Transcript Lecture 1

Physics 2220
Physics for Scientists and Engineers II
Physics for Scientists and Engineers II , Summer Semester 2009
Chapter 23: Electric Fields
• Materials can be electrically charged.
• Two types of charges exist: “Positive” and “Negative”.
• Objects that are “charged” either have a net “positive” or a net
“negative” charge residing on them.
• Two objects with like charges (both positively or both negatively
charged) repel each other.
• Two objects with unlike charges (one positively and the other
negatively charged) attract each other.
• Electrical charge is quantized (occurs in integer multiples of a
fundamental charge “e”).
q =  N e (where N is an integer)
electrons have a charge q = - e
protons have a charge
q=+e
neutrons have no charge
Physics for Scientists and Engineers II , Summer Semester 2009
Material Classification According to Electrical Conductivity
• Electrical conductors: Some electrons (the “free”
electrons) can move easily through the material.
• Electrical insulators: All electrons are bound to atoms and
cannot move freely through the material.
• Semiconductors: Electrical conductivity can be changed
over several orders of magnitude by “doping” the material
with small quantities of certain atoms, making them more
or less like conductors/insulators.
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Shifting Charges in a Conductor by “Induction”
uncharged metal sphere
Negatively charged rod
+
-
-
-
-
+
+
-
-
-
-
+
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+ -+
+
Left side of metal sphere
more positively charged
-
+
+
-
+
-
Right side of metal sphere
more negatively charged
Coulomb’s Law (Charles Coulomb 1736-1806)
Magnitude of force between two “point charges” q1 and q2 .
Fe  ke
q1 q2
r
2
Coulomb constant
Nm 2
1
ke  8.987 6 x 10

C2
4  o
9
where  o  8.854 2 x 10
12
C2
N m2
Permittivity of free space
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r = distance between
point charges
Charge
Unit of charge = Coulomb
Smallest unit of free charge: e = 1.602 18 x 10-19 C
Charge of an electron: qelectron = - e = - 1.602 18 x 10-19 C
Physics for Scientists and Engineers II , Summer Semester 2009
Vector Form of Coulomb’s Law
Force is a vector quantity (has magnitude and direction).
F 12
q1q2
 ke 2 r̂12
r
unit vector pointing from
charge q1 to charge q2
Force exerted by charge q1 on charge q2
(force experienced by charge q2 ).
Physics for Scientists and Engineers II , Summer Semester 2009
Vector Form of Coulomb’s Law
Force is a vector quantity (has magnitude and direction).
q1q2
q1q2
F 21  ke 2 rˆ21   ke 2 rˆ12   F 12
r
r
unit vector pointing from
charge q2 to charge q1
Force exerted by charge q2 on charge q1
(force experienced by charge q1 ).
Physics for Scientists and Engineers II , Summer Semester 2009
Directions of forces and unit vectors
F 12
r̂21
r̂12
F 21
+
q1
+
q2
F 12
F 21
+
q1
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-
q2
Calculating the Resultant Forces on Charge q1 in a Configuration of 3
charges
q1
+
a = 1cm
q3
-
q2
+
0.5 cm
q3 = - 2.0 mC
q1 = q2 = +2.0 mC
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0.5 cm
Forces acting on q1
F 21
q1
+
q2
+
F 31
q3
Total force on q1:
F 1  F 21  F 31
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Magnitude of the Various Forces on q1
2
12
2.0mC  2.0mC
C2
9 Nm 4.0 10
2
F21  ke

8
.
99

10

3
.
596

10
N
2
2
4
2
(0.010m)
C 1.0 10 m
2
12
2
2.0 mC  2.0 mC
Nm
4
.
0

10
C
9
2
F31  ke

8
.
99

10

7
.
192

10
N
2
4
2
2
C 0.5  10 m
(0.010  0.5  2 m)
Note: I am temporarily carrying along extra significant digits in these
intermediate results to avoid rounding errors in the final result.
Physics for Scientists and Engineers II , Summer Semester 2009
Adding the Vectors Using a Coordinate System
F 21
q1
+
y
q2
+
F 31
q3
-
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x
Adding the Vectors Using a Coordinate System
F 21  F21 iˆ  0 ˆj
F 31  
y
2
2
F31 iˆ 
F31 ˆj
2
2
F 21
F 1  F 21  F 31

 

2
2
ˆ



  F21 
F31 i   
F31  ˆj
2

  2

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x
F 31
…doing the algebra…
F 1  F 21  F 31

 

