Transcript PowerPoint Presentation - Lecture 1 Electric Charge*

Math Review -20 minutes
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Lecture 2 Electric Fields Ch. 22 Ed. 7
• Cartoon - Analogous to gravitational field
• Topics
– Electric field = Force per unit Charge
– Electric Field Lines
– Electric field from more than 1 charge
– Electric Dipoles
– Motion of point charges in an electric field
– Examples of finding electric fields from continuous charges
• Demos
– Van de Graaff Generator, workings,lightning rod, electroscope,
– Field lines using felt,oil, and 10 KV supply
• Elmo
– Electron projected into an electric field
– Electric field from a line of charge
– Dipole torque
• Clickers
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Concept of the Electric Field
•
Definition of the electric field. Whenever a chrage is present and if I bring
up another charge, it will feel a net Coulomb force from it. It is
convenient to say that there is a field there equal to the force per unit
r
positive charge.
E  F / q0
•
The question is how does charge q0 know about charge q1 if it does not
“touch it”? Even in a vacuum! We say there is a field produced by q1 that
extends out in space everywhere.
•
The direction of the electric field is along r and points in the direction a
positive test charge would move. This idea was proposed by Michael
Faraday in the 1830’s. The idea of the field replaces the charges as
defining the situation. Consider two point charges:
r̂
q1
r
q0
6
r
r̂
+ q0
q1
q1q0
r̂
2
r
r
The force per unit charge is E  F / q0
The Coulomb force is
Fk
Then the electric field at r is E  k
q1
r̂ due to the point charge q1 .
2
r
The units are Newton/Coulomb. The electric field has direction and is a vector.
How do we find the direction.? The direction is the direction a unit positive test
charge would move. This is called a field line.
r̂
r
E
If q1 were positive
q1
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Example of field lines for a point negative
charge. Place a unit positive test charge
at every point r and draw the direction that
it would move
q1
r̂
r
The blue lines are the field lines.
The magnitude of the electric field isr̂
r kq1
E= 2 r̂
r
The direction of the field is given by the
line itself
q1
r̂
Important F= Eq0 , then ma=q0E,
and then a = q0E/m
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Repeat with Positive Point Charge
F
1
E
F
2
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Moving positive charge in a field of a positive
charge
y
x
10
Moving negative charge in a field of a
positive charge
y
x
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Electric Field Lines from more
two positive charges
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Electric Field Lines from two
opposite charges (+ -)
This is called an electric dipole.
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Electric Field Lines: a graphic concept
used to draw pictures as an aid to
develop intuition about its behavior.
The text shows a few examples. Here are the drawing rules.
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•
•
•
E-field lines either begin on + charges or begin at infinity.
E-field lines either end on - charges or end at infinity.
They enter or leave charge symmetrically.
The number of lines entering or leaving a charge is proportional to
the charge
• The density of lines indicates the strength of E at that point.
• At large distances from a system of charges, the lines become
isotropic and radial as from a single point charge equal to the net
charge of the system.
• No two field lines can cross.
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Example of field lines for a
uniform distribution of positive
charge on one side of a very large
nonconducting sheet.
This is called a uniform electric field.
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In order to get a better idea of field
lines try this Physlet.
• http://webphysics.davidson.edu/Applets/Applets.html
• Click on problems
• Click on Ch 9: E/M
•
Play with Physlet 9.1.4, 9.1.7
Demo: Show field lines using felt, oil, and 10 KV supply
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One point charge
Two point charges of same sign
Two point charges opposite sign
Wall of charge
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Methods of evaluating electric fields
• Direct evaluation from Coulombs Law for a single point charge
r kq1
E  2 rц
r1
• For a group of point charges, perform the vector sum
N
E
i 1
kqi
ri
2
rцi
• This is a vector equation and can be complex and messy to
evaluate and we may have to resort to a computer. The principle of
superposition guarantees the result.
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Typical Electric Fields (SI Units)
1 cm away from 1 nC of negative charge
r̂
-q
r
.
P
E
2


