Electrostatics – magnetism - electricity – electromagnetism
2000 BC
700 BC
1600
1785
William Gilbert
Shows
electrostatics
occurs
generally
Chinese
document
magnetism
1819
1831
Hans Oersted
Deflects
compass
by electricity
Charles Coulomb
Ancient Greeks
Static electricity
Elektron (Amber)
F
1
q1q2
40 r 2
Ancient Greeks
Magnetism
Magnetite found in
Magnesia district
Dr SH Connell (76826)
082 945-7508
School of Physics,
http:www.src.wits.ac.za/~connell
Michael Faraday and
Joseph Henry
Induction of current
by relative motion of
magnet and conductor
1873
1888
James Clark Maxwell
Maxwell’s Equations
EM waves with v = c
1909
2004
Robert Millikan
Q = Ne
Heinrich Herz
Verifies ME by
experiment
New results
Active research
and applications
Electrostatics (9 lectures)
Charges Insulators and Conductors
Coulomb’s Law
Electric Field and its calculation
Lines of force
Electric flux, Gauss’s theorem and applications
Electric Potential and potential difference
Relation between electric potential and field (1D)
Equipotential surfaces
Electron volt as a unit of energy
Electric potential of a point charge and charge distributions
Van de Graaff generator
Definition of capacitance
Parallel plate cylindrical capacitors
Capacitors in series and parallel
Energy of a charged capacitor
23-1,2
23-3
23-4,5
23-6
24-1,2,3,4
25-1
25-2
25-1,6
25-1
25-3
25-8
26-1
26-2
26-3
26-4
Current Electricity (7 lectures)
Electric current, current density
Drift velocity and microscopic view of currrent
Ohm’s Law and resistance
Resistivity and conductivity; temperature variation
Electric Power
Resistors in series and parallel
Emf and terminal voltage
Kirchoff’s laws and their applications
Ammeters and voltmeters, determination of resistance
Wheatstone bridge and slidewire bridge
27-1
27-2
27-2
27-4
27-6
28-2
28-1
28-3
28-5
28-5
Electrostatics is an active field …………………………
“Stationary” charges
Insulators
Moving charges
Conductors
(semiconductors)
Due to currents
(magnetic
materials)
STATIC
ELECTRICITY
Forces  fields
Potential Energy
Capacitance
CURRENT
ELECTRICITY
Batteries & circuits
Energy conversion
Resistance
ELECTROMAGNETISM
Forces  fields
Generate currents
Inductance
Shell model of the Atom
• small +vely charged nucleus
• surrounded by e- in planetary orbits
• normal atom is neutral
+ve nucleus
N protons
Charge = +Ze
nuclear diameter
10-15 m
(Rutherford’s exp)
The atom is
mostly
empty space
The Basics
(Serway 23-1,2,3,4,5,6)
How do we know about static electricity ?
Make sense of experiments …….
1. Insulators can be charged by conduction
2. Conductors can be charged by induction
3. There are two types of charges
a) Like charges repel
b) Unlike charges attract
4. Benjamin Franklyn’s convention, when rubbed with fur
a) Glass acquires a +ve charge
b) Rubber acquires a –ve charge
5. Charge is conserved (not created or destroyed)
a) Charging results from separation and transferal of charges
6. Definition of charge
MKS/SI unit of charge is the Coulomb (C)
The charge that results from a flow of current
of 1 Amp for 1 second.
7. Charge is quantised, q = Ne, e = 1.602 x 10-19 C
8. Materials come in three types
a) Conductors (Charge moves freely : Cu, Al, ….)
b) Insulators (Charge is not mobile : glass, rubber …)
c) Semiconductors (Intermediate behaviour, Si, Ge …)
Triboelectric series
+
Rabbits fur
Glass
Mica
Wool
Cats fur
Silk
Cotton
Wood
Amber
Resins
Metals (Cu, Ni, Co, Ag, etc)
Sulphur
Metals (Pt, Au, etc)
Celluloid
_
When two materials are
rubbed against each
other, the one higher in
the chart will lose
electrons
There are two types of charges, positive and negative
Like charges repel, Unlike charges attract
Neutral
metal
sphere
Charging by induction
Remove
ground
connection
Induced
redistribution
of charge
NB : No contact !
Partial discharge
by
grounding
Excess
charge
remains
A charged object induces near-surface charge in an insulator
Atoms or Molecules near the surface are induced to become partial temporary dipoles
This is the mechanism by which the comb attracts the paper
Coulomb’s Torsion Balance
Coulomb’s Law
Coulomb showed experimentally that the electric force
between two stationary charged particles is
1. Inversely proportional to the square of the separation of the
charged particles and directed along the line joining them
2. Directly proportional to the product of the two charges.
3. Attractive for unlike charges and repulsive for like charges.
q1q2
F k 2
r
If F is in Newtons, q1 and q2 in Coulombs
and r in meters, then,
1
q1q2
F
40 r 2
0 = permittivity of free space
= 8.86 x 10-12 C2N-1m-2 or Farads.m-1
then
k = 1/40 = 9 x 109 mF-1
Twist measures repulsive force
The resultant force in a system of many charges
Force is a vector quantity.
The force between two charges is
1
q1q2
F12 
.rˆ
2
40 r
and attractive (repulsive) for an overall
-ve (+ve) sign.
+q1
-q2
q1q2 < 0  attractive force
The Principle of Superposition
If there are more than two point charges, then Coulomb’s Law holds
for every pair of charges. Let the charges be q1, q2, q3, q4, … qn, ….
Then the resultant force on qn is given by the vector sum
FnR  Fn1  Fn 2  Fn3  .....  (Fnn  0)  ....
See Examples …….
Coulomb’s Law – example 1
Find the resultant force (and direction) on q1
y
q3 = -2.0 mC
30°
q1 = -1.0 mC
q2 = +3.0 mC
x
15 cm
The Hydrogen Atom
Compare the electrostatic and gravitational the forces
(The average separation of the electron and proton
r = 5.3 x 10-11 m ≈ ½ Å)