2
2
ˆ



  F21 
F51  i   
F31  ˆj
2

  2



 3.596 10 2 N  5.086 10 2 N iˆ  5.086 10 2 N ˆj
  1.5 10 2 N iˆ  5.110 2 N ˆj
F1 has a magnitude of
F1  5.3 102 N
Physics for Scientists and Engineers II , Summer Semester 2009
Calculating the force on q2 … another example using an even more
mathematical approach
Charges
Location of charges
q1 = +3.0 mC
x1=3.0cm ; y1=2.0cm ; z1=5.0cm
q2 = - 4.0 mC
x2=2.0cm ; y2=6.0cm ; z2=2.0cm
In this example, the location of the charges and the distance
between the charges are harder to visualize 
Use a more mathematical approach!
Physics for Scientists and Engineers II , Summer Semester 2009
Calculating the force on q2 … another example using an even more
mathematical approach
q1q2
F2  F12  ke
r̂
2 12
d12
d12=distance between q1 and q2.
rˆ12  unit vecto r pointing from q1to q 2 .
Physics for Scientists and Engineers II , Summer Semester 2009
Calculating the force on q2 … mathematical approach
We need the distance between the charges.
d12 is distance between q1 and q2.
y
q1
+
r2  r1
r1
- q2
r2
z
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x
d12  (r2  r1 )  (r2  r1 )
Calculating the force on q2 … mathematical approach
Distance between charges q1 and q2 .
 2.0 cm   3.0 cm    1.0 cm 

 
 

r2  r1   6.0 cm    2.0 cm    4.0 cm 
 2.0 cm   5.0 cm    3.0 cm 

 
 

  1.0 cm    1.0 cm 

 

d12  (r2  r1 )  (r2  r1 )   4.0 cm    4.0 cm 
  3.0 cm    3.0 cm 

 

 1.0 cm 2  16.0 cm 2  9.0 cm 2  26 cm
Physics for Scientists and Engineers II , Summer Semester 2009
Calculating the force on q2 … mathematical approach
We need the unit vectors between charges. For example, the unit
vector pointing from q1 to q2 is easily obtained by normalizing the
vector pointing from from q1 to q2.
y
q1
+
r1
r̂12
d12  (r2  r1 )  (r2  r1 )
r2  r1
- q2
r2
z
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x
r2  r1
rˆ12 
d12
Calculating the force on q2 … mathematical approach
The needed unit vector:
  1.0 cm 
  1.0 


r2  r1
1 
1 
rˆ12 

 4.0 cm  
 4.0 
d12
26 cm 
14 



3
.
0
cm

3
.
0




Physics for Scientists and Engineers II , Summer Semester 2009
Calculating the force on q2 … mathematical approach
You can easily verify that the length of the unit vector is “1”.
 1

Length of rˆ12   4
 3

  1
26  
4



26
  3
26  
Physics for Scientists and Engineers II , Summer Semester 2009

26 
1 16 9
26

 

1
26 
26 26 26
26

26 
Calculating the force on q2 … another example using an even more
mathematical approach
F2  F12  ke
q1q2
rˆ
2 12
d12
  1.0 

3.0 mC  (4.0) mC 1 
 ke
4
.
0


26 cm 2
26 

  3.0 
2   1.0 
2


Nm
C  100cm  
 8.9876 109 2   0.462 10 12 2 
4
.
0
 

C 
cm  1m  

  3. 0 
  1.0 


 41.48 4.0  N
  3.0 


2
 41.48 
 0.42 




   165.92  N    1.7  10 2 N
 124.44 
 1.2 




Physics for Scientists and Engineers II , Summer Semester 2009
Calculating the force on q2 … another example using an even more
mathematical approach
…and if you want to know just the magnitude of the force on q2 :
F2  F2  F2
 41.48 N   41.48 N 

 

   165.92 N     165.92 N 
 124.44 N   124.44 N 

 

