Download 14.5-14.8

Example Problem
A particle with charge +2 nC (1 nanoCoulomb=10-9 C) is located
at the origin. What is the electric field due to this particle at a
location <-0.2,-0.2,-0.2> m?
Solution:
1. Distance and direction:

r
1 𝑞1
𝐸1 =
𝑟
2
4𝜋𝜀0 𝑟
𝑟 = 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑_𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 − 𝑠𝑜𝑢𝑟𝑐𝑒_𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛
𝑟 = −0.2, −0.2, −0.2 − 0, 0, 0
𝑟 =
(−0.2)2 + (−0.2)2 + (−0.2)2 = 0.35m
𝑟
−0.2, −0.2, −0.2
𝑟=
=
= −0.57, −0.57, −0.57
𝑟
0.35
Example Problem
2. The magnitude of the electric field:
1 𝑞1
𝐸1 =
𝑟
2
4𝜋𝜀0 𝑟
2 2 × 10−9 C
Nm
1 𝑞
N
9
= 9 × 10
𝐸 =
= 147
2
2
2
C
0.35 m
4𝜋𝜀0 𝑟
C
𝑦
3. The electric field in vector form:
N
−0.57, −0.57, −0.57
𝐸 = 𝐸 𝑟 = 147
C
𝑥
N
𝐸 = −84, −84, −84
C
𝑧
𝐸
Clicker Question 1
A penny carrying a small amount of positive charge Qp
exerts an electric force F on a nickel carrying a large
amount of positive charge Qn that is a distance d away (Qn
> Qp ). Which one of the following is not true?
A. The electric force exerted on the penny by the nickel is
also equal to F.
B. The number of electrons in the penny is less than the
number of protons in the penny.
C. 𝑭 ≈
𝟏 𝑸𝒑 𝑸𝒏
,
𝟒𝝅𝜺𝟎 𝒅𝟐
if d is small compared to the size of the
coins.
D. 𝑭 ≈
𝟏 𝑸𝒑 𝑸𝒏
,
𝟒𝝅𝜺𝟎 𝒅𝟐
coins.
if d is large compared to the size of the
Clicker Question 2
A positive and a negative charge are separated by a
distance r,
r
+q1
What are the directions of the
forces on the charges?
Choice 𝑭 𝒐𝒏 𝒒𝟏 𝑭 𝒐𝒏 𝒒𝟐
A
Left
Left
B
Right
Left
C
Left
Right
D
Right
Right
-q2
What is the magnitude of
the self-force?
Choice
A
B
C
𝑭 𝒐𝒏 𝒒𝟐
𝒅𝒖𝒆 𝒕𝒐 𝒒𝟐
infinite
k𝒒𝟏 𝒒𝟐 /𝒓𝟐
0
The Coulomb Force
1 Q1Q2
F
r̂
2
4 0 r

1 Q1Q2
F F
2
40 r
0 = permittivity constant
r
+
+
2
1
Force repulsive
F21
Force on “2” by “1”
+
r
-
F21
2
1
Force attractive
• The force exerted by one point charge on another acts along line
joining the charges.
• The force is repulsive if the charges have the same sign and
attractive if the charges have opposite signs.
How Strong is the Coulomb Force
Electric Field
 
E  F /q
 
E  E  x, y , z , t 
Electric field has units of Newton per Coulomb:
[N/C]
No ‘self-force’!
q
E
2
40 r
1
r  0, E  
Point charge does not exert field on itself!
The Superposition Principle
+q2
𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 3
-q1
+q3
𝐸1
𝐸 𝑛𝑒𝑡
𝐸2
The net electric field at a location in space is a vector sum
of the individual electric fields contributed by all charged
particles located elsewhere.
The electric field contributed by a charged particle is
unaffected by the presence of other charged particles.
The Superposition Principle
+q2
+q3
-q1
𝐸1
𝑃
𝐸2
𝐸 𝑛𝑒𝑡 𝐸 3
The E of a Uniformly Charged Sphere
Can calculate using principle of superposition:

1 Q
Esphere 
rˆ
2
40 r

Esphere  0
Recall this every
night before bed!
for r>R (outside)
for r<R (inside)
The Superposition Principle
The electric field of a dipole:
Electric dipole:
Two equally but oppositely charged
point-like objects
s
-q
+q
Example of electric dipole: HCl molecule
What is the E field far from the dipole (r>s)?
Calculating Electric Field
Choice of the origin
y
s
-q
+q
x
z
Choice of origin: use symmetry
s
1. E along the x-axis
𝐸−,𝑥
-q
+q
E1, x  E , x  E , x 
E1, x 
E1, x 
𝐸1,𝑥
𝐸+,𝑥
r
1
q
4 0 r  s 2 
2