e

r
F

+Ze

2
19
( Ze)e 
C
9 N.m  1.60  10

Fe 
. 2   8.99 10
2
2 
11
40
r
C
5
.
3

10
m


1
v
2
 8.2 108 N
Fg  G
me m p
r2

 
2
31

kg  1.67 10 27 kg
11 N.m  9.11 10

  6.7 10
2
2 
11
k
g
5.3 10 m





 3.6 1047 N
Fe/Fg = 2 x 1039  The force of gravity is much weaker than the electrostatic force
What other fundamental differences are there ?
Coulomb’s Law – example 2
A certain charge Q is to be divided into two parts, q and Q-q.
What is the relationship of q to Q if the two charges, placed a given
distance apart, are to have a maximum coulomb repulsion ?
The Electric Field
The electric field E at a point in space is defined as the electric force F,
acting on a positive test charge q0, placed at that point divided by
the magnitude of the charge.
F
E
q0
Units NC-1 or Vm-1
It follows that
F  Eq
 Note that the electric field E is produced by some other charge external
to the test charge q0.
 The existence of an electric field is a property of its source
 Every charge comes with its own electric field.
 The electric field will exist regardless of its magnitude and direction
being measured with a test charge.
Electric field lines




These are a way of visualizing the electric field.
The electric field vector E is tangential to the electric field line at any point.
The magnitude of the electric field vector E is proportional to the density of the lines.
Electric field lines can never cross.
 For a positive point charge, the lines
are directed radialy outward.
 For a negative point charge, the
lines are directed radialy inward.
An electric dipole has two nearby
point charges of equal magnitude q
and opposite sign, separated by a
distance d. The number of lines
leaving the positive charge equals
the number of lines entering the
negative charge.
 The electric field lines of two
neighbouring positive point charges
is as shown in the figure. At large
distances, they will approximate the
electric field lines a a single point
charge of magnitude 2q.
Example 3
Show that the density of electric field lines around a point charge is
consistent with the expression for the electric field arising from Coulomb’s
Law.
Calculating the Electric Field due to a point charge at a distance r
Place a point charge q0 a distance r from the charge q
1 q q0
F
r̂
2
40 r
q
r
+
+
F
q0
Now calculate the electric field from the definition
F
E
q0
1
q
E
r̂
2
40 r
The unit vector r lies along the line joining the point charge and the
test charge, with a direction indicating the direction the point charge
would move.
For the electric field due to a system of point charges, calculate the electric field
vectors individually, and add them vectorially using the Principle of Superposition.
E  E1  E2  E3     En
The electric field on the axis of an electric dipole
Consider the situation as shown in the figure, with y >> a.
The dipole moment is defined as p = 2aq
E  E1  E2
The magnitudes of the electric field contributions from each
charge at point P are the same. The y-components are
equal, so the resultant field is ...
E  2 E1 x  2 E1 cos 
1
q
a
2
40 ( a 2  y 2 ) ( a 2  y 2 )1 / 2
p

40 ( a 2  y 2 )3 / 2

p
40 y 3
The Electric Field due to a continuous charge distribution
Divide the continuous charge distribution into a large number of
small charge elements dq and calculate the (vectorial) field dE due
to each (considered as a point charge).
1
dq
dE 
rˆ
2
40 r
The resultant field is found by integration

