Transcript Electric Fields File

SACE Stage 2
Physics
Electric Fields
1
Electric Fields

Unit of Charge
Electric charge (q) is measured in units of Coulomb (C)
Definition
The coulomb is defined as the amount of charge transferred by an
electric current of 1 Ampere in 1 second.
1 coulomb = 1 ampere x 1 second
2
Electric Fields

Unit of Charge
Comparison of Charge and Mass of Electrons and Protons
The charge on an electron is exactly the same as the charge on a proton.
However, a proton has a mass approximately 2000 times the mass of an
electron.
qe = e (symbol for the
charge on an electron)
me
mp=2000me
qe= -1.6 x 10-19 C
qe= -1.6 x 10-19 C
qp= +1.6 x 10-19 C
3
Electric Fields

Size of the Coulomb
The Coulomb is a very large unit of charge.
Eg, consider the amount of charge contained in the electrons in 1 g of hydrogen
atoms.
1 g contains Avogadro's number of particles (6.02 x 1023)
 number of electrons (Ne) = 6.02 x 1023
Total Charge of electrons (qe) = Ne x charge on an electron (e)
= (6.02 x 1023 ) x (- 1.60 x 10-19)
= - 96320 Coulomb.
Similarly the total charge of the protons is + 96320 Coulomb.
4
Electric Fields

Size of the Coulomb
Consider the number of electrons which need to be moved to produce a
charge of 10-6 C.
Qe
106
13
Ne 


6
.
3

10
electrons
19
e 1.6 10
Hence to produce this charge it is necessary to shift 6.3 x1013 electrons from
1 g of hydrogen (for example). As a fraction, this means that we shift (6.3
x1013)/(6.02 x1023)  1 x10-10 or 10-8 %, ie, producing a charge of 10-6 C in
the lab would involve moving 0.00000001 % of the electrons in 1 gram of
hydrogen
5
Electric Fields

Size of the Coulomb
Hence the Coulomb unit is very large and therefore we use smaller units:
eg 1 milli-Coulomb = 10-3 = 1 mC
1 micro-Coulomb = 10-6 = 1 C
1 nano-Coulomb = 10-9 = 1 nC
1 pico-Coulomb = 10-12 = 1 pC
6
Electric Fields

Conservation of Charge
The fundamental concept in the model of electricity used here is the
concept of Conservation of Charge
In any system the total amount of charge remains constant.
7
Electric Fields

Conservation of Charge
Pair production: high energy bundle of electromagnetic radiation (gamma
ray photon) interacts with the nucleus of an atom knocking out a pair of subatomic particles, a proton and an electron.
electron
protons and electrons are
produced (created) in equal
numbers to conserve
charge.
charge before = 0
charge after = e+ + e- = 0
heavy nucleus
proton
8
Interaction Between Charges

Coulomb’s Law
F
1 pC
1 pC
F
10 C
Line joining
charges
10 C
Newton's Third Law still applies. The size of the two forces is the same,
but the directions of the forces exerted by each object on the other are
opposite.
9
Interaction Between Charges

Coulomb’s Law
NOTE : ONLY ONE FORCE ACTS ON EACH OBJECT, SO THERE IS A
NET FORCE ON EACH OBJECT .
Only one force acts on
this object
10
Interaction Between Charges

Coulomb’s Law
v
This force causes an acceleration
(eg centripetal acceleration)
It is found that the force also
depends on the charges and their
distance apart.
eFC
proton
11
Interaction Between Charges

Coulomb’s Law
Any two point charges have acting on them equal sized, oppositely
directed forces acting along the line joining their centres. The
magnitude of these equal sized forces is directly proportional to the
product of the charges and inversely proportional to the square of their
distance apart. The forces are attractive for unlike charges and
repulsive for like charges.
1 q1q2
F
4o r 2
constant
[Constant = 9 x 109 N m2 C-2 in
a vacuum and can be assumed
to be the same in air]
12
Interaction Between Charges