Nm
9
-9
8.99

10

(10
C)
2 

C 

kq1
4 N
E  2 r̂ 
 8.99  10
r̂
-4
2
r
10 m
C
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Typical Electric Fields
N
Fair weather atmospheric electricity = 100
C
downward 100 km high in the ionosphere
++++++++++
+++++++++
E
Earth
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Typical Electric Fields
Field due to a proton at the location of the electron in
the H atom. The radius of the electron orbit is
0.5  10 10 m.
r
Hydrogen atom
+
-
Note :
N
Volt

C meter
2


Nm
9
-19
8.99

10

(1.6

10
C)
2 

C 

kq1
11 N
E  2 r̂ 

5.75

10
r̂
-10
2
r
(0.5  10 m)
C
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Example of finding electric field from two charges
lying in a plane
15nC
We have q1  10nC at the origin, q2 atx=4
m and y=0.
What is the magnitude and direction of the electric field E at y=3 m and
x=0?
P
q1  10nC
x 0
q2  15nC
Find the x and y components of the electric field for
each charge and add them up.
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Find the magnitude of the
E1
field for q1
P
2
kq
9 Nm
Recall E  2 and k  8.99  10
r
C2
Nm 2
8.99  10
 10  10 9 C
2
N
C
E1 

10
in the y direction
(3m)2
C
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E1y  10
q1=10 nc
q2 =15 nc
N
C
E1x  0
Now find the magnitude of the field due to q2 ?
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Find the magnitude of
the field for q2
E1
E2
P
Nm 2
9
8.99  10

15

10
C
2
N
C
E2 

5.4
(5m)2
C
9
5


Find the x and y components
N 3
N
  1.08  3  3.24
C 5
C
N 4
N
E2 x  E2 cos   5.4 
 1.08  4  4.32
C
5
C
E 2y  E2 sin   5.4
q1=10 nc
q2 =15 nc
Now add all components
Ey  E1y  E2 y
Ex  E1x  E2 x
Ey  (10  3.24)
Ex  4.32
N
C
N
N
 13.24
C
C
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Add up the
components
E
E x  4.32
E  Ex2  Ey2
4.32 2  13.24 2
N
C

Magnitude of net electric field is
E 
E y  13.24
1N
C
 13.93N / C
Direction of the total electric field is

q1=10 nc
 Ey 
 13.24 
1  tan    tan 1 
  71.9

4.32
E 
q2 =15 nc
1
y
x
Using unit vector notation we can
also write the electric field vector as:


E  4.32 i  13.24 j N / C
-71.9

j

i
x
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Example of two identical charges of +15 nC on the x axis.
Each is 4 m from the origin. What is the electric field on the y
axis at P where y= 3 m? Example using symmetry to simplify.
y
P.
3
x
4
4
+15 nc
+15 nc
Find x and y components of the electric field at P due to each charge
and add them up. Find x component first.
Ex  E1x  E2 x  0
y
E1x
P.
3
4
+15 nc
E2 x
5
ϕ
Because E1x  E2 x
x
4
+15 nc
By symmetry
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Now find y component
y
E1
E1y 
.
3
4
+15 nc
Ey  E1y  E2 y  2E1y
Because E2 y  E1y
5

x
4
+15 nc
Nm 2
C
N
-9
E1 = 8.99  109

15

10
=
5.4
C2
(5m)2
C
y
E y =2E1y =E1 sin  = 2  5.4 
In unit vector notation
r
E = 6.5 ĵ N/C
3
5
N
= 6.5
C
ĵ
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Example of two opposite charges on the x axis. Each is 4 m
from the origin. What is the electric field on the y axis at P
where y= 3 m? Example using symmetry to simplify.
y
Ex
P
5
3
4
4

-15 nc
x
+15 nc
Find the x and y component of the electric field
Ey0 by symmetry
Nm 2
C
N
-9
E = 8.99  10

15

10
=
5.4
C2
(5m)2
C
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E x =2  E cos  = 2  5.4 
E = -8.6 î N/C
4
5
= - 8.6
N
C
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Electric Dipole
E1x  E cos and E1x  E2 x
E1
E1x  E2x  2E cos 

