Download Electric flux

PHYS 1444
Lecture #2
June 7, 2012
Ian Howley
•
Chapter 21
–
–
–
Electric Fields
Electric Dipoles
Chapter 22
–
Gauss’s Law
•Ch. 21 Homework due Tuesday Jun 12 at 11:59 pm.
(Try to finish by Monday, so you can ask questions in class Tuesday)
• Labs start Jun 11
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Vector Problems
• Calculate magnitude of vectors
• Split vectors into x and y components
and add these separately, using
diagram to help determine sign
• Calculate magnitude of resultant
|F|=(Fx2+Fy2)
• Use = tan-1(Fy/Fx) to get angle
Example 21-2 and 21-3 on board
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The Electric Field
• Both gravitational and electric forces act over a
distance without touching objects  What
kind of forces are these?
– Field forces
• Michael Faraday developed the idea of a field.
– Faraday argued that the electric field extends
outward from every charge and permeates through
all space.
• The field due to a charge or a group of charges
can be inspected by placing a small positive
test charge in the vicinity and measuring the
force on it.
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The Electric Field
• The electric field at any point in space is defined as the
force exerted on a tiny positive test charge
r divided by
r
magnitude of the test charge
F
– Electric force per unit charge
E
q
• What kind of quantity is the electric field?
– Vector quantity. Why?
• What is the unit of the electric field?
– N/C
• What is the magnitude of the electric field at a distance r
from a single point charge Q?
1 Q
F
kQq r 2
kQ
E

 2 
2
4

q
r
q
r
0
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Direction of the Electric Field
• If there are several charges, the individual fields due to each
charge are added vectorially to obtain the total field at any
r
r
r
r
r
point.
ETot  E1  E2  E3  E4  ....
• This superposition principle of electric field has been verified
experimentally
• For a given electric field E at a given point in space, we can
calculate the force F on any charge q, F=qE.
– How does the direction of the force and the field depend on the
sign of the charge q?
– The two are in the same directions if q>0
– The two are in opposite directions if q<0
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Field Lines
• The electric field is a vector quantity. Thus, its
magnitude can be represented by the length of the
vector, with the arrowhead indicating the direction.
• Electric field lines are drawn to indicate the direction of
the force due to the given field on a positive test charge.
– Number of lines crossing a unit area perpendicular to E is
proportional to the magnitude of the electric field.
– The closer the lines are together, the stronger the electric field
in that region.
– Start on positive charges and end on negative charges.
Earth’s G-field lines
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Electric Fields and Conductors
• The electric field inside a conductor is ZERO in a static
situation (charge is at rest) Why?
– If there were an electric field within a conductor, there would be a
force on its free electrons.
– The electrons would move until they reach positions where the
electric field become zero.
– Electric field can exist inside a non-conductor.
• Consequences of the above
– Any net charge on a conductor distributes
itself on the surface.
– Although no field exists inside (the material
of) a conductor, fields can exist outside the
conductor due to induced charges on either
surface
– The electric field is always perpendicular to
the outside surface of a conductor.
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Example 21-14
• Shielding, and safety in a storm. A hollow metal
box is placed between two parallel charged plates.
What is the field in the box?
• If the metal box were solid
– The free electrons in the box would redistribute themselves along
the surface (the field lines would not penetrate into the metal).
• The free electrons do the same in hollow metal boxes just
as well as for solid metal boxes.
• Thus a conducting box is an effective device for shielding.
 Faraday cage
• So what do you think will happen if you were inside a car
when the car was struck by
lightning?http://www.youtube.com/watch?v=lUUOdO6eEZ
A
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Example 21 – 15
•
Electron accelerated by electric field. An electron (mass m = 9.1x10-31 kg)
is accelerated in a uniform field E (E = 2.0x104 N/C) between two parallel
charged plates.
The separation of the plates is 1.5 cm. The electron is accelerated from rest
near the negative plate and passes through a tiny hole in the positive plate.
(a) With what speed does it leave the hole?
(b) Can the gravitational force can be ignored?
Assume the hole is so small that it does not affect the uniform field between
the plates.
The magnitude of the force on the electron is F=qE and is
directed to the right. The equation to solve this problem is
F  qE  ma
F qE

The magnitude of the electron’s acceleration is a 
m
m
Between the plates the field E is uniform, thus the electron undergoes a
uniform acceleration