Example
Consider the repulsive force between the electrons in 1 g of hydrogen atoms
placed on either side of the earth.
F
Earth
F
D = 12700 km
1 g of H
1 g of H
13
Interaction Between Charges
F
F
1
q1q2
4o r 2
1
40
1.6 10  6.02 10 
1.27 10 
23 2
19
7 2
F  5.2 105 N away from the other charge
An equivalent mass of 50,000kg would be needed to produce the
same force!
Electric forces are much stronger than gravitational forces.
14
Ratio of Gravitational Forces
and Electric forces
Can find the ratio between Electric force and Gravitational force by considering
an electron and a proton. (hydrogen atom, 1 proton, 1 electron)
R
electron
proton
15
Ratio of Gravitational Forces
and Electric forces
Fe =
1
4 0
q1 q 2
r2
since qp = +e
or =
1 ee
4 0 r 2
= 2.3x10-28r-2
Fg=
Gm1m2
r2
Fe: Fg
= 2.3x10-28r-2  1.01 x 1067r-2
 ratio of force independent
of separation of particles.
= 2.3 x 1039
 Fe = 2.3 x 1039 Fg
= 1.01 x 10-67r-2
16
Form of Constant in Coulomb's
Law
1
F =
4 0
where
1
4 0
q1 q 2
r2
is known as the Permittivity constant
of free space
= 9 x 109 N m2 C-2
Do not represent
1
4 0
as ke, it is incorrect as it has its own
meaning.
17
Electric Force due to More than
1 Charge
C
qC +5 C
4 cm
3 cm
qA+5 C
A
5 cm
B
qB +5 C
The Charge at C will experience forces due to the charge at A and the charge at B.
To find the resultant field we need to add the forces due to A and B as vectors.
18
Electric Force due to More than
1 Charge
The Force on C Due to B:
The Force on C Due to A:
F =
4 0
= 9 109
F=
q A qC
r2
1
6
5 10  5 10
(4 102 ) 2
= 140.6 N
Away from A
6
1
4 0
q B qC
r2
6
6
5

10

5

10
9
= 9  10
(3  10 2 ) 2
= 250 N
along the line BC
and away from B
19
Electric Force due to More
than 1 Charge
ie. By VECTOR ADDITION: ie adding FA and FC – shown opposite.
By Pythagoras’ Theorem
FR 
FA 2  FB 2
FR  140.62  2502
FR  286.8N
FB= 250 N
FResultant
q
FA = 140.6 N
20
Electric Force due to More
than 1 Charge
tan q =
250
140.6
q = 60.6°
ie. Fresultant = 286.8 N at 60.6° to the line joining A and C. (As shown)
21
Electric Fields
The field concept is needed to explain how charges exert "forces at a distance"
on other charges.
Field model for Electric Charges
large force
Source Charge
+qT
test charge
strong field
small force
+qT
test charge
weak field
22
Electric Fields
An electric field exists in a region of space if electric charges experience
forces in that space.
F
-qT
F
+qT
The direction of the field is the direction of the force on POSITIVE CHARGES
23
Example
A Calculate the electric field strength at a point P 10 cm from a point charge
of + 10 C.
qs
E
40 r 2
1
6
10

10
 E  9 109
(0.1) 2
 E  9 106 NC 1 away from q s
24
Electric Fields
If we use lines to represent these electric fields, then the spacing of the
lines represents the size of E and the direction of the lines represents
the direction of E.
E
E
25
Derivation of the Electric Field Strength
at a point in an Electric Field
Consider a small test charge qt placed a distance r from a source charge qs
Force on test charge
E
test charge
F