1
q
4 0 r  s 2 
2
1 qr 2  qrs  qs 2 / 4  qr 2  qrs  qs 2 / 4
4 0
1
r  s 2  r  s 2
2
2qrs
4 0 r  s 2 2 r  s 2 2
2
Approximation: Far from the Dipole
E1, x 
1
2 srq
2
2
40 
s 
s
r

r


 

2 
2

2
2
s 
s

if r>>s, then  r     r    r 2
2 
2

E1, x
1 2 sq

40 r 3

E1 
1 2sq
,0,0
3
40 r
While the electric field of a point charge is proportional to 1/r2,
the electric field created by several charges may have a different
distance dependence.
2. E along the y-axis
𝐸+

E
𝐸2
1
q
rˆ
2
40 r

s
s
r  0, y ,0  ,0,0   , y ,0
2
2

s
s
r  0, y ,0   ,0,0  , y ,0
2
2
𝐸−
y
𝑟+
s
y2   
2
-q
s
+q
2

E 

E 
1
q
40
s
y2   
2
1
q
40
s
y2   
2


 r  r 





s
 , y ,0
2
2
s
y2   
2
2
s
, y ,0
2
2
s
y2   
2
2
s
y2   
2
2
2. E along the y-axis
𝐸+

E 
𝐸2
𝐸−
y
𝑟+
1
q
40
s
y2   
2
s
 , y ,0
2
2
s
y2   
2
2



1
E2  E   E  
40

E 
1
q
40
s
y2   
2
qs
q
33
2 2
 2 s
y   
2

s
+q
2
s
y2   
2
 1s,0,0



1 qs
,0,0
if r>>s, then E2 
3
4 0 r
-q
s
 , y ,0
2
at <0,r,0>
2
3. E along the z-axis
y

E1 

E2 
 1 qs
,0,0
3
40 r
at <0, r, 0>
or <0, 0, r>
z
1 2sq
,0,0
3
40 r
at <r, 0, 0>
x
Due to the symmetry E along the z-axis must
be the same as E along the y-axis!
Other Locations
The Electric Field
Point Charge:

E1 
1
q1
rˆ
2
40 r
Dipole:
for r>>s :
y
+q
z
1 2qs
E
,0,0
3
4 0 r
x
-
at <r,0,0>
1 qs
,0,0
3
4 0 r
at <0,r,0>
1 qs
E
,0,0
3
4 0 r
at <0,0,r>
E
s
-q
+
Example Problem
y
A dipole is located at the origin, and is composed of
E=? particles with charges e and –e, separated by a distance
210-10 m along the x-axis. Calculate the magnitude of
the E field at <0, 210-8, 0> m.
1 sq
Since r>>s: E1, x 
200Å
40 r 3
2 
10
19 


Nm
2

10
m

1.6

10
C
9
E1, x   9  10


3
2 
C  


2  10 8 m
4 N
x
E1, x  7.2  10
C
2Å
Using exact solution:
4 N
E1, x  7.1999973  10
C


Interaction of a Point Charge and a Dipole
+q

Edipole +Q
-q
s
F
𝑑≫𝑠


F  QEdipole  Q

F
 1 2qs
,0,0
3
40 d
• Direction makes sense?
- negative end of dipole is closer, so its net contribution is larger
• What is the force exerted on the dipole by the point charge?
- Newton’s third law: equal but opposite sign
Dipole Moment
x:
y, z:

E1 

E2 
1 2qs
p
,0
,0,0
,0
33
40 rr
r>>s
 1 qs
p
,,00,,00
33
40 r
The electric field of a dipole is proportional to the
Dipole moment: p = qs

p  qs, direction from –q to +q
Dipole moment is a vector pointing
from negative to positive charge
Dipole in a Uniform Field


F  qE
Forces on +q and –q have the same
magnitude but opposite direction



Fnet  qE  qE  0
It would experience a torque about its
center of mass.
What is the equilibrium position?
Electric dipole can be used to measure
the direction of electric field.
Choice of System
Multiparticle systems: Split into objects to include into system
and objects to be considered as external.
To use field concept instead of Coulomb’s law we split the
Universe into two parts:
• the charges that are the sources of the field
• the charge that is affected by that field
A Fundamental Rationale
• Convenience: know E at some
r location – know the electric
force on any charge: F  qE
• Can describe the electric properties of matter in terms of
electric field – independent of how this field was produced.
Example: if E>3106 N/C air becomes conductor
• Retardation
Nothing can move faster than light c
c = 300,000 km/s = 30 cm/ns
Coulomb’s law is not completely correct – it does not
contain time t nor speed of light c.

F
1 q1q2
rˆ
2
40 r

E
1
q
rˆ
2
40 r
v<<c !!!