 
E   E n   dE

1/ 2
2
2

E   dE   dE cos    dE sin  


The Electric Field on the axis of a charged straight rod
(line of charge)
There is a charge per unit
length of l, so dq = ldx
and q = ll.
Note y-components
cancel
dE 
1
ldx
40 r
E
2
l a
a
rˆ
1 ldx
l

2
40 x
40

l a
a
dx
l

2
x
40
l 1
1 
l  l 



 

40  a l  a  40  a(l  a) 
1
q

40 a(l  a)
l a
 1
 x 
 a
The Electric Field on the axis of charged ring
The resultant field on axis must have no perpendicular component – by symmetry.
The charge element is
q
dq 
dl
2a
1
dq x
1
x
dE x  dE cos  

dq
2
2
2 3/ 2
40 r r 40 ( x  a )
1
x
1
x
Ex  
dq 
dq 
2
2 3/ 2
2
2 3/ 2 
40 ( x  a )
40 ( x  a )
1
x

40 ( x 2  a 2 ) 3 / 2
2a

0
q
1
xq
dl 
2a
40 ( x 2  a 2 ) 3 / 2
The cathode ray tube
This device is used to display electronic
information from oscilloscopes, radar
systems, television receivers and computer
monitors.
An electron beam is produced by an electron
gun, which consists of a biased hot filament
together with intensity control, acceleration
and focusing electrodes.
The monoenergetic electron beam (usually a
few keV of energy) will strike the screen
producing a luminous point at a location
determined by the electric field of parallel
deflection plates, arranged to control both the
x and the y deflection of the beam.
The signals on the electrodes of the x and y
deflection plates, as well as the intensity
control electrode, determine the image that is
visualised.
The flux of E
(Serway 24.1-24.4)
We can imagine drawing electric field lines so that the number of lines per
unit area perpendicular to the field is proportional to the field strength.
Flux density : The number of field lines per unit area, perpendicular to the
field direction (a vector quantity). This quantity is proportional to Electric field
strength E.
Flux (symbol FE) : is the total number of field lines passing through a
particular area (a scalar quantity).
It follows ….
F E  EA  E.A  EA cos
Area elements
We assumed a uniform E field and a flat area in the previous discussion.
If we take an infinitesimally small area element, then locally, the field will be
uniform and the area element flat. The flux through this area element
(number of field lines through this area element) is ….
dA
E

dF E  E.dA  EdA cos
An area element is therefore a vector with an orientation of its surface
normal and a magnitude of its area. (Actually, the orientation could also be
the opposite direction. This is explained later.)
Total flux through a finite surface
Integrate the infinitesimal contributions …
(example of a surface integral of a vector field)
F E   E.dA
A
Gauss’s Law
The total outward flux passing through a closed surface (Gaussian Surface) is
equal to q/0, where q is the net charge enclosed by the surface.
F E   E.dA 
A
q
0
Example : Find E at a distance r from a point charge.
F E   E.dA   EdA  E.4r 2
A

By symmetry, E is constant in
magnitude over the gaussian
surface, and E is parallel to dA
z
A
dA
q
0
r
+q
1
q
E 
40 r 2
Now, the
electrostatic
force is
E
x
qq0
F  Eq0 
40 r 2
1
y
Choose a spherical
gaussian surface
centered on the
charge
Gauss’s law is consistent
with Coulomb’s Law
Note : The orientation of the vectorial area element is always outwards by
convention for a closed surface
Note : The presence of outside charges has no effect on the total flux
passing though the gaussian surface. Inward flux is (-ve), outward flux is
(+ve).
dA
+q
Gaussian
surface
Note : Under conditions of symmetry, Gauss’s Theorem is extremely
powerful, allowing conclusions to be drawn and solutions to be calculated
very efficiently.
E near a charged straight wire
• The wire has a charge l per unit length, so the charge element is dq = ldl
and q=ll.
• Choose a cylindrical gaussian surface placed symmetrically about the wire.
• By symmetry, the field is the same in magnitude over the curved surface and
E and dA point radially outward, i.e. the angle between them is zero.
• There is no flux through the top and bottom of the cylinder.
Plan
view
gaussian surface
+
x
+
E
+
+
dA
+
+
dA
E
+
l
l
l
+
x
F E  E.dA  E.2 x.dl 
+
0
A
0
+
+
+
E 
(as before)