 1721 N 2  27529 N 2  15485 N 2
 2.110 2 N
Physics for Scientists and Engineers II , Summer Semester 2009
23.4 The Electric Field
Remember from Physics 2210 :
The gravitatio nal field g 
Fg
m
at the location of the mass m.
g is defined as the force Fg acting on a test particle of mass m divided by that mass.
Similarly, Faraday defined :
The electric field vector E 
Fe
qo
E is defined as the force Fe acting on a test charge q o divided by q o .
It is convenient to use positive test charges. Then, the direction of
the electric force on the test charge is the same as that of the field
vector. Confusion is avoided.
Physics for Scientists and Engineers II , Summer Semester 2009
23.4 The Electric Field
N
SI units of electric field E :
C
Note : The measured E is NOT produced by the test charge itself.
The test charge only detects E produced by source charge Q.
Of course, the test charge also produces an additionalelectric field. However, there is no force acting on
the test charge due to the field produced by the test charge. Therefore, the force on the test charge is solely
due to the electric field of the source charge Q alone.The field measured is the actual field that exists once
the test charge is removed again.
Q
+ +
+ +
+ +
+ +
qo
E
+
test charge
Source charge
Physics for Scientists and Engineers II , Summer Semester 2009
23.4 The Electric Field
Force on a charge q in an electric field :
Fe  q E
For positive charge q :
Fe is in the same direction as E .
For negative charge q :
Fe and E are in opposite directions.
Once you know
E at some point in space, you can calculate the
force on any charged particle placed at that point.
Physics for Scientists and Engineers II , Summer Semester 2009
23.4 The Electric Field of a “Point Charge” q
Fe  ke
Force exerted on test charge q 0 by q :
qq0
rˆ
2
r
Fe
q
Electric field created by q at the place of q 0 : E 
 ke 2 rˆ
q0
r
where r is the distance between q and q 0 and
r̂ is the unit vecto r pointing from q to q 0 .
q
r̂
r
Physics for Scientists and Engineers II , Summer Semester 2009
q0
23.4 The Electric Field of a Positive “Point Charge” q
(Assuming positive test charge q0)
Fe
q0
+
Force on test charge
E
+
P
Electric field where test charge
used to be (at point P).
The electric field of a positive point charge points away from it.
Physics for Scientists and Engineers II , Summer Semester 2009
23.4 The Electric Field of a Negative “Point Charge” q
(Assuming positive test charge q0)
Fe
-
q0
Force on test charge
E
-
P
Electric field where test charge
used to be (at point P).
The electric field of a negative point charge points towards it.
Physics for Scientists and Engineers II , Summer Semester 2009
23.4 The Electric Field of a Collection of Point Charges
q
E   ke 2 rˆi
ri
i
Physics for Scientists and Engineers II , Summer Semester 2009
23.4 The Electric Field of Two Point Charges at Point P
y
P
E ?
y
q1
q2
a
x
b
q
q1
q2
E   ke 2 rˆi  ke 2 rˆ1  ke 2 rˆ2
ri
r1
r2
i
Physics for Scientists and Engineers II , Summer Semester 2009
23.4 The Electric Field of Two Point Charges at Point P
y
P
r1
q1
r2
y
a
Pythagoras:
r12  a 2  y 2
b
q2
x
q1
q2
ˆ
E  ke 2
r  ke 2
rˆ
2 1
2 2
a y
b y
Physics for Scientists and Engineers II , Summer Semester 2009
r22  b 2  y 2
23.4 The Electric Field of Two Point Charges at Point P
y
rˆ1 
r1 p
P
rˆ2 
r1 p
r1 p
r1

r2 p
r2


1
1  0    a  1   a 
1
R  C 1            
r1
r1   y   0   r1  y 
a2  y2


R
q1
 0
R   
 y
r̂2
r̂1
C1
C2
q1
E  ke 2
a  y2

1
1  0  b 1  b 
1
R  C 2            
r2
r1   y   0   r1  y 
b2  y 2
q2
 a
C1   
 0 
  a
 
 y 
b
 
 y
b
C 2   
 0
x
 a
q2
   ke 2
2
2
2  y 
b

y
a y  
1
Physics for Scientists and Engineers II , Summer Semester 2009
b
 
2
2  y
b y  
1
23.4 The Electric Field of Two Point Charges at Point P
E  ke
a
q1
2
E x  ke
E y  ke
y
a
a
2

3
 a
q2
   ke
 y 
b2  y 2

2
q1a
2
y
2
q1 y
2
y
2

3

3
 ke
2
2
 ke
Physics for Scientists and Engineers II , Summer Semester 2009
b
b
q2b
2
y
2
q2 y
2
y
2

3

3

3
2
2
2
b
 
 y
23.4 The Electric Field of Two Point Charges at Point P
Special case: q1= q and q2 = -q AND b = a
E from + charge
total E
(has no y - component)
E from - charge
+
-
q
-q
E x  k e
E y  ke
a
a
q1a
2
 y2
q1 y
2
y
2

3

3
2
2
 ke
 ke
b
q2b
2
b
 y2

3
q2 y
2
y
2
2

3
 k e
2
 ke
a
qa
2
a
Physics for Scientists and Engineers II , Summer Semester 2009
 y2

3
qy
2
y
2
2

3
 ke
2
a
 ke
qa
2
 y2
a

3
2
qy
2
y
 ke
2

3
0
2
a
2qa
2
 y2

3
2
23.4 The Electric Field of Two Point Charges at Point P
Special case: q1= q and q2 = q AND b = a
total E
E from other
+ charge
E from + charge
+
+
q
q
E x  k e
E y  ke
(has no x - component)
a
a
q1a
2
y
2
q1 y
2
 y2

3

3
2
2
 ke
 ke
b
q2b
2
b
y
2

q2 y
2
 y2
3
2

3
 ke
2
 ke
Physics for Scientists and Engineers II , Summer Semester 2009
a
qa
2
a
y
2

qy
2
 y2
3
2

3
 ke
2
a
 ke
qa
2
y
a
2

3
0
2
qy
2
 y2

3
2
 ke
a
2qy
2
 y2

3
2
This is called an electric DIPOLE
Special case: q1= q and q2 = -q AND b = a
E from + charge
total E
(has no y - component)
E from - charge
+
-
q
-q
E x  ke
a
2qa
2
y
2

3
Ey  0
2
For large distances y (far away from the dipole), y >> a:
E x  ke
2qa
y 
2
3
2
 ke
2qa
y3
E falls off proportional to 1/y3
Fall of faster than field of single charge (only prop. to 1/r2).
From a distance the two opposite charges look like they are
almost at the same place and neutralize each other.
Physics for Scientists and Engineers II , Summer Semester 2009