1.6  1019 C 2.0  104 N / C
eE
15
2
a


3.5

10
m
s
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PHYS
me
9.1  1031
kg1444 Ian Howley


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Example 21 – 15
Since the travel distance is 1.5x10-2m, using one of the kinetic eq. of motions,
v2  v02  2ax v  2ax  2  3.5 1015 1.5 102  1.0 107 m s
Since there is no electric field outside the conductor, the electron continues
moving with this speed after passing through the hole.
(b) Can the gravitational force can be ignored? Assume the hole is so
small that it does not affect the uniform field between the plates.
The magnitude of the electric force on the electron is



Fe  qE  eE  1.6  1019 C 2.0  104 N / C  3.2  1015 N
The magnitude of the gravitational force on the electron is


FG  mg  9.8 m s 2  9.1 1031 kg  8.9  1030 N
Thus the gravitational force on the electron is negligible compared to the
electromagnetic force.
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Electric Dipoles
• An electric dipole is the combination of two equal charges
of opposite sign, +Q and –Q, separated by a distance l,
which behaves as one entity.
• The quantity Ql is called the electric dipole moment and
is represented by the symbol p.
–
–
–
–
The dipole moment is a vector quantity
The magnitude of the dipole moment is Ql. Unit? C-m
Its direction is from the negative to the positive charge.
Many of diatomic molecules like CO have a dipole moment. 
These are referred as polar molecules.
• Symmetric diatomic molecules, such as O2, do not have dipole moment.
– The water molecule also has a dipole moment
which is the vector sum of two dipole moments
between Oxygen and each of Hydrogen atoms.
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Dipoles in an External Field
• Let’s consider a dipole placed in a uniform electric
field E.
• What do you think will happen to the dipole
in the figure?
– Forces will be exerted on the charges.
• The positive charge will get pushed toward right
while the negative charge will get pulled toward left.
– What is the net force acting on the dipole?
• Zero
– So will the dipole move?
• Yes, it will.
– Why?
• There is torque applied on the dipole.
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Dipoles in an External Field, cnt’d
• How much is the torque on the dipole?
– Do you remember
the formula for torque?
r
r r
•   r  F I thought you could!
– The magnitude of the torque exerting on each of the charges
with respect
to the rotational axis at the center is
r
•
•
r
l
l
  Q  r  F  rF sin      QE  sin   QE sin 
2
2
r r
 l
 Q  r  F  rF sin       QE  sin   l QE sin 
 2
2
– Thus, the total torque is
•
l
l
 Total   Q   Q  QE sin   QE sin   lQE sin   pE sin 
2
2
r r
– So the torque on a dipole in vector notation is   p  E
• The effect of the torque is to try to turn the dipole so that
the dipole moment is parallel to E.
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r
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Potential Energy of a Dipole in an External Field
• What is the work done on the dipole by the electric field to
change the angle from 1 to 2?
r
Why negative?
2
2 r
2
W   dW     d    d
Because  and  are opposite
1
1
1
directions to each other.
• The torque is   pE sin  .
• Thus the
work
done
on
the
dipole
by
the
field
is
2
2
W    pE sin  d  pE cos   pE  cos2  cos1 
1
1
• What happens to the dipole’s potential energy, U, when
positive work is done on it by the field?
– It decreases.
• If we choose U=0 when 1=90 degrees, then the potential
r r
energy at 2= becomes U  W   pE cos   p  E
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Electric Field from a Dipole
• Let’s consider the case in the picture.
• There are fields due to both charges.
r
r Ther total E
ETot  E Q  EQ
field from the dipole is
• The magnitudes of the two fields are equal
EQ  EQ 
1
Q
1
Q
1
Q


2
4 0  2
4 0 r 2   l 22 4 0 r 2  l 2 4
2 
 r  l 2 


• Now we must work out the x and y components
of the total field.
– Sum of the two y components is
• Zero since they are the same but in opposite direction
– So the magnitude of the total field is the same as the
sum of the two x-components:
E • 2E cos 
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1
p
l
Q

2
2
2
2
2 0 r PHYS
l 1444
4 2 Ianr Howley
 l 4 4 0 r 2  l 2 4
1


32
15
Example 21 – 17
•
Dipole in a field. The dipole moment of a water molecule is 6.1x10-30C-m. A water
molecule is placed in a uniform electric field with magnitude 2.0x105N/C.
(a) What is the magnitude of the maximum torque that the field can exert on the molecule?
(b) What is the potential energy when the torque is at its maximum?
(c) In what position will the potential energy take on its greatest value? Why is this
different than the position where the torque is maximized?
(a) The torque is maximized when =90 degrees. Thus the magnitude of
the maximum torque is
  pE sin   pE 