l
20 x


l

E near a charged isolated conductor, any shape.
• There is no further motion of charge within the conductor once equilibration
has been attained, in an isolated conductor.
• Hence there is no field inside the conductor (otherwise free electrons would
move, F = qE).
• Also, E must be perpendicular to the surface, as there can be no component
of E lying in the surface.
charged isolated conductor
gaussian surface
• Choose a gaussian surface just under the charged conductor surface as shown.
• Since E is zero inside, fE=0 for this gaussian surface. Hence by Gauss’s
Theorem, there is no charge within the gaussian surface.
• If an isolated conductor is charged, then all the charge is on the outside of
the conductor
E near the surface of a charged conductor
Choose a cylindrical
gaussian surface
perpendicular to the charged
conductor and intersecting
its surface as shown.
The charge element is
g.s.
E
Surface
charge
density s
+ + +
A + + + + + +
+ + + +
+ + + + +
+ + + + + + + + + + + +
+ + + + + + + + + + + +
E=0
dq = sdA
(s = charge/unit area) and
q = sA.
E ┴ surface & E = 0 inside
the conductor.
So all the flux is through the
surface A.
Note … for a thin sheet insulator
sA
F E   E.dA  EA 
0
A
s
E 
0
E
s
2 0
Hollow conductors (Electrostatic shields)
For a gaussian surface inside a hollow charged
conductor, fE = 0, since there is no charge
inside the g.s.
Therefore E = 0
(No field inside a hollow conductor)
+
+
+
g.s.
+
+
+
+
+
+
+
+
+ +
A hollow conductor is an electrostatic shield,
even if it is full of holes. (Faraday Cage)
+
+
+
Electric Potential and potential difference
(Serway 25-1,2,3,6,8)
If the work done by an external agent in moving a charge +q0 from point
A to point B is WAB, then the potential difference between points A and
B (the potential at point B w.r.t. the point A) is defined as
Ending at point B,
starting at point A
VBA
WBA
 VB - VA 
 VAB
 q0
Ending at point A,
starting at point B
Note
 If WAB is +ve, then B is at a higher potential than A.
 Units are work/charge = joule/coulomb ≡ volt (V).
 Potential difference is a scalar quantity
 Potential energy refers to a charge-field system (work done to
introduce a charge to a field). Electric potential is a scalar
characteristic of an electric field, independent of any other charges.
 WAB depends only the endpoints, not the path. Therefore the E-field is
a conservative field.
 If we take point A to be at infinity, then it will feel no E-field and we set
the potential here to be zero.
WB
VB 
 q0
Work done in bringing q0 in from
infinity to B.
So VB is the work done per unit
charge to bring a positive test
charge to point B
Electric potential and electric field
The work done by an external agent in moving a charge +q0 from point
A to point B is WAB,. (This would be the negative of the work done by
the field on the charge.)
B
B
A
A
WBA    F.dl   q0  E.dl
Therefore,
B
VBA
B
WBA
1
 VB  VA 
   F.dl    E.dl
q0
q0 A
A
Similarly,
B
VB  VB  V    E.dl

From the above
definitions we see,
J
Q
N
V
E - field  1  1
Q
m
Units : Electric potential  1Volt  1
For a uniform field,
Consider two cases ….
E is parallel to dl, A is at a
higher potential than B.
+
E is anti-parallel to dl, A is
at a lower potential than B.
A
A
ds
E
+
B
E.dl  0
So, VBA is positive (negative) when B is
at a higher (lower) potential than A
B
VBA    E.dl  ( / ) El
A
+
E
B
+
E.dl  0
The change in potential energy …..
U  q0 V  q0VBA  q0E.l
So a positive charge loses potential energy
when it moves in the direction of the field.
Equipotential surfaces
In the figure, points B and C are at the same
potential.
E
Points B and C therefore lie on an equipotential
surface
C
Equipotential Surface : Any surface consisting
of a continuous distribution of points having the
same electric potential.
An equipotential surface is perpendicular to
the lines of electric field.
s
A
B
Energy unit  the electron volt
The electron volt (eV) : The energy an electron (or proton) gains or loses by
moving through a potential difference of 1 volt.
1eV  1.6 10-19 C  V  1.6 10-19 J
Example : An electron in a TV tube has a velocity v = 3.5x107 m/s which is a kinetic
energy of Ek = ½mv2 = 5.6x10-16 J which is equivalent to 3.5x103 eV. This corresponds
to the electron being accelerated from rest through a potential difference of 3.5 kV.
The potential at a distance r from a point charge q
We can choose a straight line integral path (since the potential is path independent)
q
dx = -dl
P
X=0
dl
E
r
P
P
P