So, Enet  2E cos  î
kq
E 2
a
E1x
L
cos   
a 2a
ĵ
r
L
a  r 2   
2
p
2


L 2
2

3
2
a
a

kq L
So, Enet  2 2
î
a 2a

Enet  k

E2
L
2
r
kqL ˆ
E net   3 i
a
y
k
iˆ 
2

p 
1

3 
L 2 
r  (1  ( 2r ) 
3
2
r
kp
Enet ; 3
r
For large r
p  qL
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This is called an Electric Dipole: A pair of
equal and opposite charges q separated by a
displacement L. It has an electric dipole
moment p=qL.
r
kp
Enet ; 3
r
P
when r is large compared to L
p=qL = the electric dipole moment
r
Note inverse cube law
-q -
+ +q
L
The dipole moment p is defined as a vector directed
from - to +
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Electric Dipoles in Electric fields
A uniform external electric field exerts no net force on a dipole, but it does

 exert
torque that tends to rotate the dipole in the direction of the field (align p with Eext )

F1
x
Torque about the com = 
 FL sin
r r
 qEL sin  pE sin  p  E

F2
So,
  
t  pE
When the dipole rotates through , the electric field does work:
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Water (H2O) is a molecule that has a permanent
dipole moment.
GIven p = 6.2 x 10 - 30 C m
And q = -10 e and q = +10e
What is d?
d = p / 10e = 6.2 x 10 -30 C m / 10*1.6 x 10 -19 C = 3.9 x 10 -12 m
Very small distance but still is responsible
for the conductivity of water.
When a dipole is an electric field, the dipole
moment wants to rotate to line up with
the electric field. It experiences a torque.
Leads to how microwave ovens heat up food
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Why do Electric Dipoles align with Electric
fields ?
Work done equals
dW  td  pEsin d
The minus sign arises because the torque opposes any increase in  
Setting the negative of this work equal to the change in the potential
energy, we have

dU  dW  pE sin d
Integrating,
U   dU    dW    pE sin d   pE cos   U0
We choose U  0 when   90
 
Potential Energy U  pE cos   p  E
So, U  -p  E

Ans. The energy is minimum when p aligns with E
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Problem 22-50
An electric dipole consists of charges +2e and -2e separated by 0.61 nm.
It is in an electric field of strength 3.8 x 106 N/C. Calculate the magnitude
of the torque on the dipole when the dipole moment has the following
orientations.
(a)parallel to the field
(b) perpendicular to the field
(c) antiparallel to the field
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In Nonuniform Electric fields with a gradient,
the dipole gets attracted towards regions of
stronger fields
• When a dipole is in an electric field that varies with position,
then the magnitude of the electric force will be different for the
two charges. The dipole can be permanent like NaCl or water or
induced as seen in the hanging pith ball. Induced dipoles are
always attracted to the region of higher field. Explains why wood
is attracted to the teflon rod and how a smoke remover or
microwave oven works.
• Show smoke remover demo.
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Smoke Remover
Negatively charged central wire has electric field that varies as 1/r (strong
electric field gradient). Field induces a dipole moment on the smoke
particles. The positive end gets attracted more to the wire.
In the meantime a corona discharge is created. This just means that
induced dipole moments in the air molecules cause them to be
attracted towards the wire where they receive an electron and get
repelled producing a cloud of ions around the wire.
When the smoke particle hits the wire it receives an electron and then is
repelled to the side of the can where it sticks. However, it only has to
enter the cloud of ions before it is repelled.
It would also work if the polarity of the wire is reversed
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Motion of point charges in electric
fields
• When a point charge such as an electron is placed in an electric
field E, it is accelerated according to Newton’s Law:
a = F/m = q E/m for uniform electric fields
a = F/m = mg/m = g for uniform gravitational fields
If the field is uniform, we now have a projectile motion problemconstant acceleration in one direction. So we have parabolic
motion just as in hitting a baseball, etc except the magnitudes
of velocities and accelerations are different.
Replace g by qE/m in all equations;
For example, y =1/2at2 we get y =1/2(qE/m)t2
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Example illustrating the motion of a
charged particle in an electric field:
An electron is projected perpendicularly to a downward
uniform electric field of E= 2000 N/C with a horizontal
velocity v=106 m/s. How much is the electron deflected
after traveling a distance d=1 cm. Find the deflection y.
y
v
•
electron
d
E
E
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y
v
•
electron
d
E
E
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Finding Electric Fields for Continuous
Distributions of Charge
• Electric Field due to a long wire
• Electric field due to an arc of a circle of uniform
charge.
• Electric field due to a ring of uniform charge
• Electric field of a uniform charged disk
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What is the electric field from an infinitely long wire
with linear charge density of +100 nC/m at a point 10
away from it. What do the lines of flux look like?
++++++++++++++++++++++++++++++++
Typical field for the electrostatic smoke remover
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Continuous distribution of charges
• Instead of summing the charge we can imagine a
continuous distribution and integrate it. This distribution
may be over a volume, a surface or just a line. Also use
symmetry.
y
dE
.r
.
+x
-x
- L/2
dq
E   dE   kdq / r 2
L/2
dq  rdV volume
dq  s dA area
dq   dl line
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Find electric field due to a line of uniform + charge of length L
with linear charge density equal to  t  t  te y s
tt bsects te rd
   ++++++++++++++++++++++++++++++++
y
dE
dE
Ex 
L /2
 dE
x
0
L / 2
-x
- L/2

