Transcript PHYS113 Electricity

PHYS113 Electricity
and Electromagnetism
Semester 2; 2002
Professor B. J. Fraser
1
1. Electric Charge

What is charge?






700 BC - Greeks write of effects
of rubbing amber (Electrum)
1600’s - Gilbert shows
electrification is a general
phenomenon
1730 - C. Dufay concludes “there
are 2 distinct Electricities”
1750 - Ben Franklin shows +ve & ve charges
Electrostatics involves the
forces between stationary
charges.
Charge is a basic atomic
property



forces between electrons & nuclei
unlike charges attract
like charges repel
2
Transfer of Charge




Charge transfer
touching  charge sharing
(conduction)
only the electrons move
Can appear as though positive
charge has moved
+ ++
++
+





++
+
Unit of charge: Coulomb, C
1 Coulomb = 1 Ampere  second
Electronic charge, e = 1.602 x
10-19 C
i.e. 1 C = 6.3 x 1018 electrons
a small number!!
3
Conservation & Quantisation

Charge is always conserved


it cannot be created or destroyed
Charge only comes in fixed
packets



the packet size is ± e
It cannot wear off
The light from distant quasars
(billions of years old) shows
evidence of exactly the same
atomic charge.
4
Forces Between Charges

Coulomb’s Law





1785: Coulomb experimentally
determines force law between 2
charged point sources, q1 and q2.
1
F 2
F  q1q2
r
Thus: F  k q1q2
r2
where k = 8.99 x 109 N m2 C-2
Electric force has direction
(vector)
Hence Coulomb’s Law is:
qq
F 12  k  1 2 2  r̂ 12
 r 


F12 is the force on q1 due to q2
r12 is a unit vector from q2 to q1
along the line that joins them.
5
Hints for Problem Solving


Draw a clear diagram
Forces are vectors




include coordinates
i.e. Fx and Fy or i and j components
add vectorially
Shortcuts due to symmetry?
Example: Electric Forces in a Plane
 Calculate the forces on q1 and q3
-3 nC
q3
2.0 m
q1
q2
+2 nC
+2 nC
2.0 m
6
Solution: Forces in a plane
Force on q1
 This is due to q2 and q3
q3
F1
F13
F12
q1
j
i
q2
F 1  F 12  F 13
 q q 
q q  
 k  1 2 2 rˆ12   1 2 3 rˆ13 
 r13  
 r12 
 q 

q 
 k q1  22  iˆ    23  ˆj 
 r13 
 r12 

9
9





3
x
10
9
 9  2 x 10  
ˆ
 9 x 10 2 x 10 
 i   

2
2
2



 2
  9.0 x 109 N  iˆ   13.5 x 10 9 N  ˆj
F1 
 Fx 2  Fy 2
 16.2 x 10  9 N
7

ˆj 

Solution: Forces in a plane
Force on q3
 This is due to q2 and q3
q3
F31
F32
F3
q1
j
i

q2
 q 
q  
F 3  k q3  21 rˆ 31   22 rˆ 32 
 r32  
 r31 
 2 x 10 9 

ˆ


 j
 2 2




 9 x 109  3 x 10 9 
   2 x 10 9 

ˆ
ˆ
 cos i  sin  j 
 
2
2 
  2 2 

  13.5 x 10 9 N  ˆj  4.8 x 10 9 N iˆ  4.8 x 10 9 N  ˆj
 4.8 x 10 9 N iˆ   18.3 x 10 9 N  ˆj

Find magnitude as before
8
2.The Concept of the Electric
Field






Why is there a force between
charged particles?
How does each particle know
that the other one is there?
What happens in space between
charged particles?
This is an example of an
action-at-a-distance force.
E.g. Gravitation, Magnetism
These forces are described in
terms of a field in space
surrounding the particle or
object.
9
Electric Field Strength

Test an invisible force field?