q
1 qs qt
40 r 2

qt
qs
E 
40 r 2
qt
r
qs
1
26
Example
A Calculate the electric field strength at a point P 10 cm from a point charge
of + 10 C.
27
Example
B
Calculate the force on an electron placed at point P. (Charge on an
electron, qe = 1.6 x 10-19C.)
28
Example
B
Calculate the force on an electron placed at point P. (Charge on an
electron, qe = 1.6 x 10-19C.)
F
E
q
 F  qE
 F  1.6 10 19  9 106
 F  1.44 10 12 N towards the positive charge
29
Example
C
Calculate the acceleration of the electron.
30
Example
C
Calculate the acceleration of the electron.
F
a
m
1.44 10 12
a 
9.1110 31
 a  1.58 1018 ms  2 towards positive charge
31
Rules For Drawing Electric Field
Lines
1. Electric field lines do not cross
Incorrect
correct
2. begin on positive charges and end on negative charges
NOT
3. The direction of the field is the direction of the force on a
positive test charge
F
+q
+q
F
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Rules For Drawing Electric Field
Lines
Strong
field
4. The number of field lines per unit area
represents the strength of the field.
5. Field lines cut conductors at 90o.
Weak
field
F
6. No field exists in a conductor.
33
Electric Fields due to Point
Charges
-
34
Electric Fields in Nature
In household wires
10-2
In radio waves
10-1
In the atmosphere
102
In sunlight
103
Under a thundercloud
104
In a lightning bolt
104
In an X-ray tube
106
At the electron in a hydrogen atom
6 x 1011
At the surface of a uranium nucleus
2 x 1021
35
Electric Fields due to more than
one charge
Consider the electric field at a point P due to two positive charges of
magnitude +5C positioned as shown below.
P
4 cm
3 cm
qA = +5 C
5 cm
qB = +5 C
At point P there are electric fields due to A and B. Any charge at P will
experience forces due to the charge at A and the charge at B.
To find the resultant field we need to add the fields due to A and B as vectors.
36
Electric Fields due to more than
one charge
Electric field due to qA
1
qA
EB 
40 r 2
1
qA
EA 
40 r 2
6
5 10
 E A  9 10
(4 10  2 ) 2
9
 E A  2.8 10 NC
7
Electric field due to qB
1
along AP and away from A
6
5