0
dE = k dq /r2
dq =  dx
r
+x
dq
L/2

Apply Coulombs law to a slug of charge dq
dEy= k  dx cos  r2
dE = k  dx /r2
dEy= dE cos θ
E y  k 
L /2
cos dx /r 2
L / 2
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y
dEy
dE
r =y sec 
y
-x
-L/2
0
r
E y  k 
dx = y sec2 
d
x= y tan
+x
x
L/2
dx / r2 = d / y
dq
L /2
r2 y2sec2 
cos dx /r 2
L / 2
E y  k 
0
 0
0
cos d / y  k / y  cos d  k / y 2sin 0
2k
Ey 
sin  0
y
 0
sin  0 
L /2
y  L /4
2
2
k
Ey 
y
L
y 2  L2 /4 43
What is the field at the center due to arc of
charge uniformly distributed along arc?
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What is the field at the center due to arc of
charge uniformly distributed along arc?
dq=λds
s=r 
ds=r d
0
Ey 
 dE
y
0
 0
0 due to symmetric element of charge above and below the x axis
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What is the field at the center due to arc of
charge uniformly distributed along arc?
dq=λds
0
Ey 
 dE
y
0
s=r 
ds=r d
 0
0 due to symmetric element of charge above and below the x axis
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Now find the x component
dEx= k dq cos  r2
dq=λds
s=r 
ds=r d
dEx= k  ds cos  r2
0
0
 0
 0
E x  k  rd cos /r 2  k /r  d cos
2k
Ex 
sin 0
r
47
Find the electric field on the axis of a uniformly charged
ring with linear charge density   Q/2R
Ez 
 dE cos
k cos
Ez 
r2

 ds

2
2
0
0
 ds   Rd  R  d  2R

dq =  ds

kQz
Ez  2
2 3/2
(z  R
)


dq
ds
dE  k 2  k 2
r
r
s  R
k cos
Ez 
2R
2
r
r2 =z2+R2
cos  z/r
=0 at z=0
=0 at z=infinity
=max at z=0.7R48
Chapter 22 Problem 30
A disk of radius 2.7 cm has a surface charge density of 4.0 µC/m2
on its upper face. What is the magnitude of the electric field
produced by the disk at a point on its central axis at distance z = 12
cm from the disk?
4
Chapter 22 Problem 44
At some instant the velocity components of an electron moving
between two charged parallel plates are vx = 1.7 x 105 m/s and vy =
2.8 x 103 m/s. Suppose that the electric field between the plates is
given by E = (120 N/C) j.
(a) In unit vector notation what is the electron's acceleration in the
field?
(b) What is the electron's velocity when its x coordinate has changed
by 2.7 cm?
50
Chapter 22 Problem 50
An electric dipole consists of charges +2e and -2e separated by
0.74 nm. It is in an electric field of strength 3.4 x106 N/C.
Calculate the magnitude of the torque on the dipole when the
dipole moment has the following orientations.
(a) parallel to the field
(b) perpendicular to the field
(c) antiparallel to the field
51