Download 1 o = 8.55 x10 12 C2 / Nm2 F = 1 4 0 Q1Q2 r2 ˆr

Lecture 2
Question from Last Class
Simple Answer: Yes!
Chapter 13. Electric Field
Complicated Answer: Depends!
Pauli Exclusion Principle: No two electrons in an atom can be
in the same quantum state (n, l, ml, ms)
*Superposition of Electric Fields
http://www.hydropole.ch/hydropole/hydrogen/about.htm
Coulomb’s Law of Electro-static Force
!
F=
1 Q1Q2
r̂
4!" 0 r 2
r
+
!
F =F=
+
!
F21
2
Force on “2” by “1”
1
Force repulsive
and the permittivity constant is
How Strong is the Coulomb’s Force?
1 Q1Q2
4!" 0 r 2
r
+
Compare Coulomb and gravitational forces for a proton
and electron in a hydrogen atom
-
!
F21 2
1
Force attractive
! o = 8.55x10 "12 C 2 / Nm 2
1
Electric Field
Electric field due to a Point Charge
Observation: The net Coulomb force on a given charge is
always proportional to the strength of that charge.
F1
-
+
q1
F
q0
test charge
For a Point Charge:
!
E1 =
1 q1
rˆ
4!" 0 r 2
F2
q2
! ! !
F = F1 + F2
!
#
q &q
q
F = 0 $$ 21 r̂1 + 22 r̂2 !!
4'( 0 % r1
r2 "
Define the electric field, which is independent of the
test charge, q0, and depends only on position in space:
!
! F
E=
q0
Electric Field
q1 and q2 in F are the sources
of the electric field, q0 is the test charge
The Superposition Principle
With this concept, we can “map” the electric field anywhere
in space produced by any arbitrary:
Charge Distribution
Bunch of Charges
E=
F
1
qi
r̂i
#
4!" 0
ri2
+
+
+
+
-
-
+
+
E=
1
4!" 0
dqi
#r
2
r̂
+
+ + +
+ + + + ++
+
“Net” E at origin
-
These charges or this charge distribution
“source” the electric field throughout space
2
Electric Field due to an Electric Dipole
!
E+
Electric Field due to an Electric Dipole
!
E+
P
P
!
E!
!
E!
r+
r+
x
r-
qd
2!" o r 3
E=
The electric dipole moment (C-m):
x
!
!
p = qs
r-
+
q+
s
- q-
+
Dipole
center
q+
s
- q-
+
Dipole
center
!
p
Direction of p is from
negative to positive
-
Rewrite magnitude of electric field at P:
E=
Electric Field due to an Electric Dipole
!
E+
!
E!
r+
x
E!
q+
s
- q-
!
E=
+
Dipole
center
!
p
1
r3
1 q
rˆ
4!" 0 r 2
!
$+
s
s
r+ = 0, y ,0 %
, ,0,0 = %
, , y ,0 !(
2
2
!(
#* ' r+ = r, =
!
s
s
r%, = 0, y ,0 %
, %
, ,0,0 = , y ,0 !(
2
2
"!)(
In general:
r-
+
E along the y-axis
p
E=
2!" o r 3
P
p
2!" o r 3
&s#
y2 + $ !
%2"
2
!
E+ =
1
4() 0
!
E' =
1
4() 0
-
2
&s#
y2 + $ !
%2"
2
s
, y ,0
2
'q
&s#
y2 + $ !
%2"
2
s
' , y ,0
2
q
&s#
y2 + $ !
%2"
&s#
y2 + $ !
%2"
2
&s#
y2 + $ !
%2"
2
3
E along the y-axis
&s#
y2 + $ !
%2"
E along the z-axis
!
!
!
E2 = E + + E ! =
1
4./ 0
!
!
!
E2 = E + + E ! =
1
4./ 0
2
!
if r>>s, then E2 !
Other Locations
q
3
! s,0,0
' 2 - s *2 $ 2
%y + + ( "
, 2 ) #"
&%
qs
3
! 1,0,0
' 2 - s *2 $ 2
%y + + ( "
, 2 ) #"
&%
"1 qs
, 0, 0
4#$ 0 r 3
at <0,r,0>
Due to the symmetry E along the z-axis must
be the same as E along the y-axis!
Electric field
For a Point Charge:
!
E1 =
1 q1
rˆ
4!" 0 r 2
Dipole:
for r>>s :
y
s
-q
z
+q
x
!
E=
1 2qs
, 0, 0
4!" 0 r 3
at <r,0,0>
!
E=
!1 qs
, 0, 0
4"# 0 r 3
at <0,r,0>
!
E=
!1 qs
, 0, 0
4"# 0 r 3
at <0,0,r>
4
Electric Dipole Moment
Example Problem
y
E=?
A dipole is located at the origin, and is composed of particles with
charges e and –e, separated by a distance 2×10-10 m along the xaxis. Calculate the magnitude of the E field at <0,2×10-8,0> m.
Since r>>s: E1, x =
200Å
1 sq
4!" 0 r 3
s
!"
Dipole moment: p = q ! Ls s is the separation of the
two charges
2Å
The direction of the dipole moment is taken to be from
the negative to the positive end of the dipole.
x
p≈ 6×10-30 C·m∼0.04 e·nm
Electric Dipole in a uniform electric field
Choice of System
Multi-particle systems: Split into objects to include into
system and objects to be considered as external.
•the charges that are the sources of the field
• the charge that is affected by that field
F = qE
5
A Point Charge in a Uniform Field
•
DEMO
An particle of charge q injected in a E field will experience a force given
by F = q E. The resulting acceleration can be found from Newton's
second law.
!
!
F = ma
•
! q !
a= E
m
In a region of uniform E field, the particle will experience a constant
acceleration and the resulting parabolic trajectory.
x
-----------------P
v0
E
+ + ++ + + + + + + ++ +
6