10
 EB  9 109
2 2
(3 10 )
 EB  5 107 NC 1
along BP and away from B
37
Electric Fields due to more than
one charge
Vector Diagram
EB
Er
To calculate the length and
direction of the resultant Electric
Field, will need to use the cosine
rule (if right angled triangle, use
Pythagoras) and your trig ratios.
EA
38
Electric Field due to point charges of
equal magnitude but opposite sign:
-
39
Electric Field due to point charges of
equal magnitude and the same sign:
+
40
An Infinite Conducting Plate
The electric field for an infinite conducting plate must be uniform.
A positive charge leaving a positively charged surface would be equally
repelled from all directions.
The positive charge must move at right angles to the surface. A similar
charge at a different position on the plate must also have the same force
exerted on it from all directions so it too must move away from the plate at
right angles.
41
An Infinite Conducting Plate
For a positively charged infinite plate:
42
An Infinite Conducting Plate
Two Parallel Infinite Conducting Plates
Field due to
Negative
Plate
Field due to
Positive Plate
Net Field here
is ZERO!
Negative
Plate
Positive
Plate
Net Field here
is ZERO!
Net field outside the plates must be zero as the uniform field from the infinite
negative plate means the field strength must be the same everywhere, and the
same applies to the positive plate ie. at a given point outside the plates the
fields from each plate cancel by the principle of superposition.
43
Electric Fields and Conductors
There cannot be a parallel component to
the electric field at the surface of a
conductor as this would imply that there is
a force on the charges on the surface and
as they are static charges and that the field
is uniform at the surface, therefore parallel
component = 0.
+
_
There is no electric field inside the conducting material.
44
Electric Field Inside a Hollow
Conductor
No field inside the hollow
conductor
A Hollow Spherical Uncharged
Conductor Placed in the Region
Between Two Oppositely
Charged Parallel Plates:
+
_
45
Electric Fields and Sharp Points:
The electric field around a Pear Shaped Object:
Stronger field is in the vicinity of
the sharpest curvature (at points
of maximum curvature).
46
Electric Fields and Sharp Points
A test charge is placed at position P as shown below. If the net force at P is
zero, then there must be a greater charge on the smaller sphere. Now this
corresponds to the fact that if the charged object is pear shaped then
considering the position P inside the object, hence net force on a test charge at
P will only be zero if there are more charges at the sharper end of the object.
P
P
47
Corona Discharge
At the sharp end of a pair shaped conductor, the build up of charge can be
large enough to create an electric field large enough to accelerate ionic
particles in the air, which, when accelerating, collide with other particles in
the air and they to, become ionised. This charges the air in the vicinity of
the sharp end of the conductor to draw away any excess charge.
48
Application: Photocopier and Laser
Printers
Photocopying
Materials, such as selenium, that conduct better in light than the dark are
known as photoconductive.
If a selenium surface was positively charged in the dark, light will cause
electrons to be drawn to it to neutralise the charge only where light was
incident.
Because selenium is not a good conductor the charge remains on the surface
in a static position.
49
The Process of Photocopying
Important mechanical part of a photocopier is the cylindrical aluminium drum
with a coating of Selenium. Aluminium is a good conductor of electricity
and is therefore earthed. All processes occur as the drum rotates.
1. Charging the Drum
A highly charged wire (Corona Wire)
at 5000V to 8000V extends the
length of the drum and close to it.
As the corona wire is highly
charged, the corona effect allows
positive charges to ‘jump’ on to the
drum.
50
The Process of Photocopying
2. Projecting an image onto the drum.
The original image is placed onto a flat glass platten of the photocopier. A
strip of bright light is passed over the image and then through a series of
mirrors on to the drum. Where there is black on the original image, no
light is reflected, were there is white, lot of light is reflected and where
there is grey, some light is reflected.
Where the light hits the selenium
drum, it becomes conductive and
draws electrons from the aluminium
drum to neutralize it. This leaves an
invisible latent image on the drum.
51
The Process of Photocopying
3. Applying toner to the Drum
The drum then passes a position where negatively charged toner (a black
powder) is dusted onto the drum. The toner consists of fine toner particles
(negative) stuck to larger iron particles (positive).
There is a magnetic roller (near the main drum) attracts the iron carrier
particles containing the toner to its surface. As the positive charge is
much larger on the selenium drum than the iron particles, the toner
particles are attracted on to the main drum.
The toner is the attracted to the charged areas of
the selenium only. This leaves an image on the
selenium of the original image.
52
The Process of Photocopying
4. Putting the image onto paper using the corona wire.
Paper is now passed next to the selenium coated drum. Just before the
paper is passed the selenium drum, it passes another positively charged
corona wire (the transfer corona) which charges the underside of the
paper. This charge is stronger than the charge on the drum so that as the
paper passes the selenium drum, the image is transferred onto the paper.
53
The Process of Photocopying
5. Discharging the Paper with a Corona Wire.
As soon as the paper has passed the selenium drum it passes another
corona wire which is negatively charged to neutralise the paper.
6. Fixing the Image onto the Paper
At this stage the paper is carrying toner
particles. The paper is then fed between
two rollers (fusing rollers) which melts the
toner to the paper. The upper roller is a
hot roller used to melt the toner and the
lower roller is a presure roller so that when
the toner has melted the presure causes
the toner to melt into the fibre of the paper.
54
The Process of Photocopying
The drum is then scraped to removed any excess toner and then exposed to
bright light to drain off any residual charge. The next cycle of copying can
begin.
55
The Process of Laser Printing
The process of laser printing is similar to that of photocopying. Instead of
using a strip of white light, a laser beam is focused onto the drum.
The corona wire charges the drum, a laser beam is focussed onto it which
selectively discharges the drum. The area of the drum is the same as the
area of the paper onto which the image will appear.
The laser (controlled by a computer) scans the drum (which spins very
quickly) and selectively discharges the drum in the spots that will appear
white in the final copy.
56
The Process of Laser Printing
The toner is applied, as
in photocopying, and
the whole process from
then on is the same as
in photocopying.
57