Test for an electric field by
measuring force experienced by a
positive test charge.
We know that E  F and: E  1/q0





Q
See if a test object experiences a force!
where q0 = charge of test charge
E = electric field, N C-1
Hence: E  F
q0
and since: F  k
qq0
r
2 ˆ
r
Then:
q
E  k 2 rˆ
r
Electric field seen by q0 due to q.
+ +
+ + +
++ + +
+ + +
Charged
Object
Electric
Field
Region
q0
F
+
Test
Charge
E
F
q0
10
Electric Field Lines








Electric field strength is a
vector quantity.
Much easier to represent using
vectors pointing in field
direction - electric field lines.
Concept due to M. Faraday
“lines of force”
Electric Field lines point away
from positive charges
Field lines point in the direction
of the force or electric field
Density (spacing) of field lines
depends upon magnitude of E.
Field lines never intersect.
11
Electric Fields in Nature
Field Description
Interplanetary space
Strength
(NC-1=Vm-1)
10-3 – 10-2
At Earth’s surface in clear
weather
In a thunder storm
100 - 200
Electrical breakdown of dry
air
Van der Graaff generator
3 x 106
Fermilab accelerator
Atom at electron orbit radius

103
106
1.2 x 107
109
All charges (fixed & moving) produce
an electric field that carries energy
through space at the speed of light.
12
Field Due to Point Charges


Electric fields add vectorially:
E = E1 + E2 + E3 + …
Thus:
qi
E  k  2 rˆi
i ri
Worked Example
 Find the electric field at point A for
j
the dipole shown
+
6 cm
4 cm
A
q1 = +12 nC
-
i
q2 = -12 nC
Field at A due to q1.
 q1 
E A1  k  2  rˆ
 r1 
 12 x 10 9 
4
-1
ˆ
ˆ




 9 x 10 
i

3
.
0
x
10
i
NC

2 2
 6 x 10  
9
13
Field Due to Point Charges
Field at A due to q2
E A2

 q1 
 k  2  rˆ
 r2 
9



12
x
10
9
4
-1
ˆ
ˆ




 9 x 10 

i

6
.
8
x
10
i
NC

2 2
 4 x 10  
Total electric field at A:
E A  9.8 x 10 4  iˆ NC -1
EA1
A
+q1




EA2
- q2
No component in the j direction
Example of an electric dipole
Often found in nature (e.g. molecules)
For more: See Section 21.11
14
Field due to a line segment

Charge, Q, distributed uniformly
along length, L, with charge density:
l = Q/L
Worked Example
 What is the electric field at a
distance R from a rod of length 2L
carrying a uniform charge density, l?
 Consider an infinite collection of
charge elements, dQ. j
L
^
r
dE
y

R
O
i
dy
P
-L
r
dE
dQ
15
Field due to line segment
(contd)
L
dQ
E  k  2 rˆ
r
L
But l = Q/L and
thus dQ = l dy
L
dy
 kl  2 rˆ
r
L
Centre rod at origin
 For every charge at + y, there is
another corresponding charge at -y
 Thus, fields in j component add to 0.
L
L
dy
dy
E  kl  2 cos î - sin  ĵ  kl  2 cos î
r
r
L
L



L
dy  R 
î 
2 
r r 
L
 kl 
L
 kl 
R
dy î
3
r
L
R
L
 kl 
L
R
2
y


cos = R/r
r2 =(R2 + y2)

2 32
dy î
Can you do
this integral?
16
Solution to Field Due to a Line
Segment



The solution to the field due to
a line segment is:
2lL
Ek
î
2
2
R L R
So, what is the big deal?
Well, what happens if L >> R?
2kl
E
î
R
Thus, the field
from a line
charge is
proportional to
1/R and not 1/R2.
17
Field Due to a Surface


Consider a charge Q uniformly
distributed across surface of area A
Surface charge density is: s = Q/A
Worked Example
 Find the electric field at distance R
from an infinite plane sheet with
surface charge density s.
Divide the sheet
into an infinite
P
collection of line
segments, L, long
and, dx, wide
R
k^
^j
r

L
^i
x
dx
18
Field Due to a Surface



Charge on each strip:
dQ = s dA = s L dx
Charge per unit length:
l = dQ/L = s dx
From previous example, each strip
sets up electric field:
E = 2kl/r = 2 k s dx/r
 Summing for all the strips:

dx
E  ks 
sin  k̂ - cos î
But i
r

components

sin  dx
sum to 0
 2ks 
k̂
r


 2ks


 R


R
2
x
 R
2 12
R
 2ks  2
dx
2
R  x 

dx
2
x

2 12
k̂
Can you do
this integral?
19
Solution to Field Due to a
Surface

The solution to the field due to
a surface is:
E  2 ks k̂


So, what’s the big deal this
time?
How does the field vary with R?
Thus, the field
from a surface in
independent of
the distance R!
20
Particles in an Electric Field

A particle of charge, q, in an electric
field, E, experiences a force:
E = F/q
 F = qE = ma





The particle accelerates at a
+ve particle moves in direction of E
I.e. from +ve to -ve charge regions.
Thus an electron will be deflected
toward a +ve charged plate as its
moving past it.
Examples: operation of CRT’s, TV
tubes, etc.
+ + + + + + + + +
screen
e- - - - - - - - 21
Particles in an Electric Field
Worked Example
 An electron in near-Earth space is
accelerated Earthward by an electric
field of 0.01 NC-1. Find its speed
when it strikes air molecules in the
atmosphere after travelling 3 Earth
radii (19 000 km).

