Coulombs Law and Electric Fields

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Coulombs Lawand Electric Fields
Physics 102 Lecture 02

• Today we will
• get some practice using Coulombs Law
• learn the concept of an Electric Field

Note the change in the Daily Planner
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But first, a leftover from last lectureACT
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
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Recall Coulombs 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
• qs 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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Three Charges
Example

• Calculate force on 2mC charge due to other two
  charges
• Calculate force from 7mC charge
• Calculate force from 7mC charge
• Add (VECTORS!)

Q2.0mC
4 m
6 m
Q7.0mC
Q-7.0 mC
10
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Three Charges
Example

• Calculate force on 2mC charge due to other two
  charges
• Calculate force from 7mC charge
• Calculate force from 7mC charge
• Add (VECTORS!)

Q2.0mC
4 m
6 m
Q7.0mC
Q-7.0 mC
10
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Three Charges
Example

• Calculate force on 2mC charge due to other two
  charges
• Calculate force from 7mC charge
• Calculate force from 7mC charge
• Add (VECTORS!)

F7
Q2.0mC

• F k q1q2/r2

5 m
F-7
4 m
6 m
Q7.0mC
Q-7.0 mC
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Example
Adding Vectors F7F-7
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Example
Adding Vectors F7F-7
y components cancel x components F F7,x
F-7,x F7,x (3/5)F7 F-7,x (3/5)F-7 F
(3/5)(55)x10-3 N6 x 10-3 N
F
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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

Example
E
E (9?109)(1.6?10-19)/(10-10)2 N 1.4?1011 N/C
(to the right)
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Preflight 2.3

• What is the direction of the electric field at
  point B?
• Left
• Right
• Zero

72 16 9
it is closer to the negative charge, and the
field lines point toward negative charges .
B only has the charge from the negative which is
pushing away from itself .
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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ACT E Field

• What is the direction of the electric field at
  point C?
• Left
• Right
• Zero

Red is negative Blue is positive
Away from positive charge (right)
Towards negative charge (right)
Net E field is to right.
y
C
x
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ComparisonElectric 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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E Field from 2 Charges
Example

• Calculate electric field at point A due to
  charges
• Calculate E from 7mC charge
• Calculate E from 3.5mC charge
• Add (VECTORS!)

A
Note this is similar to (but a bit harder than)
my earlier example.
4 m
6 m
Q7.0mC
Q-3.5 mC
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E Field from 2 Charges
Example

• Calculate electric field at point A due to
  charges
• Calculate E from 7mC charge
• Calculate E from 3.5mC charge
• Add

E7

• E k q/r2

E3
5 m
4 m
6 m
Q7.0mC
Q-3.5 mC
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Adding Vectors E7E3
Example

• Decompose into x and y components.

E7
E7y
E7 (4/5)
Q2.0mC
E7x
E7 (3/5)
5
4 m
4 m
6 m
Q7.0mC
Q-3.5 mC
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Adding Vectors E7E3
Example

• Decompose into x and y components.
• Add components.

E7
Etotal
E3
4 m
Ex 2.25?103 N/C Ey 1.0?103 N/C
6 m
Q7.0mC
Q-3.5 mC
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Preflight 2.2

• What is the direction of the electric field at
  point A?
• Up
• Down
• Left
• Right
• Zero

Red is negative Blue is positive
6 8 3 58 25
y
A
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?
• Up
• Down
• Right
• Left
• Zero

Red is negative Blue is positive
y
A
B
x
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Electric Field of a Point Charge
0.8?1011 N/C
Example
32?1011 N/C
25?1011 N/C
2.9?1011 N/C

E
This is becoming a mess!!!
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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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Preflight 2.5
Charge A is 1) positive 2) negative 3) unknown
Field lines start on positive charge, end on
negative.
93 4 3
44
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Preflight 2.6
X
A
B
Y
Compare the ratio of charges QA/ QB 1) QA 0.5QB
2) QA QB 3) QA 2 QB
lines proportional to Q
12 9 63
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Preflight 2.8
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 2) False
Density of field lines gives E
14 86
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ACT E Field Lines
B
A
Compare the magnitude of the electric field at
point A and B 1) EAgtEB 2) EAEB 3) EAltEB
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E inside of conductor

• Conductor ? electrons free to move
• Electrons feels electric force - will move until
  they feel no more force (F0)
• FEq if F0 then E0
• E0 inside a conductor (Always!)

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Preflight 2.10
"Charge A" is actually a small, charged metal
ball (a conductor). The magnitude of the electric
field inside the ball is (1) Negative (2)
Zero (3) Positive
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Recap

• E Field has magnitude and direction
• E?F/q
• Calculate just like Coulombs 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 ? E0

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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 600 AM Wednesday.

Have a great weekend. No classes Monday. See
you next Wednesday!