Download Electric Fields

Physics 102: Lecture 02
Coulomb’s Law
and Electric Fields
Today we will …
• get some practice using Coulomb’s Law
• learn the concept of an Electric Field
Note the change in the
Daily Planner
1
But first, a leftover from last lecture
ACT: Induced Dipole
• An uncharged conducting sphere is hung next to a
charged sphere. What happens when the
uncharged sphere is released?
1) Nothing
2) Attracted to charged sphere.
3) Repelled from charged sphere.
Van de Graaff demo
2
Recall Coulomb’s Law
• Magnitude of the force between charges q1 and q2
separated a distance r:
F = k q1q2/r2
k = 9x109 Nm2/C2
• Force is
– attractive if q1 and q2 have opposite sign
– repulsive if q1 and q2 have same sign
• Units:
– q’s have units of Coulombs (C)
• charge on proton is 1.6 x 10-19 C
– r has units of m
– F has units of N
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5
Three Charges
• Calculate force on +2mC charge due to other two charges
F7
Q=+2.0mC
4m
– Calculate force from +7mC charge
– Calculate force from –7mC charge
– Add (VECTORS!)
Q=+7.0mC
6m
Q=-7.0 mC
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10
Three Charges
• Calculate force on +2mC charge due to other two charges
F+7
Q=+2.0mC
4m
– Calculate force from +7mC charge
– Calculate force from –7mC charge
– Add (VECTORS!)
Q=+7.0mC
6m
Q=-7.0 mC
5
10
Three Charges
• Calculate force on +2mC charge due to other two charges
• F = k q1q2
/r2
(9 109 )(2 10 6 )(7 10 6 )
F7 
N
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3
F7  5 10 N
(9 109 )(2 106 )(7 106 )
F7 
N
25
3
F7  5 10 N
F+7
Q=+2.0mC
4m
– Calculate force from +7mC charge
– Calculate force from –7mC charge
– Add (VECTORS!)
Q=+7.0mC
6m
F-7
Q=-7.0 mC
6
14
Adding Vectors F+7+F-7
F+7
4m
Q=+2.0mC
Q=+7.0mC
6m
F-7
Q=-7.0 mC
14
7
Adding Vectors F+7+F-7
y components cancel
x components:
F+7
Q=+2.0mC
F = F+7,x + F-7,x
4m
F+7,x = (3/5)F+7
F-7,x = (3/5)F-7
F
F-7
F = (3/5)(5+5)x10-3 N=6 x 10-3 N
Q=+7.0mC
6m
Q=-7.0 mC
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8
Electric Field
• Charged particles create electric fields.
– Direction is the same as for the force that a + charge
would feel at that location.
– Magnitude given by:
E  F/q = kq/r2
Qp=1.6x10-19 C
+
r = 1x10-10 m
E
E = (9109)(1.610-19)/(10-10)2 N = 1.41011 N/C (to the right)
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21
Preflight 2.3
What is the direction of the electric field at point B?
“it is closer to the negative charge, and the field lines point toward
negative charges .”
Right “B only has the charge from the negative which is pushing away from
itself .”
72% 1) Left
16% 2)
9% 3) Zero
“electric fields of equal magnitudes but opposite directions are
present due to the positive and negative charges .”
Since charges have equal magnitude, and point B is
closer to the negative charge net electric field is to the
left
y
A
B
x
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23
ACT: E Field
What is the direction of the electric field at point C?
A. Left
Red is negative
B. Right
Blue is positive
C. Zero
Away from positive charge (right)
Towards negative charge (right)
Net E field is to right.
y
C
x
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25
Comparison:
Electric Force vs. Electric Field
• Electric Force (F) - the actual force felt by a
charge at some location.
• Electric Field (E) - found for a location only – tells
what the electric force would be if a charge were
located there:
F = Eq
• Both are vectors, with magnitude and direction.
Add x & y components. Direction determines
sign.
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Preflight 2.2
What is the direction of the electric field at point A?
6%
1) Up
Red is negative
8%
2) Down
Blue is positive
3%
3) Left
58% 4) Right
25% 5) Zero
A
y
B
x
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ACT: E Field II
What is the direction of the electric field at point A, if
the two positive charges have equal magnitude?
A. Up
Red is negative
B. Down
Blue is positive
C. Right
D. Left
A
y
E. Zero
B
x
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39
321011 N/C
Electric Field of a Point Charge
+
This is becoming a mess!!!
2.91011 N/C
E
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40
Electric Field Lines
•
Closeness of lines shows
field strength
- lines never cross
•
# lines at surface  Q
•
Arrow gives direction of E
- Start on +, end on -
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42
Preflight 2.5
X
A
Charge A is
B
Y
Field lines start on positive charge, end on negative.
1) positive
93%
2) negative
4%
3) unknown
3%
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44
Preflight 2.6
X
A
Y
B
Compare the ratio of charges QA/ QB # lines proportional to Q
1) QA= 0.5QB
12%
2) QA= QB
9%
3) QA= 2 QB
63%
22
45
Preflight 2.8
X
A
Y
B
The electric field is stronger when the
lines are located closer to one another.
The magnitude of the electric field at point X is greater than at point Y
1) True
14%
2) False
86%
Density of field lines gives E
23
46
ACT: E Field Lines
B
A
Compare the magnitude of the electric field at
point A and B
1) EA>EB
2) EA=EB
3) EA<EB
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47
E inside of conductor
• Conductor  electrons free to move
– Electrons feels electric force - will move until
they feel no more force (F=0)
– F=Eq: if F=0 then E=0
• E=0 inside a conductor (Always!)
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48
Preflight 2.10
X
A
Y
B
"Charge A" is actually a small, charged metal ball (a conductor). The
magnitude of the electric field inside the ball is:
(1) Negative
9%
(2) Zero
74%
(3) Positive
18%
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50
Recap
• E Field has magnitude and direction:
– EF/q
– Calculate just like Coulomb’s law
– Careful when adding vectors
• Electric Field Lines
– Density gives strength (# proportional to charge.)
– Arrow gives direction (Start + end on -)
• Conductors
– Electrons free to move  E=0
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To Do
• Read Sections 17.1-17.3
• Extra problems from book Ch 16:
– Concepts 9-15
– Problems 11, 15, 23, 27, 29, 39
• Do your preflight by 6:00 AM Wednesday.
Have a great weekend.
No classes Monday.
See you next Wednesday!
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