Download Physics for Scientists & Engineers 2

Review -Field Lines from a Point Charge
 The electric field lines
from a point charge
extend out radially
Physics for Scientists &
Engineers 2
 For a positive point
charge, the field lines
point outward
Spring Semester 2005
• Terminate at infinity
Lecture 5
 For a negative charge,
the field lines point
inward
• Originate at infinity
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Physics for Scientists&Engineers 2
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Electric Field from a Point Charge
January 19, 2005
Physics for Scientists&Engineers 2
Electric Field from Three Point Charges
 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
 Consider three charges
 We can consider q0 to be a test charge, and we can define
the electric field from charge q as
 The three charges are placed at
q1 : ( a, 0 )
q2 : ( 0, 0 )
q3 : ( 0,b )
 What is the electric field at point
P?
E=
q1 = 1.50 µC
q2 = 2.50 µC
q3 = !3.50 µC
F
1 q
=
q0 4!" 0 r 2
 The electric field is a vector and to add electric fields we
must add the components separately
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Electric Field from Three Point Charges (2)
Electric Field from Three Point Charges (3)
 Now we add the components
 The electric field at P due to q1 is
!
q !
E1 = k 12 ex
b
Ex = k
 The electric field at P due to q3 is
q
" kq sin ! %
" q sin ! q3 %
Ey = $ 22 2 ' + k 32 = k $ 22
+
# a +b &
# a + b 2 a 2 '&
a
!
q !
E3 = k 32 ey
a
# a&
# 8.00 m &
! = tan "1 % ( = tan "1 %
= 53.1°
$ b'
$ 6.00 m ('
 The electric field at P due to q2 is
a 2 + b 2 = ( 8.00 m ) + ( 6.00 m ) = 100 m 2
2
! " kq cos! % ! " kq sin ! % !
E2 = $ 22 2 ' ex + $ 22 2 ' ey
# a +b &
# a +b &
2
! is given by tan -1 ( a / b )
January 19, 2005
# ( 2.50 µC ) sin ( 53.1° ) "3.50 µC &
Ey = 8.99 !10 9 %
+
= "311 N/C
100 m 2
( 8.00 m )2 ('
$
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 Now we use the components
to get the magnitude
( 509 N/C)
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 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
E = Ex2 + Ey2
E=
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Electric Field from an Electric Dipole
Electric Field from Three Point Charges (4)
2
2
" 1.50 µC ( 2.50 µC ) cos ( 53.1° ) %
Ex = 8.99 !10 9 $
2 +
' = 509 N/C
100 m 2
# ( 6.00 m )
&
a + b is the distance from q2 to point P squared
2
q1 " kq2 cos! %
" q1 q2 cos! %
+$
' = k $# 2 + 2
'
b2 # a2 + b2 &
b
a + b2 &
+ ( !311 N/C ) = 597 N/C
2
 We have shown the electric field lines from a dipole
 Now the direction
# Ey &
# "311 N/C &
! = tan "1 % ( = tan "1 %
$ 509 N/C ('
$ Ex '
! = "31.5°
electric field points to the right and down
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Electric Field from an Electric Dipole (2)
Electric Field from an Electric Dipole (3)
 Let’s get an expression for the electric field of a dipole
along a line including both charges
 Let’s put both charges on the x-axis a distance d
apart
 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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Physics for Scientists&Engineers 2
• 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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Physics for Scientists&Engineers 2
Electric Field from an Electric Dipole (4)
Electric Field from an Electric Dipole (5)
 The principle of superposition tells us that the
electric field at any point x is the sum of the
electric field from +q and -q
 Rearranging we get
E = E+ + E! =
E=
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 Let’s look at this equation far away (x >> d)
1
q
1
q
2 #
2
4!" 0 $
4!" 0 $
1 '
1 '
&% x # d )(
&% x + d )(
2
2
Physics for Scientists&Engineers 2
#2
#2
q $$
d '
d ' '
$
& 1 # )( # &% 1 + )( )
4!" 0 x 2 &% %
2x
2x (
 This equation gives the electric field everywhere on the x axis except for x = ±d
1 q
1 !q
+
4"# 0 r+2 4"# 0 r!2
 Replacing r+ and r- we get
E=
10
E!
q %%
d
d
( %
((
' 1 + $ ...*) $ '& 1 $ + ...*) *)
4"# 0 x 2 '& &
x
x
E!
q $ 2d '
qd
& )=
4"# 0 x 2 % x ( 2"# 0 x 3
or
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Definition of Electric Dipole
Example - Electric Dipole Moment of Water
 We can define the vector electric dipole moment as a
vector that points from the negative charge to the positive
charge
!
!
p = qd
• 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
p
E=
2!" 0 x 3
Physics for Scientists&Engineers 2
 The charge distributions of the
atoms are approximately spherical
and the hydrogen atoms are
arranged such that they make an
angle of 105° with each other
 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?
• p is the magnitude of the dipole moment
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 Everyone knows that water is H2O
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Electric Dipole Moment of Water (2)
 We assume that the center of mass of the two hydrogen
atoms is halfway between the two atoms
and that the two positive charges are
effectively located there
 The distance between the these two positive charges and
the two negative charges assumed at the center of the
oxygen atom is
d = (10 !10 m)cos 52.5° = 0.6 "10 !10 m
 Our result for the electric dipole moment of water is then
p = 2ed = (3.2 !10 "19 C)(0.6 !10 "10 m) = 2 !10 "29 C m
 This oversimplified result is close to the known value of 6.2⋅
10-30 C ⋅ m
• Assumed charge distribution not what actually happens
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