Transcript + E - Purdue Physics

Question 1 (Chap. 14)
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=KQpQn/d2 , if d is small compared to the size of the
coins.
D. F=KQpQn/d2 , if d is large compared to the size of the
coins.
Question 2 (Chap. 14)
Which one of the following is not true?
The electric force exerted by an electron on an electron:
A. decreases by a factor of 25 if the distance is
increased by a factor of 5.
B. has the same magnitude as the electric force exerted
by a proton on a proton at the same distance.
C. has the same direction as the electric force exerted by
a proton on a proton at the same distance.
D. is much weaker than the gravitational force between
them.
The Superposition Principle
The E of a Uniformly Charged Sphere
Can calculate using principle of superposition:

Esphere 
1
Q
rˆ
2
40 r

Esphere  0
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 of the dipole?
Calculating Electric Field
Choice of the origin
y
s
-q
+q
x
z
Choice of origin: use symmetry
1. E along the x-axis
E1, x  E , x  E , x 
E1, x 
E1, x 
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 
r = 1,0,0
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
1
q
rˆ
2
40 r


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
s
y2   
2
2

E 

E 
1
q
40
s
y  
2
s
 , y ,0
2
2
2
1
q
40
s
y2   
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 
1
q
40
s
y  
2
2
s
 , y ,0
2
2
s
y2   
2
2



1
E2  E   E  
40
s
y2   
2
2

E 
1
q
40
s
y  
2
2
qs
q
33
2 2
 2 s
y   
2

s
 , y ,0
2
2
s
y  
2
2
2
 1s,0,0



1 qs
,0,0
if r>>s, then E2 
3
4 0 r
at <0,r,0>
3. E along the z-axis

E1 

E2 
 1 qs
,0,0
3
40 r
1 2sq
,0,0
3
40 r
at <0,r,0>
or <0,0,r>
Due to the symmetry E along the z-axis must
be the same as E along the y-axis!
at <r,0,0>
Other Locations
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.
sq
Since r>>s:
40 r 3
2 
10
19 


Nm
2

10
m

1.6

10
C
9
E1, x   9  10


3
2 
8
C  


2  10 m
N
E1, x  7.2  104
C
E1, x 
200Å
1

x
2Å
Using exact solution:
4 N
E1, x  7.1999973  10
C

Interaction of a Point Charge and a Dipole

Edipole

F

F


F  QEdipole  Q
 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.
Neutral Matter in the Electric Field
A wooden dowel rod is balanced on a sharp
needle and placed between a pair of parallel
plates connected to an electrostatic
generator.
When the plates are charged, the dowel rod:
A) Could not care less
B) Will orient perpendicular to the direction of the E – field
C) Will orient parallel to the direction of the E – field
D) Will jump out of the area with E-field
Polarization of Atoms
E
+
-
+
Force due to E created by positive charge shifts electron
cloud and nucleus in opposite directions: electric dipole.
An atom is said to be polarized when its electron cloud has
been shifted by the influence of an external charge so that
the electron cloud is not centered on the nucleus.
Induced Dipole
An applied electric field creates induced dipoles!
E
• it is not a permanent dipole
• an induced dipole is created when a neutral object is polarized
by an applied 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 location – know the electric
force on any charge.
• 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 !!!