VP    E  dl    Edl cos 180   Edl
q
1
 
dx 
2
40 x
40

r

1
r
q
x
 
1
q
40 r
In general
V
q
40 r
X=∞
The potential on the axis of a dipole
Since the two charges are in magnitude but opposite in sign, the potential on the
axis of the dipole must be zero, although there is an electric field.
(Explain)
V=0
+q
+
-q
-
The potential of any number of point charges
Calculate the potential for each charge separately and add them together.
VTot = + V1 + V2 + V3 + ….
Equipotential surfaces for the point charge and the dipole
equipotential
surface
E-field line
No work is done when moving a charge along an equipotential surface
Potential
plots for
the point
charge
and the
dipole
Example 4 : Energy and potential
A 100 eV electron is fired directly towards a metal plate that has a surface charge
density of -200 nCm-2. From what distance must the electron be fired if it should
just reach the plate ?
Example 5 : Energy and potential
Two electrons are stationery when 0.05 cm apart. They are allowed to move apart
under the influence of their mutual repulsion. What are the velocity and energy in
electron volts of each electron when they are 1cm apart ?
Example 5 : Energy and potential
A beam of alpha particles accelerated from rest through a potential difference of
2.6 x 106 V in a vacuum is scattered from a fixed gold target a large distance away.
Very occasionally an alpha particle collides head on with a gold nucleus and is
scattered straight back. Calculate the distance of the alpha particle’s closest
approach to the gold nucleus, assumed to be stationery (neglect recoil) ?
Applications of Electrostatics
Van de Graaff Generator
Electrical breakdown of the air due to corona discharge
occurs at a field Emax
Emax
Q

2
40 Rmin
Vmax 
Q
40 Rmin
Rmin 
Vmax
Emax
Emax = 3 x 106 Vm-1, so if we want Vmax = 1 x 106 V,
then
Rmin
1106

 0.33m
6
3 10
Note : E (and s) are more concentrated where r is
small – especially at sharp points  Applications ….
Electro-static precipitator
A corona discharge is induced in a column by applying a high voltage (40 –
100 kV) between a co-axial wire and the walls of the column. As air with
particulates passes up the column, the particulates acquire a negative charge
due to interactions with the corona discharge. These particulates are the
extracted from the flow electro statically and they collect on the walls of the
column.
With precipitators
Without precipitators
The Xerox process
Definition of capacitance
(Serway 26-1,2,3,4)
Two conductors, connected to the terminals of a battery, will acquire equal and
opposite charges +Q and -Q, even if the conductors are not of the same size,
since a charge Q will flow through the battery. The potential difference between
the conductors will be equal to V, the terminal voltage of the battery.
-Q
+Q
Experimentally for such a system,
V
-
Q V
 Q  CV
+
The constant C is called the
capacitance and depends on the
geometry of the system,
For simple cases, we can calculate the
capacitance.
We will do this for two simple cases ….
parallel plate capacitors and cylindrical
capacitors,
Units of C 
Q Coulomb

 Farad (F)
Volt
V
1 micro farad (mF) = 10-6 F
1 pico farad (pF) = 10-12 F
Since the concept of capacitance involves potential difference, there are
allways two parts to a capacitive system.
Method of calculation of capacitance
1. Assume charges +Q and -Q are on the system.
2. Calculate the E-field between the two parts of the system using Gauss’s
Theorem, or other method.
3. Calculate the potential difference between the two parts using
4. Find C from
Q  CV
V    E.dl
Parallel plate capacitors
Assume charges +Q and -Q on the plates with area A, i.e. a surface charge
density of s = Q/A.
g.s.
Total area A, s = +Q/A
Neglect end effects
Total charge = +Q
X=0
+
+
+
+
d
-
-
-
-
E
+
+
+
+
+
+
+
-
-
-
-
-
-
-
X=d
Total charge = -Q
s = -Q/A
dA
F E   E.dA  EA 
Choose a g.s. as shown
A
Work is done to charge
the capacitor, as
dl charge is brought to
the plates against the
field
sA
s
E 
0
0
Note : the same result would hold for any g.s. taken anywhere between the
plates, as the field is uniform (same magnitude and direction).
The Potential Difference between the plates is
s
Qd
V    E  dl    E (dx) cos180  Ed  d 
A
0
d
0
0

