertMillikanduring
(1909-1913).Inthisexperiment,oilissprayedinveryfinedrops(around10
-4
mmindiameter)
intothespacebetweentwoparallelhorizontalplatesseparatedbyadistanced.Apotential
differenceV
AB
ismaintainedbetweenthem,causinganupwardelectricfieldbetweenthem.
Thedropsarethenobservedthroughamicroscope.Usingthewell-knownMillikanOil-Drop
Experimentwewillshowthatthecharge,qisalwaysamultipleofthefundamentalchargee.
Charge is quantized
Charge on a proton
?5=
Coulombs
?5=
C
(where is an integer multiple of the elementary charge)

17. Proof:
Consider an oil drop of mass mand charge q (with radius r) that is in free fall. Now, consider
the following Free Body diagram for the oil drop
????????????
??????
????
Oil drop
x
y
The well-known Millikan Oil-Drop Experiment shows that charge qis always a
multiple of the fundamental charge e.

21. Charge is quantized
•Usingthedefinitionthat e=1.602177×10
−19
C,
otheelectron’schargeis−e.
oTheprotonhascharge+e.
•The particles found in nature all have charges which are integral
multiples of the elementary charge e:
oq = ne where n = 0, ±1, ±2 . . ..
oBecause of this, we say that charge is quantized
•The mass of the electron ism
e
= 9.109 ×10
−31
kg
•The mass of a proton is m
p
= 1.673 ×10
−27
kg

22. silk
glass rod
silk
glass rod
-
--
-
+
+
+
+
Conservation of Charge
Consideraglassrodandapieceofsilkcloth(both
uncharged)shownintheupperfigure.Ifwerubthe
glassrodwiththesilkclothweknowthatpositivecharge
appearsontherod(seelowerfigure).Atthesametime
anequalamountofnegativechargeappearsonthesilk
cloth,sothatthenetrod-clothchargeisactuallyzero.
Thissuggeststhatrubbingdoesnotcreatechargebut
onlytransfersitfromonebodytotheother,thus
upsettingtheelectricalneutralityofeachbody.Charge
conservationcanbesummarizedasfollows:Inany
processthechargeatthebeginningequalsthechargeat
theendoftheprocess.
Net charge before = Net charge after
i f
Q Q

23. Technological Application of electric charge
The Laser printer

24. The Laser printer revision
1. The laser writes negative charges on the positively charged drum only at those places where
the information/image to be copied/printed appears and leaves the other blank spaces as
positive charges.
2. The positively charged toner adheres itself only to the areas where there are negative
charges, because of the attraction between positive and negative charges. While the areas on
the drum where there are positive charges is not affected, since the positive areas on the drum
and the positively charged toner repel each other.
3. As the paper on which the information/image is to be printed/copied is fed into the machine
as indicated on the diagram a more negative charge is sprayed by negative wires, such that the
positively charged toner can adhere itself on the paper.
4. Fuser rollers heat the toner and presses it onto the paper so that the information/image
produced by the toner could be permanently fixed on the paper.
5. A lamp discharges the drum and leaves it uncharged.

25. 1 2
2
0
1

4
q q
F
r

1 2
Consider two charges and placed at a distance .
The two charges exert a force on each other that has the
following characteristics:
The force acts along the line connecting the tw
q q r
Coulomb's Law
1. o charges.
The force is attractive for charges of opposite sign.
The force is repulsive for charges of the same sign.
The magnitude of the force, known as Coulomb
force, is given by the equatio
2.
3.
1 2
0
-12 2
2
0
0
2
n
The constant is known as the
=8.85 10 N m /C .
The Coulomb force has the same form as Newton's
gravitational force. The two differ in one as
1
.
4

pect:
The gravi
q q
F
r



 

permittivity constant
tational force is always attractive.
Coulomb's force on the other hand can be either
attractive or repulsive depending on the sign
of the charges involved.
1 2
2

mm
F G
r


26. The unit of charge in the SI unit system is the "Coulomb"
In principle we could use Coulomb's law for two equal charges as follows:
Place the two charges at a distance
(symbol C )