The electron experiences a force:
F = ma = qE
a = qE/m
 For motion at constant acceleration:
v2 = u2 + 2as = 2as
1
19
2
7
2qEs  21.6 x 10 10 1.9 x 10  2
v

31

m
9
.
1
x
10


v = 2.6 x 108 ms-1
 i.e. 0.8 x speed of light
22
Fishnets and Flux:
The Gaussian Surface


Consider a fishnet with water
flowing through it.
The rate of flow through net is the
flux.
fw = vA





v = velocity of flow
A = area of net
If the net is angled at  to the flow:
fw = vA cos
A

In vector form:
fw = v  A
where the direction of A is normal to
net
23
Defining Electric Flux


For an irregular shape, area A is sum
of infinitesimal elements dA.
Thus, summing over 2-D surface S:
fw   v  d A
S


Now, replace water with electric
field, i.e. there is no physical motion.
The electric flux through a surface
of area A is:
fE   E  d A
S


The electric flux through a surface
is proportional to the number of
field lines passing through a surface.
If the fishnet is formed into a
closed shape (e.g. lobster pot) its
called a Gaussian surface.
24
The Gaussian Surface
dA

E
The total electric flux (number of
field lines) passing through this
surface is:
f E   E  d A




where A points perpendicularly away
from each element dA.
If the flux in one side is the same as
that out then the total flux is zero.
If there is no net charge inside a
Gaussian surface the electric flux
through it adds to zero.
Gaussian surfaces are imaginary
constructions!
25
3. Gauss’ Law




Consider a point charge surrounded
by a Gaussian sphere.
The electric field is:
q
1 q
E  k  2  rˆ 
rˆ
2
4e 0 r
r 
where e0 = permittivity of free space
= 8.85 x 10-12 C2 N-1 m-2
The electric flux through the
surface is then:
f E   E  d A
Radial field lines are
always normal to sphere
S
Flux lines
  E dA
S

q
4e 0 r
2
 dA
sphere
 q 
2



 
4

r
2
 4e 0 r 
Gaussian surface
26
Gauss’ Law in General

f E   E  d A 
S
Qencl   qi
i
Qencl
e0

Gauss’ Law states
that the electric
flux through any
closed surface
enclosing a point
charge Q is
proportional to Q.
The surface need
not be centred on
Q and can be any
shape.
Example: Coulomb’s Law from Gauss’
Law
 What is the electric field due to a
point charge?
 Consider a Gaussian sphere of radius
r centred on a charge q.
 Only interested in radial field
direction.
 All fields in other directions cancel.
27
Coulomb from Gauss



Consider surface elements dA
If E is along dA then:
E.dA = E dA cos(0º) = E dA
Hence:
E
 E  d A   E dA
S
S
 E  dA
+q
S
 E 4 r 2 


dA
From Gauss’ Law:
E 4 r 2   q

r
e0
Gaussian surface
Rearranging:
q
E
4 e 0 r 2
Which, since F = qE, gives Coulomb’s
Law, where we put E radially outward
from the charge q.
28
Applications of Gauss’ Law




Use Gauss’ law to find electric flux
or field in a symmetrical situation.
Shape of the Gaussian surface is
dictated by the symmetry of the
problem.
Worked Example
Find the electric field due to an
infintely long rod, positively charged,
of constant charge density, l.
+ + + + + + + + + + + +
P
29
Electric Field of Long Rod