 6.1  1030 C  m 2.5  105 N C  1.2  10 24 N  m
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Example 21 – 17
(b) What is the potential energy when the torquer is at its maximum?
r
Since the dipole potential energy is U   p  E   pE cos
And  is at its maximum at =90 degrees, the potential energy, U, is
U   pE cos   pE cos  90   0
Is the potential energy at its minimum at =90 degrees? No
Why not?
Because U will become negative as  increases.
(c) In what position will the potential energy take on its greatest value?
The potential energy is maximum when cos= -1, =180 degrees.
Why is this different than the position where the torque is maximized?
The potential energy is maximized when the dipole is oriented so
that it has to rotate through the largest angle against the direction
of the field, to reach the equilibrium position at =0.
Torque is maximized when the field is perpendicular to the dipole, =90.
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Ch 22 Gauss’s Law
•
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Chapter 22:
- Gauss’s Law
- Gauss’s Law with many charges
- What is Gauss’s Law good for?
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Gauss’ Law
• Gauss’ law states the relationship between electric
charge and electric field.
– More general and elegant form of Coulomb’s law.
• The electric field from a distribution of charges can be
obtained using Coulomb’s law by summing (or
integrating) over the charge distributions.
• Gauss’ law, however, gives an additional insight into
the nature of electrostatic field and a more general
relationship between the charge and the field
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Electric Flux
• Let’s imagine a surface of area A through which a uniform
electric field E passes
• The electric flux is defined as
– FE=EA, if the field is perpendicular to the surface
– FE=EAcos, if the field makes an angle  with the surface
r r
• So the electric flux is defined as F E  E  A.
• How would you define the electric flux in words?
– Total number of field lines passing through the unit area
perpendicular to the field. N E  EA  F E
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Example 22 – 1
• Electric flux. (a) Calculate the electric
flux through the rectangle in the figure
(a). The rectangle is 10cm by 20cm
and the electric field is uniform with
magnitude 200N/C. (b) What is the flux
if the angle is 30 degrees?
The electric flux is
r r
F E  E  A  EA cos
So when (a) =0, we obtain
F E  EA cos  EA   200 N / C   0.1 0.2m 2  4.0 N  m 2 C

And when (b) =30 degrees (1,2,3) we obtain



2
o
2
200
N
/
C

0.1

0.2m
cos30

3.5
N

m
C
F E  EA cos30  

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Generalization of the Electric Flux
• Let’s consider a surface of area A that has
an irregular shape, and furthermore, that
the field is not uniform.
• The surface can be divided up into
infinitesimally small areas of DAi that can
be considered flat.
• And the electric field through this area can
be considered uniform since the area is
r r
very small.
F E   Ei  dA
• Then the electric flux through the entire
n r
r
surface is
FE 
open surface
 E  DA
i
i
i 1
• In the limit where DAi  0, the discrete
summation becomes an integral.
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FE 
r r
Ei  dA
Ñ

closed
surface
22
Generalization of the Electric Flux
• We define the direction of the area
vector as pointing outward from the
enclosed volume.
– For the line leaving the volume, </2, so cos>0. The flux is positive.
– For the line coming into the volume, >/2, so cos<0. The flux is
negative.
– If FE>0, there is a net flux out of the volume.
– If FE<0, there is flux into the volume.
• In the above figures, eachr field
r that enters the volume also leaves
the volume, so F E  Ñ
 E  dA  0.
• The flux is non-zero only if one or more lines start or end inside the
surface.
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Generalization of the Electric Flux
• The field line starts or ends only on a charge.
• Sign of the net flux on the surface A1?
– Net outward flux (positive flux)
• How about A2?
– Net inward flux (negative flux)
• What is the flux in the bottom figure?
– There should be a net inward flux (negative flux)
since the total charge inside the volume is
negative.
• The flux that crosses an enclosed surface is
proportional to the total charge inside the
surface.  This is the crux of Gauss’ law.
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Gauss’ Law
• The precise relation between flux and the enclosed charge is
r r Qencl
given by Gauss’ Law
Ñ
 E  dA 
0
• 0 is the permittivity of free space in the Coulomb’s law
• A few important points on Gauss’ Law
– Freedom to choose!!
• The integral is performed over the value of E on a closed surface of our choice
in any given situation.
– Test of existence of electrical charge!!
• The charge Qencl is the net charge enclosed by the arbitrary closed surface of
our choice.
– Universality of the law!
• It does NOT matter where or how much charge is distributed inside the
surface.
– The charge outside the surface does not contribute to Qencl. Why?
• The charge outside the surface might impact field lines but not the total number
of lines entering or leaving the surface
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Gauss’ Law
q’
q
• Let’s consider the case in the above figure.
• What are the results of the closed integral of the
gaussian surfaces A1 and A2?
– For A1
– For A2
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r r q
E  dA 
Ñ