C
Q A 0

V
d
(Units of 0 must be Fm-1)
Parallel plate cylindrical capacitors (co-axial cable)
This consists of two co-axial cylinders of radius a and b and length L. We
assume that L >> b, (long capacitor). Assume charges +l and -l per unit length.
F E   E.dA  E 2xl 
A
E 
ll
0
b
a
-
l
20 x
-
The Potential Difference between the plates is
a
V    E  dl    E (dx) cos180
b
l
l
ln xba
dx  
20 x
20
b
l
a
l
b

ln 
ln
20 b 20 a
a
 
l
-
+
+
+
x +
+
+
+
+
+
+
+
-
-
+E
+
+
+
+
g.s.
dl
Q (ll )20 20l
C 

b
V
l ln a
ln ab
Capacitors in series and parallel
Series
Since the inner pair of plates of two
capacitors connected in series were
originally uncharged, they must carry
an equal and opposite charge after
charging, so that the total charge on the
inner plates is still zero.
Therefore
and
V1 
but
therefore
Qtot  Q1  Q2  Q3  
C1
+Q1
V1
C2
C3
-Q2 +Q3
-Q1 +Q2
V2
-Q3
V3
Qtot
Vtot
Ctot
Q
Q1
Q
, V2  2 , V3  3 , 
C1
C2
C3
Vtot  V1  V2  V3  
Qtot Q1 Q2 Q3
 

 
Ctot C1 C2 C3
1
1
1
1
 

 
Ctot C1 C2 C3
Parallel
The potential difference
across the capacitors is the
same and equal to the
source.
Vtot
C1
+Q1
-Q1
C2
+Q2
-Q2
+Q3
C3
-Q3
Therefore
Vtot  V1  V2  V3  
And the total charge on the capacitors connected in parallel is the sum of the
charges on the individual capacitors.
Qtot  Q1  Q2  Q3  
therefore
VtotCtot  V1C1  V2C2  V3C3  
Ctot  C1  C2  C3  
Energy of a charged capacitor
The electrical potential energy stored in a charged capacitor corresponds to the
work done in charging the capacitor. The energy is stored in the electric field
between the plates of the capacitor.
Suppose that a charge q has been transferred during the charging of a capacitor.
Potential difference across capacitor
V
q
C
Therefore the work done by the battery in (external agent) in transferring a small
additional charge dq is
dW  Vdq 
q
dq
C
The total work required to charge the capacitor from q=0 to some charge q=Q
is
Q
q
Q2
W   dW   dq 
C
2C
0
This work required to charge the capacitor appears as a stored electrical
potential energy
2
U
Q
1
1
 QV  CV 2
2C 2
2
C
0 A
d
For a parallel plate capacitor
V  Ed
Q2 1
1
U
 QV  CV 2
2C 2
2
U
1 0 A
Ed 2  1  0 AdE 2
2 d
2
1
uE   0 E 2
2
Energy density
in an E-field
Uses of Capacitors
Capacitors are used in most electronic circuits.
(Usually related to the storage of electrical energy.)
 Smoothing the direct current output of rectifier circuits.
 Switching of high currents in electric power applications.
When capacitors and resistors are used in combination, the circuit has a
characteristic time constant.
 Noise filters when the signal and the noise are in different frequency bands.
 Tuning a radio to a particular frequency (using the idea of resonance).
Modern switching devices (e.g. computer keyboards).
Current Electricity (Serway : 27-1,2,4, 6
and 28-1,2,3,5)
Electric current, current density
When electric charge moves, we have an electric current.
Definition – Electric Current
Current is the rate at which charge flows
through a surface (e.g. the cross-sectional area
of a conductor).
Units : 1 Amp = 1C/1s
Average current
Q
I 
t
Instantaneous current
I
dQ
dt
Surface
Definition – Charge Carriers
A mobile charge.
• Metal conductor (electrons)
• Semiconductor (electrons, holes)
• Molten / dissolved ionic solid (anions, cations)
• Plasma (electrons, nuclei)
• Beam (any charged particles – e, p, AXq, m, …
Convention – Direction of flow
This is the direction of flow of a positive charge
carrier
The microscopic model of current
Consider a cylindrical conductor with a charge carrier density of n in which a
current I is flowing. This constitutes an average drift velocity, vd, of each
charge carrier.
x  vd t
S,
area A
V  xA  vd tA
How many charge carriers passes through S at a given point in a time t ?
Each charge carrier moves on average a distance x  vd t
The volume swept out in this time is therefore V  xA  vd tA
The number of charge carriers is N  nV  nvd tA
Therefore, as each charge carrier carries a charge q
q Nq
I