Consider motion of
a test charge.
Only field lines
radially away from
the rod are
important.
Consider a
Gaussian cylinder
around part of the
rod, radius r,
height, h.
Total flux through cylinder is:
ftotal   E  d A 
top


h
+ + + + + + + + + +

dA
dA
r
dA
 E  d A   E  d A
bottom
side
But, @ top & bottom E  dA
E.dA=0
For the side E is parallel to dA
E.dA=E dA
30
Field due to a Long Rod
fE 
 E  d A   E dA  E  dA
total
side
side
 dA  side area of a cylinder of height, h
side
 2π rh
lh
fE  2 rh  
e0 e0
q
Gauss’ Law
Rod Charge Density
l
E 
2 e 0 r
Compare this with
our previous result.
E varies as 1/R
31
Charged Spherical Shell
Worked Example
 E-field inside & outside a charged
spherical shell (e.g. plane, car)
Outside the shell
 Use a Gaussian sphere of radius r
centred on the shell. Then:
E.dA = E dA
(since E ||dA)
Q
e0
  E  d A
 E  dA
r
 E 4 r 2 
R
E
E 
Q
4 e 0 r 2
32
Inside a Charged Spherical
Shell
Inside the shell
 r < R so consider a Gaussian sphere
inside the shell.
 no net charge enclosed by sphere
 Qencl = 0, so
f E 

Q
e0
0
Inside the shell the field is zero: a
physically important result.
No field inside
the shell
Faraday Cage!!
E
33
Solid Polarisable Sphere
Worked Example
 What is the electric field outside &
inside a solid nonconducting sphere
of radius R containing uniformly
distributed charge Q.
Outside the sphere:
 r > R
 consider spherical Gaussian surface
fE   E  d A 
Qencl
E 
4 e 0 r 2
As before
Qencl
e0
0
+
34
Inside the Solid Sphere
Inside the sphere
 r < R
 Charge enclosed by a Gaussian
sphere of radius r<R is:
Q  charge density volume 
 Q  4

3



r 

3
 4  R  3

 3

r3
Q 3
R
r
R
35
Field Inside Solid Charged
Sphere
Hence, from Gauss’ Law:

Q
r3
1
E
Q 3
2
4 e 0 r
R 4 e 0 r 2
E
 Q 
E  
r
3
 4 e 0 R 
Q
4 e 0 R 2
~r
1
~ 2
r
r
R

The same behaviour is found for
other forces, e.g. gravity.
36
Behaviour of Charges & Fields
Near Conductors






The electric field is zero
everywhere inside a conductor.
Electrons move to create an E
field which opposes any
external field.
Free charges move to the
outside surfaces of conductors
A result of Gauss’ law.
The electric field near a
conductor is perpendicular to
its surface.
A parallel component would
move charges and establish an
electric field inside.
37
Why Doesn’t My Radio Work


The electric field outside
a charged conductor is:
where:
Q
s
s
E
e0
area
Proof
 Consider a Gaussian cylinder
straddling the conductor’s surface.
fC   En dA 
++++++
+
++
+E = 0 +
+
+
+
+
+ + +++



Q
e0
 En A
En
dA
Q
s
 En 

e0 A e0
Closed hollow conductors admit no
electric field
EM shielding  “Faraday Cages”
Car Radios and biomagnetics
38
Importance and Tests of
Gauss’ Law




Coulomb’s law  experimental
evidence of Gauss’s law
1/r2 law is the key prediction
Gauss’ law is so basic that its
essential to test its validity
Tests of F  1/r2±d
Robinson
1769
d = 06
Cavendish
1773
0.02
Coulomb
1785
0.10
Maxwell
1873
5 x 10-5
Plimpton & Lawton
1936
2 x 10-9
Williams, Faller &
Hill
1971
3 x 10-16
39
4. Electric Potential: Technology
Can’t Live Without It!





Technology relies on using
energy associated with
electrical interactions
Work is done when Coulomb
forces move a charged particle
in an electric field.
This work is expressed in terms
of electric potential (energy)
Electric potential is measured
in Volts.
Basic to the operation of all
electric machines and circuits.
40
Mechanical Analogue
In mechanics

Work done in moving from point
ab
b
Wab   F  d S
a



results in a change in potential
energy:
W a  b= Ua - Ub
When W
>0
Ua > Ub
e.g. a mass falling under gravity
ab
41
What is Electric Potential?
In electricity


Consider a test charge q0
moving with respect to a
charge, q, fixed at the origin.
The work done is:
b
Wab   q0 E  d S
a

When integrated along the path
and thus:
b
U  q0  E  d S
a

This is the change in electric
potential energy, for a charge
q0 moving from a  b.
42
Electric Potential Energy

Since:
b
U    F  d S
a
qq0 d S
qq0  1 1 
U 

  
2

4 e 0 a rab 4 e 0  ra rb 
b


By definition, a charge infinitely far
away has zero potential energy.
The electric potential energy
between 2 charges is then:
1
qq0
qq0
U r  
k
4 e 0 r
r

Since this is a scalar the total
potential energy for a system of
charges is:
 q1q2 q1q3

