Post on 07-Oct-2020
Lecture 4 1. Review superposi.on principle and dipole field pa5ern
+ -E
E-
E+
E+ + E- = EE-
E+
E- + E+
+ -Field Pa5ern of Charge:
Method of Parallelogram:
+ -
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Clicker 4-‐1
Lecture 5 Dipole field pattern and dipole interactions
Lec5-1 Sketch a dipole field pattern
Consider a dipole centered at the origin. A parallelogram of E fieldvectors due to the dipole charges has been constructed at point A.
x
y
~E+
~E�
~Enet
A
B
Please sketch in your notebook the parallelogram formed by the fieldvectors at B. Wait. At the appropriate time discuss the parallelogramconstruction with your neighbour. Complete the parallelogramconstruction and label the three field vectors. Enter your clicker answer.
Choice Have you completed your parallelogram construction at B?1 Yes2 No
Clicker 401: Dipole field pa5ern and dipole interac.on
Field at a point P
xP
-q +q
+s
2�s
2
s
r
EP = E+ + E�
=kq
(r � s2 )
2� kq
(r + s2 )
2
=kq
r2
1
(1� ✏)2� 1
(1 + ✏)2
�
where: ✏ =s
2r
Digression on small epsilon expansion
General Formula:
1
1� ✏= 1 + ✏Where a is nega.ve:
Product:
So the general formula above works for both posi.ve and nega.ve!
(1� ✏)a ⇡ 1� a✏
(1� ✏)a = 1� a✏+O(✏2) ⇡ 1� a✏
(1� ✏)2 = 1� 2✏+ ✏2 = 1� 2✏+O(✏2) ⇡ 1� 2✏
(1� ✏)3 = 1� 3✏+ 3✏3 + ✏3 = 1� 3✏+O(✏2) ⇡ 1� 3✏
(1� ✏)(1 + ✏) ⇡ (1� ✏2) ⇡ 1
Digression on small epsilon expansion con4nued... Examples:
1
(1� ✏)2=
1
1� (2✏)⇡ 1 + 2✏
1
(1 + ✏)2=
1
1 + 2✏⇡ 1� 2✏
At P (from slide 2):
EP =kq
r
2
1
(1� ✏)2� 1
(1 + ✏)2
�
[ ] = (1 + 2✏)� (1� 2✏) = 4✏ = f
⇣s
2x
⌘=
2s
x
✏ =(s/2)
x
p = qsDipole:
xP
-q +q
+s
2�s
2s
r
E =kq
x
2· 2sx
=2kqs
x
3=
2kp
x
3
More on dipoles...
P1
E+
E�
~Edipole
P1 = ~E+P1 +
~E�P1
dir of EP1 : (�x̂)
~E = 2E+cos↵
= 2⇥ kq
r2· s/2
r=
kqs
r3
Dipole in Uniform E:
~Fdipole
= ~F+q
+ ~F�q
= q ~E � q ~E = 0
Uniform dipole in uniform E-‐field con.nued...
F+
F� ↵
⌧ = Fd = Eqd = Eqs sin↵
~⌧ = ⌧ n̂
= (Ep sin↵)n̂
= ~p⇥ ~E Direc.on: CW
Is electric field real? Source creates field: field spreads out with the speed of light.
Time taken to spread out 1 foot:
�t ⇡ 0.3
3⇥ 108= 10�9 sec = ns
e+ + e� ! ��Dipole field r
e+e� system
2kp
r3
At t=0, reac.on occurs à dipole system dissolves At P: Electric field due to the dipole con.nues to be present for t=0 à t1 = r/c