 nqvd A
t t
Current
Current density
I  nqvd A
I
J   nqvd
A
Density of charge carriers
(Number of charge carriers
per unit volume)
Average drift speed
Charge on each carrier
Drift velocity explained
 Consider an isolated conductor in which the charge
carriers are free electrons. These electrons undergo
random motion, with many collisions amongst each
other and with the metal ions, somewhat analogous to
the molecules in a gas.
 The motion is rapid and erratic but does not lead to
any overall motion of charge.
 When a potential difference is applied to the ends of
the conductor, there is an overall tendency for the
electrons to move to the region of lower potential.
 The average motion is the drift velocity, vd.
Question : Why is each intercollision segment parabolic ?
Example 7 : Drift speed in a copper wire
A current of 10 A flows in a wire of cross sectional area 3.31 x 10-6m2. What is the
drift speed of the electrons ?
(Assume each Cu atom contributes two free electrons. The density of Cu is 8.95
g/cm3).
Now show that it would take an electron more than 1 hour to travel one meter.
Reconcile this with your perception of a light turning on following the flick of a
switch almost instantaneously.
(Hint : The E-field that causes the electrons to move travels at about 66% the
speed of light.)
Ohm’s Law and resistance
The current density is usually considered a vector quantity
J  nqv d
Ohm’s Law (established by experiment)
For most metals, and some other materials, the ratio of current density to
electric field is a constant s, where s is independent of the electric field
producing the current.
Current density
J  sE
Conductivity - the
constant of proportionality
Electric field
Ohmic Materials are materials which obey Ohm’s Law
Consider the E-field E to be due to a
potential difference V over a length l,
and note the relation of current I to current
J density. (Recover the form V=IR.)
I
ΔV
 σE  σ
A
l
 l 
 V  I    IR
 sA 
Resistivity is the inverse of
conductivity
1
ρ
σ
J
So
R
l
l

sA A
and

RA
l
Resistance
From Ohm’s Law …..
V  IR
We see that the unit of resistance is
Note
 For any uniform conductor
of cross section A and
length l, the resistance is
given by R   l    l 
 sA   A 
 The conductivity and the
resistivity are characteristics
of a particular material …
see table. ρ  1 / s
 For any conductor obeying
Ohm’s Law V  IR
 The unit of resistivity is
Ohm.m (W.m)
1V
1W 
1A
Resistivity and conductivity; temperature variation
It is found that resistivity  changes with temperature T.
To a good approximation, resistivity  varies linearly with temperature T.
  0  T
  0 1   T  T0 
 is called the temperature
coefficient of resistivity.
0 and T0 represent the reference
resistivity and temperature
usually Room Temperature
values.
Rearranging the equation to
show  ...
1 
1    0 


 0 T  0 T  T0 
For metals, (T) is linear until very low
temperatures where it levels off
For the three Group IV semiconductors in
the previous table,  is negative,
showing that (T) decreases with
increasing temperature as more
charge carriers become available
The units of  are °C-1 or K-1
The resistance of a resistor also varies linearly with temperature according
to a similar equation (show this) …
R  R0 1   T  T0 
Application
The platinum resistance thermometer – when immersed in a temperature
bath, the resistance changes by a known amount, allowing the temperature
to be calculated.
Electric Energy and Power
Suppose that we have a resistor with a potential difference V=VB-VA across it.
VA
VB
A current flows
through a resistor
I
By definition, I = Q/t, so the charge Q loses potential energy U = VQ in
the time t.
The rate at which the charge Q loses potential energy is the power that is
dissipated in the resistor.
U
Q
V
 VI
t
t
P  VI
Electric Energy and Power
The energy that is lost by the charge U = VQ is dissipated in the resistor and
appears as heat energy.
Using Ohm’s Law.
V  IR
P  VI
We can show …..
P  I 2R
V2
P
R
Low resistance copper cable is
expensive. Power utilities therefore
use higher resistance cheaper
cables.
There will be less power dissipated as
heat if electrical energy is
transported at high voltage (up to
765 kV).
This is also known
as Joule Heat that
is dissipated in R
Conceptual Example
Suppose that there are two light bulbs, A and B, connected in parallel across a
voltage source. Bulb B has a resistance that is twice that of A. Which bulb will
give more light ?
The amount of light is related to the temperature,
(higher temperature, more light) and this will
depend on the power dissipated (more power,
higher temperature).
Bulb A
Resistance R
Bulb B
Resistance 2R
Both of the light bulbs have the same voltage, V
across them (they are connected in parallel).
The power dissipated in A is
V2
PA 
R
V 2 PA
The power dissipated in B is PB 

2R 2
-
The light bulb with the lower resistance will dissipate more
power and glow more brightly
+
Resistors in series and parallel
As we saw in the case of capacitors, resistors can be combined in series
or in parallel. In each case a sequence of resistors can be replaced by
an equivalent resistor
R2
R1
R2
R1
+
-
Two resistors in parallel
+
-
Two resistors in series
There are also more complicated combinations
Combining resistors in parallel
We wish to find an equivalent resistor that will have the same effect
as two resistors in parallel.
R2
Req
R1
+
=
+
-
What is Req in terms of R1 and R2 ?
-
Combining resistors in parallel
Note
• The potential difference across both resistors must be the same.
• The current must branch at the parallel junction.
• Ohm’s law applies to each resistor, as well as to the circuit.
V2 , R2
V  V1  V2
I2
V1 , R1
I1
V
I = I1 + I2
+
I  I1  I 2
V V1 V2