0
r r  q
Ñ
 E  dA   0
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Coulomb’s Law from Gauss’ Law
• Let’s consider a charge Q enclosed inside our
imaginary Gaussian surface of sphere of radius r.
– Since we can choose any surface enclosing the charge, we choose the
simplest possible one! 
• The surface is symmetric about the charge.
– What does this tell us about the field E?
• Must have the same magnitude at any point on the surface
• Points radially outward ( or inward) parallel to the surface vector dA.
• The Gaussian integral can be written as
r r
E  dA 
Ñ

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Ñ

Ñ



EdA  E dA  E 4 r 2 
Qencl
0
PHYS 1444 Ian Howley

Q
0
Solve
for E
Q
E
4 0 r 2
Electric Field of
27
Coulomb’s Law
Gauss’ Law from Coulomb’s Law
• Let’s consider a single static point charge Q
surrounded by an imaginary spherical surface.
• Coulomb’s law tells us that the electric field at a
1 Q
spherical surface is
E
4 0 r 2
• Performing a closed integral over the surface, we obtain
r
r r
1 Q
1 Q
rˆ  dA 
dA
E  dA 
2
2
4 0 r
4 0 r
1 Q
1 Q
Q
2

dA 
4 r 
2
2
4 0 r
4 0 r
0
Ñ

Ñ

Ñ

Ñ

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
Gauss’ Law

28
Gauss’ Law from Coulomb’s Law
Irregular Surface
• Let’s consider the same single static point
charge Q surrounded by a symmetric spherical
surface A1 and a randomly shaped surface A2.
• What is the difference in the number of field lines passing through
the two surface due to the charge Q?
– None. What does this mean?
• The total number of field lines passing through the surface is the same no matter
what the shape of therenclosed
r surface
r is.r
– So we can write:
Ñ

Ñ

A1
A2
E  dA 
E  dA  Q
0
– What does this mean?
• The flux due to the given enclosed charge is the same no matter what the shape of
r r
5/7/2017the surface enclosing it is. 
PHYS
1444 Ianlaw,
Howley E  dA  Q , is valid for any surface
29
Gauss’
Ñ

0
surrounding a single point charge Q.
Gauss’ Law w/ more than one charge
• Let’s consider several charges inside a closed surface.
• For each charge, Qi, inside the chosen
r closed surface,
r Qi
r
Ei  dA 
Ñ

0
What is
Ei ?
The electric field produced by Qi alone!
• Since electric fields can be added vectorially, following the
superposition principle, ther total field
E is equal to the sum of the
r
fields due to each charge E   Ei . So
What is Qencl?
r r
r
r
Qi Qencl

total
Ñ
 E  dA  Ñ
  Ei  dA   0   0 The
enclosed charge!
• The value of the flux depends on the charge enclosed in the
surface!!  Gauss’ law.

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
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So what is Gauss’ Law good for?
• Derivation of Gauss’ law from Coulomb’s law is only
valid for static electric charge.
• Electric field can also be produced by changing
magnetic fields.
– Coulomb’s law cannot describe this field, but Gauss’ law is
still valid
• Gauss’ law is more general than Coulomb’s law.
– Can be used to obtain electric field, forces, or charges
Gauss’ Law: Any differences between the input and output flux of the electric
field over any enclosed surface is due to the charge within that surface!!!
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Announcements
•
•
•
•
5/7/2017
HW Due Tuesday 11:59pm
Ryan Hall Lecture on Tuesday/Thursday next week
Test 1 three weeks from today
This lecture will be posted on the class site
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