U  k

 
r13
 r12

43
Uranium Nucleus Example
Worked Example

Calculate the electrostatic
potential energy between 2
protons in a Uranium nucleus
separated by 2 x 10-15 m.
q1q2
U r   k
r

1.6 x 10 
 9.0 x 10 
-19 2
9
2 x 10
-15
13
~ 10 J
44
Electric Potential
Definition:
 Electric potential is potential
energy per unit charge:
V r   U r 




q0
where U(r) is the potential
energy of test charge q0 due to
a charge distribution.
V(r) is a property of the
charges producing it, not q0.
Volt = unit of electric potential
1 V = 1 volt = 1 J/C
Note also that 1 V/m = 1 N/C
45
Potential & Charge Distribution



For a single point charge; q, a
distance r away, the electric
potential is:
q
q
V r   k 
r 4 e 0 r
Potential is zero if r = 
For a collection of charges:
n
qi
V  k
i 1 ri

For a charge distribution:
dq
V  k
r
46
Electric Potential Difference

Difference in electric potential for a
charge q between points a and b.
1 1
V  Vb  Va  kq  
 rb ra 
U Ua
1
V  b
   F ds
q0
q0 a
b
b
V    E  d s   Ed
a



For a uniform
field, d || E
i.e potential difference can be
expressed as a path-independent
integral over an electric field.
All charge distributions have an
electric potential
The potential difference Va - Vb is
the work/unit charge needed to
move a test charge from a  b
without changing its kinetic energy.
47
The electron volt




For the definition of volt, 1J of work
is needed to move 1 C of charge
through a potential difference of 1V
A more convenient unit at atomic
scales is the electron-volt:
The energy gained by an electron (or
proton) moving through a potential
difference of 1 volt:
1 eV = (1.6 x 10-19 C)(1 V)
= 1.6 x 10-19 J
Not an SI unit but a very useful one!
Worked Example
 In a hydrogen atom the e- revolves
around the p+ at a distance of 5.3 x
10-11 m. Find the electric potential at
the e- due to the p+, and the
electrostatic potential energy
between them.
48
Worked Examples
e-
A very
simplistic
picture

p+
r
Electric potential due to proton:
q 9 x 109 1.6 x 10-19 
V r   k 
 27 V
-11
r
5.3 x 10

Electrostatic p.e. is given by:
q1q2
U12  k
r12
  e V p   1.6 x 10-19 27   4.3 x 10-18 J
49
Forces on Charged Particles
Worked Example
 In a CRT an electron moves 0.2
m in a straight line (from rest)
driven by an electric field of 8
x 103 V/m. Find:
(a) The force on the electron.
(b) The work done on it by the
E-field.
(c) Its potential difference
from start to finish.
(d) Its change in potential
energy.
(e) Its final speed.
50
Worked Examples
(a) Force is in opposite direction
to the E-field, magnitude:
F  qE   1.6 x 10-19 8 x 103   1.3 x 10-15 N
(b) Work done by force:
Work  Fs   1.3 x 10-15  0.2  2.6 x 10-16 J
(c) Potential difference is
defined as work/unit charge:
W  2.6 x 10-16
3
V  

1.6
x
10
V
-19
q  1.6 x 10
Alternatively (e- opposite to p+):
b
d
a
0
V    E  d s    E dx   Ed 
 8 x 103 0.2   1.6 x 103 V
51
Worked Examples
(d) Change in potential energy:
b
U  q0  E  d s
a
 q0 V
 - 1.6 x 10-19 1.6 x 103 
 - 2.6 x 10-16 J
 work done 
(e) Loss of PE = gain in KE = ½mv2
2KE 
v
m
22.6 x 10-16 

9.1 x 10-31 
 2.4 x 107 ms -1
52
Worked Examples
Worked Example
 A proton is accelerated across a
potential difference of 600 V. Find
its change in K.E. and its final
velocity.
By definition, 1 eV = 1.6 x 10-19 J.
 Acceleration across 600 V
 Proton gains 600 eV.
K.E. = 600(1.6 x 10-19) = 9.6 x 10-17 J


Final velocity is:
29.6 x 10-17 
v
1.7 x 10-27 
 3.4 x 105 ms -1

If it started from rest
53
Equipotentials






Regions of equal electric potential
may be joined by contour lines.
These are equipotentials.
In 3-D these can form equipotential
surfaces where the potential is the
same at each point on the surface.
Field lines and equipotentials are
always perpendicular.
No work is done in moving a charge
along an equipotential surface
because there is no change in
potential.
The surface of a conductor is an
equipotential since charge is
uniformly distributed across the
surface of conductors.
54
Obtaining E from the Electric
Potential