Req R 1 R 2
-
1
1
1


Req R 1 R 2
1
1
1
1



 ......
Req in terms of parallel resistors ….
Req R 1 R 2 R 3
Combining resistors in series
We wish to find an equivalent resistor that will have the same effect
as two resistors in series.
R1
R2
Req
=
+
-
+
What is Req in terms of R1 and R2 ?
-
Combining resistors in series
Note
• In this case, the current through each resistor must be the same.
• The potential difference at each resistor must equal the total voltage.
• Ohm’s law applies to each resistor, as well as to the circuit.
V1
V2
R1
I  I1  I 2
R2
V  V1  V2
I1
I2
IR  I1R1  I 2 R2
V
I
+
-
Req in terms of series resistors ….
R  R1  R2
R  R1  R2  R2  ......
Electromotive force and terminal voltage
A source of electrical energy in a circuit, such as a battery (or a
generator) is said to provide an emf or electromotive force (although it’s
a source of energy, not force)
The battery, so to speak, pumps the charge up to a higher potential.
Then, if there is a path for it, the charge flows around the external circuit
as an electric current.
The emf is the work done per unit charge. It has the same units as
electric potential.
The unit of emf is therefore the volt (V).
Internal Resistance
From the circuit it is clear that the current
flows through the battery too, and that a
battery also has an internal resistance.
A real battery therefore has a source emf
called  and an internal resistance r.
When connected to a load resistance R, so
that a current I flows through the circuit, the
terminal voltage of the battery is therefore
V    Ir
The emf called  and can be measured in the
so-called open circuit voltage mode, when the
battery is not connected to a load resistance,
and a voltmeter which requires negligible
current to operate is used.
Real battery
Source emf

Internal
resistance r
Internal Resistance
The terminal voltage across the battery depends on the current that is being drawn.
R is the resistance
of the load
resistance (a
resistor, or a device
like a toaster, TV,
heater etc …).
Clearly
Terminal voltage
V    Ir
Battery with emf  and
internal resistance r.
V  IR
I
V  IR

Rr
Power and Internal Resistance
Electrical power generated or dissipated is the product
of voltage and current
P  IV
The power that is generated by our battery (or
generator) is therefore
Ptot  I  I R  I r
2
Total power
generated
Power dissipated
in load as joule
heat
2
Power dissipated
inside battery
Power transferred to the load resistance
In this circuit, some of the power P is delivered to the
load, while some is lost inside the battery.
We would like to deliver as much power as possible to
the load.
Is there an optimal value of load resistor R to achieve
this ?
2
  
Pload ( R)  I R  
R

 (R  r) 
2
Answer
Kirchoff’s laws and their applications
Often it is not possible to reduce a circuit to a single loop, using the rules for the
combination of series and parallel resistors alone.
We will restrict ourselves
to “resistive circuits” which
They can however be further analysed using the
contain only resistors and
Kirchoff’s voltage and current Laws.
sources.
Kirchoff’s Current Law (Kirchoff’s First Law)
The algebraic sum of the currents entering any node (junction) must equal the sum
of currents leaving the junction.
I2
I1
I1  I 2  I 3
I in  I out


I3
This is a statement of the fact that charge is conserved. In a steady state circuit,
there can be no build-up of charge, therefore the current is conserved.
+
+
-
-
Redrawn equivalent circuit showing two joined
nodes are actually a single node with four legs
Kirchoff’s Voltage Law (Kirchoff’s Second Law)
The algebraic sum of the changes in potential across all of the elements around
any closed loop must equal zero.
 V
i
0
closed loop
This is a statement of the conservation of energy. When charge has moved
around a circuit to arrive back at the same point it must have the same potential
energy. As U=Vq, the potential must also be the same.
Potential changes when traversing an element in the direction from a to b,
when the current direction is as shown
I
a
V  IR
b
I
a
+
b
V  
b
V  
-
a
+
a
V   IR
b
Direction of I
is important,
it must be
consistently
applied.
JIW Watterson ‘02
JIW Watterson ‘02
JIW Watterson ‘02
JIW Watterson ‘02
JIW Watterson ‘02
JIW Watterson ‘02
JIW Watterson ‘02
Ammeters and voltmeters, determination of resistance
Serway 28-5
JIW Watterson ‘02
JIW Watterson ‘02
JIW Watterson ‘02