.
=
q
r
Units of Charge
9
0
2 2
9
2 -12 2
0
1
1 . = 1 C if 8.99 10 N:
4
1 1 1
8.99 10 N
4 4 3.14 8.85 10 1
For practical reasons that have to do with the accuracy of the definition, the
electric current is used instead. The
m q F
q
F F
r


  
    
  
electric current in the circuit of the figure
is defined by the equation i.e., the amount of charge that flows
through any cross section of the wire per unit time. The unit of current in S
,
I

i
dq
i
dt
i

s the ampere and it can be defined very acc(symbo uratel A) ly.
If we solve the equation above for we get .
Thus if a current = 1A flows through the circuit,
a charge = 1C passes through any cross section
of the wire in one second.
dq dq idt
i
q


27. The net electric force exerted by a group of
charges is equal to the vector sum of the contribution
from each charge.
Coulomb's Law and the Principle of Superposition
1 1 2 3
12 13 1 2 3
1 12 13
For example, the net force exerted on by and is equal to
Here and are the forces exerted on by and , respectively.
In general, the force exerted o
.
n
F q q q
F F q q q
F F F 
 
 

1
1 12 13 14 1 1
2
12 13
by charges is given by the equation
...
One must remember that , , ... are vectors and thus
we must use vector addition. In the example of fig. we have
n
n i
i
q n
F F F F F F
F F
f

      
     
 
:
1 12 14
F F F 
  

28. Electric Fields
We will introduce the concept of an electric field. As long ascharges are stationary, Coulomb’s
law describes adequately the forces among charges. If the charges are not stationarywe must
use an alternative approach by introducing the electric field (symbol ).
E

In connection with the electric field, the following topics will be
covered:
•Calculating the electric field generated by a point charge.
•Using the principle of superposition to determine the electric field created by a collection
of point charges as well as
•continuous charge distributions.
•Once the electric field at a point Pis known, calculating the electric force on any charge
placed at P.

29. Previously we discussed Coulomb’s law, which gives the force between two pointcharges.
The law is written in such a way as to imply that q
2
acts on q
1
at a distance rinstantaneously
(“action at a distance”):
1 2
2
0
1

4
q q
F
r

 
1 2
generates electric field exerts charge a force on
  
FE E qq
Electric interactions propagate in empty space with a large but finitespeed (c= 310
8
m/s).
In order totake into accountcorrectly the finite speed at which these interactions propagate,
we have toabandon the “action at a distance” point of view and still be able to explain how q
1
knows about the presence of q
2
.
The solution is to introduce the new concept of an electric fieldvector as follows: Point
charge q
1
does not exert a force directly on q
2
. Instead, q
1
creates in its vicinity an electric
field that exerts a force on q
2
.

30. Consider the positively charged rod shown in the figure.
For every point in the vicinity of the rod we define
the electric field vector as follows:
We place
P
E
Definition of the Electric Field Vector
1.

0
0
0
a test charge at point .
We measure the electrostatic force exerted on
by the charged rod.
We define the electric field
.
vector at point

as:
SI Units :
q P
F q
E P
F
E
q

2.
positiv
3
e
.




0

From the definition it follows that is parallel to .
We assume that the test charge is small enough
so that its presence at point does not affect the charge
distribution on the
N/C

d

ro
E F
q
P
Note:
 
and thus alter the electric
field vector we are trying to determine.E

0
F
E
q




31. q
q
o
r
P
E

2
0
1

4
q
E
r

0
0 0
Consider the positive charge shown in
the figure. At point a distance from
we place the test charge . The force
exerted on by is equal to:
q
P r
q q
q q
Electric Field Generated by a Point Charge
0
2
0
0
2 2
0 0 0 0

1

4
1 1

4 4
The magnitude of is a positive number.
In terms of direction, points radially
as shown in the figure.
If were a negative charge the magnitude
of
q q
F
r
q q qF
E
q qr r
E
E
q