Recall:
b
b
a
a
Vb  Va   dV    E  d s
b
b
a
a
  dV    E  d s



If the direction s is parallel to E for
infinitesimal elements ds from a to b
dV   E ds
dV
E
ds
Electric field is the rate of change
of potential V in the direction ds.
In 3-D space we use x, y & z
components to express in terms of
partial derivatives.
dV
Ex  
,
dx
dV
Ey  
,
dy
dV
Ez  
dz
55
Vector Notation

In vector notation:
V ˆ V ˆ V ˆ
E i
j
k
x
y
z
ˆ  ˆ  ˆ  
  i  j  k  V
y
z 
 x
 V

Where  is the gradient
operator.
56
Vector Notation Example
Worked Example
 A potential distribution in space is
described by:
V = Axy2 - Byz
where A and B are constants. Find
the electric field.
E  V
dV
 Ay 2
dx
dV
 2Axy - Bz
dy
dV
 -By
dz
E   Ay 2  iˆ  2 Axy  Bz  ˆj   By  kˆ
57
Potential Due to Charge
Distributions

If E is known, use:
b
V  Vb  Va    E  d s
a

If E is not known, use:
n
qi
V  k
i 1 ri

for continuous charge
distributions:
dq
V  k
r
58
Parallel Plates
Worked Example
 Two parallel metal plates have an
area A = 225 cm2 and are l=0.5 cm
apart, with a p.d. of 0.25 V between
them. Calculate the electric field.
ds
V  Vleft  Vright
0V
0.25V
b
  E  d s
a
l
  E dx
0
l
 E  dx
0.1V 0.2V
0
 El
x =0

x =0.5m
V 0.25

 50 Vm -1
l
0.5
This is obvious from the definition
of units of electric field = V/m.
E 
59
Uniformly Charged Disc
Worked Example
 Find the electric potential and
electric field along the axis of a
uniformly charged disc of radius R
and total charge Q.
R
y
(y2 + x2)½
x
P
dy

Consider the disc divided into rings
of radius,r, width, dr.
dq
k
V  k 
dq
2
2 
r
y x
60
Uniformly Charged Disc
(contd)


For the ring shown: dV 
For the total potential we integrate
over all rings:
R
V  k
0


k dq
y2  x2
dq
y2  x2
By definition of charge density:
Q
Q
s

for disc
2
area  R
For the ring:
dq = s 2 y dy
s 2
y dy
V
4e 0 0 y 2  x 2
R
s
2
2

y x 0
2e 0
s

R2  x2  x
2e 0
R



 
61
Uniformly Charged Disc

The field is only in the x direction.
 E  Ex  
dV
dx
1

  Q  2
2 2


E 
R  x   x 
2 

x  2e 0 R 


x


 1
1
2
2e 0 R  R 2  x 2  2 
Q
62
Why Sparks Occur at Pointed
Tips








Recall: Conducting objects contain
zero electric field.
Charge resides on outer surface
This surface is an equipotential.
Equipotential surfaces outside the
conductor are parallel to its surface.
For curved conductors, surface
1
charge density:
s
r
1
Hence: E 
(radius of curvature)
r
Small radius implies s and E are large
E.g. at points and tips
63
St. Elmo’s Fire

Regions of strong E-field


Corona discharge



Ionisation of air
greenish glow (St. Elmo’s Fire)
E > 3 x 106 V/m
Ionisation


Current flow
Carry away excess charge
Lightning conductors
 Do not attract
lightning
 Introduce a lower
potential
difference region
close to clouds.
+
++
+++
+ ++
+
+
+
+
+
++
++
++
++
+
64
Uses in Technology
Accelerators (1929)
(Giancoli Section 44.2, p1115)
 Van der Graaf HV Accelerator
 Works because E-field inside
Gaussian sphere is zero
 1m sphere  3 x 106 V
 Up to 20 MV produced
Precipitators (See Figure shown)
 Remove dust and particles from coal
combustion
 -ve wire @ 40 - 100 kV
 E-field  particles to wall
 > 99% effective.
Photocopiers (1940)
(Giancoli Example 21.5, p555)
 Image on +ve photoconductive drum
 Charge pattern  -ve toner pattern
 Heat fixing  +ve paper.
65