Transcript PHy 184 lecture 5 - Home Page | MSU Department of Physics

PHY 184
Spring 2007
Lecture 5
Title: Electric Field Examples
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184 Lecture 5
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Announcements
 Homework Set 1 is done. The average score was
9.6/10.
 Homework Set 2 is open - we will open Homework
Sets now already on Thursday.
 Helproom coverage is posted in LON-CAPA.
 Honors option work in the SLC:
• Please sign up for time slots! Use the sign-up sheet after
class.
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Review - The Electric Field
 A charge creates an
electric field around itself
and the other charge feels
that field.
+
+
Test charge q
Test charge: point object with a very
small positive charge so that it does
not modify the original field
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The electric field at a
specified point: place a
positive test charge q at
the point and measure the
electrostatic force that
acts on the test charge,

  F
E (x ) 
q
3
Review -Field Lines from a Point Charge
 The electric field lines
from a point charge
extend out radially.
 For a positive point
charge, the field lines
point outward
• Terminate at infinity
 For a negative charge,
the field lines point
inward
• Originate at infinity
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Electric Field from a Point Charge
 Suppose we have two charges, q
and q0, separated by a distance r.
The electric force between the
two charges is
1 qq0
F
4 0 r 2
 We can consider q0 to be a test
charge, and determine the electric
field from charge q as
F
1 q
E

q0 4 0 r 2
 The electric field is a vector, so
to add electric fields we must add
the components separately.
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Electric Field from 4 Point Charges (1)
 Four charges q1=10 nC, q2=-20 nC, q3=20 nC
and q4=-10 nC form a square of edge length
5 cm. What electric field do the particles
produce at the square center?
 Idea: Use the superposition principle.
  


E  E1  E 2  E3  E 4
 Step 1: Choose your coordinate system and
stick with it!
 Step 2: Look at the x and y coordinates
separately.
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Electric Field from 4 Point Charges (2)
q1 = 10 nC, q2 = -20 nC
q3 = 20 nC, q4 = -10 nC
x component (at 0)
E1x  0
and
E2x
q2
 k 2
r
E3x
q3
 k 2
r
E4x  0
E2x is positive!
E3x is positive!
9
40

10
N
9
5 N
E x  9  10
 2.88  10
2
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(0.05) / 2 C
C
7
Electric Field from 4 Point Charges (3)
q1 = 10 nC, q2 = -20 nC
q3 = 20 nC, q4 = -10 nC
y component (at 0)
E 2y  0
and
E 3y  0
E1y
q1
 k 2
r
E1y is negative!
E 4y
q4
 k 2
r
E4y is negative!
9
(

)
20

10
N
9
5 N
E y  9  10
 1.44  10
2
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(0.05) / 2 C
C
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Electric Field from 4 Point Charges (4)
E at the center pt.
N
magnitude E  E  E  3.5  10
C
direction   arctan(Ey /E x )
2
x
2
y
5
 degrees below thex axis
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Electric Field from Three Point Charges
 Consider three charges
q1  1.5  C
q2  2.5  C
q3  3.5  C
 The three charges are
placed at
q1 : (0, a )
q2 : (0,0)
q3 : (b,0)
 What is the electric field
at point P ?
P: (b,a)
a = 8.0 m ; b = 6.0 m
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Electric Field from Three Point Charges (2)
 The electric field at P due
to q1 is

q1
E1  k 2 xˆ
b

q
E1  k 12 xˆ
b
 The electric field at P due

q3
to q3 is
ˆ
E3  k
a
2
y
 The electric field at P due
to q2 is
  kq cos  
 kq 2 sin   ˆ
2
ˆ
E2   2
x  2
y
2 
2 
 a b 
 a b 
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Note: tan = a/b
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Electric Field from Three Point Charges (3)
 Now we add the components, to obtain
E x  509 N/C
E y  311N/C
Ex  509 N/C
Ey  311 N/C
 Magnitude of E
E  Ex2  Ey2  597 N/C
 Direction of E
anglewith x axis  arctan(Ey / E x )  31.5
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Electric Field from an Electric Dipole
 A system of two oppositely charged point particles is called an
electric dipole.
 The vector sum of the electric field from the two charges
gives the electric field of the dipole (superposition principle).
 We have shown the electric field lines from a dipole
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Electric Field from an Electric Dipole (2)
 Expression for the electric field of a dipole along a line
including both charges …
 We will derive a general expression good anywhere along the
dashed line and then get an expression for the electric field
a long distance from the dipole.
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Electric Field from an Electric Dipole (3)
 Two charges on the x-axis a distance d apart
• Put -q at x = -d/2
• Put +q at x = +d/2
 Calculate the electric field at a point P a distance x
from the origin
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Electric Field from an Electric Dipole (4)
 Principle of superposition:The electric field at
any point x is the sum of the electric fields
from +q and -q
q
1 q
E  E  E 

2
4 0 r 4 0 r2
1
 Replacing r+ and r- we get

q 
1
1


E

2
2
4 0  x  1 d  x  1 d  


2
2
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Electric Field from an Electric Dipole (5)
 This equation gives the electric field everywhere on
the x-axis (except for x = d/2)

q 
1
1


E

2
2
4 0  x  1 d  x  1 d  


2
2
 Let’s look at this equation far away along the positive
x-axis (x >> d)
Taylor series
 d
  d

E
1      1     
2 
4 0 x  x
  x

q
1
1   2
 1  2  
q
qd
 2d 


2

4 0 x  x  2 0 x 3
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Definition of Electric Dipole Moment
 We define the vector electric dipole moment
as a vector that points from the negative
charge to the positive charge
• p is the magnitude of the dipole moment
• q is the magnitude of one of the opposite
charges
• d is the distance between the charges
 Using this definition we can write the electric
field far away from an electric dipole as
E
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

p  qd
p
2 0 x
3
18
Functional Dependence E(x)
Point charge
Dipole
E(x)=
E(x)=
E(x)
1/4
1/8
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Electric Dipole Moment of a Water Molecule
 Chemistry reminder - the H2O molecule
The distribution of electric charge in a H2O molecule
is non-uniform. The more electronegative oxygen atom
attracts electrons from the hydrogen atoms. Thus,
the oxygen atom acquires a partial negative charge
and the hydrogen atoms acquire a partial positive
charge. The water molecule is "polarized."
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Example - Electric Dipole Moment of Water
 Suppose we approximate the water molecules as two positive
charges located at the center of the hydrogen atoms and
two negative charges located at the center of the oxygen
atom. What is the electric dipole moment of a water
molecule?
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Electric Dipole Moment of Water (2)
Our result for the electric dipole moment of water is then
d  1010 m  cos 52.5  0.6  1010 m
p  2ed  2  1029 C m
This oversimplified result is comparable to the measured value
of 6.2x10-30 C  m.
(The assumed charge distribution not precise.)
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Demo - Polarization
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Math Reminder (1)
d
a
y
a
x
c
a
b
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Math Reminder (2)
Binomial theorem
:
(Taylor series)
In our case: x=d/2x, n=-2 and x=-d/2x, n=-2
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