 

  
outward


would remain the same. The direction
of would point radially instead.
E
E inward



32. O O
The net electric electric field generated by a group of point charges is equal
to the vector sum of the electric field vectors ge
E
Electric Field Generated by a Group of Point Charges. Superposition

1 2 3
1 2 3 1 1 3
1 2 3
nerated by each charge.
In the example shown in the figure, .
Here , , and are the electric field vectors generated by , , and ,
respectively.
, , and No mte:
E E E E
E E E q q q
E E E
  
   
  
  
1 2 3 1 2 3 1 2 3
,
ust
,
be added as vectors:
x x x x y y y y z z z z
E E E E E E E E E E E E        

33. •Canberepresentedby“fieldlines”/“forcelines”
•TheE-fieldateachpointisrepresentedbyatangenttothelineatthatpoint
•ThedensetheE-fieldlinesthegreatertheE-field
•E-fieldlinesdonotcross(iftheydothenthereexiststwodifferentE-fieldsin
differentdirections).
Motion of an electric charge in a uniform electric field
Showing that the path of a charged particle moving through a uniform E-field is
parabolic. Consider the case as shown in the figure below.

36. Consider the continuous charge distribution shown in the
figure. We assume that we know the volume density of
the electric charge. This i

Electric Field Generated by a Continuous Charge Distribution
3
s defined as
Our goal is to determine the electric field generated
by the distribution at a given point . This type of problem
can be solved using t
(
he
Uni
principle of superpos
ts: C/m ).
dq
dV
dE
P


ition
as described below.
Divide the charge distribution into "elements" of volume . Each element
has charge . We assume that point is at a distance from .
Determine the electric field generated b
dV
dq dV P r dq
dE

1

.
2.

2
0
2
0
y at point .
The magnitude of is given by the equation .
4
ˆ1
Sum all the contributions: .
4
dq P
dq
dE dE dE
r
dVr
E
r





3.


P
dq
r
dE
dV
ˆr

37. C
A
dq
Determine the electric field generated at point
by a uniformly charged ring of radius and total charge .
Point lies on the normal to the ring plane that passes through
the ring cent
E P
R q
P
Example:

2 2
er , at a distance . Consider the charge element
of length and charge shown in the figure. The distance
between the element and point is .
The charge generates at an electric fie
C z
dS dq
P r z R
dq P
 

2
0
3/23
2 2
0
0
ld of magnitude
that points outward along the line :
. The -component of is given by
4
cos . From triangle PAC we have: cos /
.
4
4
z
z z z
dE AP
dq
dE z dE
r
dE dE z r
zdq zdq
dE E dE
r
z R

 



 
   


 
3/2 3/2
2 2 2 2
0 0

4 4
z
z zq
E dq
z R z R 
 
 



38. electric field lines
P
Q
P
E

Q
E

In the 19th century Michael Faraday introduced the
concept of electric field lines, which help visualize the electric field vector
without using mathematics. For the relatio
E
Electric Field Lines.

n between the electric field lines and : E
1. At any point the electric field vector is tangent to the electric field lines.


P E
P
P
E

electric field line
2. The magnitude of the electric field vector E is proportional
to the density of the electric field lines.

P Q
E E

39. q
q
negative charg
3. Electric field lines extend away from (where they originate)
and toward (where they terminate).
Example 1: Electric field lines of a negative
pos
po
itive cha
int charg
rge
e
s
s
e - :q
2
0
1

4
q
E
r

-The electric field lines point toward the point charge.
-The direction of the lines gives the direction of .
-The density of the lines/unit area increases as the
distance from decreases.

E
q


In the case of a positive point charge
the electric field lines have the same
form but they point

out
No
w
te:
ard.

40. In the next chapter we will see that the
electric field generated by such a plane has
Example 2: Electric field lines of an electric field generated by an infinitely
large plane uniformly charged.
the form shown in fig. .
The electric field on either side of the plane has a constant magnitude.
The electric field vector is perpendicular to the charge plane.
The electric field vector
b
E
1.
2.
3.

points away from the plane.
The corresponding electric field lines are given in fig. .
For a negatively charged plane the electric field lines point inward.
c
Note:

41. Electric field lines generated by
an electric dipole (a positive
and a negative point charge of
the same size but of opposite
sign)
Example 3:
Electric field lines generated by
two equal positive point charges
Example 4:

42. Electric Potential
A charged particle placed in an E-field experiences an E-force, therefore the
charged particle has electric potential energy.
To move the charge from one location to another work must be done.
Definition:
The electric potential V at a point in an E-field is defined as the electric potential energy
per unit charge at that point.
V is a scalar because energy and charge are both scalars
Denoting the electric potential energy U
p
we may write
?
?
?
or
?
Units of potential
?5 66?5
volt named after
Alessandro Volta (1745 –1827)

43. The E –V relationship
The relationship between the Eand the scalar potential Vcan be shown by
considering a charge moving from a point A to a point B in the presence of an E-field.
SupposeachargeqmovesfromapointatoapointBinthepresenceofanE-field,thechargein
factmovesfromapointwheretheelectricpotentialisV
A
toapointwherethepotentialisV
B
and
itsenergychangesfrom
?
?
to
?
?
.Sothatfrom
?
wecanwrite
? ?
?
?
?
? ? ?
?
?
?
? ?
where the RHS is the work done in moving the charge q from A to B
??
.
?? ? ?
But and
??
?
?
from first principles we get
??
?
?
? ? ? ?
?
?

44. or
? ?
?
?
, now since
? ?
?
?
?
?
? ?Hence,
?
?
?
?
where and
? ? ?
. Alsowe can write –dVcomponent-
wise as
?
,
?
and
?
. So that
?
?
?
where,
!
!?
!
!?
!
!?
.
The E –V relationship

45. Definition:
!
!?
!
!?
!
!?
thus Gradient of V (points in the increasing
direction of V)
“Such that the E-field is the negative of the gradient of the scalar potential”
For a uniform E-field, as shown below along x-axis
x-axis
then with at
?
4
?
4
?
4
( Eis uniform/constant)
and
?
?
?5
is in the direction of decreasing V

46. Electric Potential of a point charge at R
For a positive point charge the E-field is radial as shown in the figure below
q
q
o
r
P
E

Hence,
!?
!?
where r = radial distance
but
5
8
,
?
?
.
, therefore,
5
8
,
?
?
.
!?
!?
4
6
?
4
4
6
?
4
4
4
Note:
?
because the potential far from a source charge is negligible (zero)

47. That is, 5
8
,
?
?
for some point charge q
However, for several point charges
5
,
6
,
7
,
8
….
?
the net potential at a point P is
the scalar sum of the individual potentials
4
5
5 4
6
6 4
7
7 4
?
? 4
?
?
?
?@5
Definition:
A surface joining points with the same electric potential is called
“an equipotential surface” . No work is done between two points are at the same
potential. The E-field lines are always perpendicular to equipotential surfaces.

48. Electrostatic Energy
A particle with mass mand charge qthat moves with velocity vin an E-field has total
energy
? ?
, where
?
is kinetic energy and
?
is potential energy
6
moving from point
5
(with potential
5
) to point
6
(with potential
6
) the particle
has velocity
5
(at point
5
) and
6
(at point
6
) such that the
Conservation of Energy gives
5
6
5 6
6
6
6
6
5
6
5 6

49. Now recall the Work –Energy Principle , which says, the total work done changes
a particle’s kinetic energy
6
6
5
6
56
6
6
5
6
Note:1 volt == potential difference through which qof 1 C has tomove to gain 1 J
of energy.
From
5
6
6
6
5
6
5
6
56
energy gained we see that:
•If , then
56
(meaning
5 6
to get
•If , then
56
(meaning
5 6
to get

50. That means: A positive charge must go from a larger potential to a smaller potential
to gain kinetic energy, while a negative charge must go from a lower potential to a
higher potential to gain kinetic energy.
If a charged particle starts from rest (
5
) and moves to a point with zero potential
(
6
) , then we get
6
6
5
Or simply
6
“used in electrostatic accelerators”