Transcript Electric Field

Physics 121: Electricity &
Magnetism – Lecture 3
Electric Field
Dale E. Gary
Wenda Cao
NJIT Physics Department
Electric Force and Field Force
What? -- Action on a
distance
 How? – Electric Field
 Why? – Field Force
 Where? – in the
space surrounding
charges

September 18, 2007
Fields

Scalar Fields:




Temperature – T(r)
Pressure – P(r)
Potential energy – U(r)
Vector Fields:




 
Velocity field – v (r )
 
Gravitational field – g (r )
 
Electric field – E (r )
 
Magnetic field – B (r )
September 18, 2007
Vector Field Due to Gravity

When you consider the
force of Earth’s gravity in
space, it points
everywhere in the
direction of the center of
the Earth. But remember
that the strength is:

Mm
F  G 2 rˆ
r

m
M
This is an example of an
inverse-square force
(proportional to the
inverse square of the
distance).
September 18, 2007
Idea of Test Mass
Notice that the actual
amount of force depends
on the mass, m:

GMm
F   2 rˆ
r
 It is convenient to ask
what is the force per unit
mass. The idea is to
imagine putting a unit test
mass near the Earth, and
observe the effect on it:

F
GM
  2 rˆ   g (r )rˆ
m
r
 g(r) is the “gravitational
field.”

September 18, 2007
Electric Field


Electric field is said to exist in
the region of space around a
charged object: the source
charge.
Concept of test charge:



Small and positive
Does not affect charge
distribution
Electric field:



 F
E
q0
+ +
+ +
+
+
+
+ +
Existence of an electric field is
a property of its source;
Presence of test charge is not
necessary for the field to exist;
September 18, 2007
Electric Field
1.
A test charge of +3 µC is at a point P where an
external electric field is directed to the right and has a
magnitude of 4×106 N/C. If the test charge is
replaced with another test charge of –3 µC, what
happens to the external electric field at P ?
A. It is unaffected.
B. It reverses direction.
C. It changes in a way that cannot be determined.
September 18, 2007
Electric Field

 F
E
q0



Magnitude: E=F/q0
Direction: is that of the force that acts on the
positive test charge
SI unit: N/C
Situation
Value
Inside a copper wire of household circuits
10-2 N/C
Near a charged comb
103 N/C
Inside a TV picture tube
105 N/C
Near the charged drum of a photocopier
105 N/C
Electric breakdown across an air gap
3×106 N/C
At the electron’s orbit in a hydrogen atom
5×1011 N/C
On the suface of a Uranium nucleus
3×1021 N/C
September 18, 2007
2. Which diagram could be considered to show the
correct electric force on a positive test charge due to a
point charge?
B.
A.
C.
D.
E.
September 18, 2007
Electric Field due to a Point Charge Q

F
1 Qq0
rˆ
2
40 r

 F
1 Q
E

rˆ
2
q0 40 r
Direction is radial: outward for +|Q|
inward for -|Q|
 Magnitude: constant on any spherical
shell
 Flux through any shell enclosing Q is
the same: EAAA = EBAB
B

r
Q
A
q0

September 18, 2007
Electric Field due to a group of
individual charge




F0  F01  F02  ...  F0n




 F0 F01 F02
F
E


 ...  0 n
q0 q0 q0
q0
 

 E1  E2  ...  En

1
E
40
qi
i r 2 rˆi
i
September 18, 2007
Example: Electric Field of a Dipole

Start with
E  E   E 



q
1 q

40 r2 40 r2
1
q
1
q

40 ( z  d / 2) 2 40 ( z  d / 2) 2
q
40 z
[(1 
2
d 2
d
)  (1  ) 2 ]
2z
2z
If d << z, then,
[(1 

1
So
d 2
d
2d
2d
2d
)  (1  ) 2 ]  [(1 
 ...)  (1 
 ...)] 
2z
2z
2 z (1!)
2 z (1!)
z
E
q
2d
1 qd

40 z 2 z
20 z 3
p  qd
E
1
p
20 z 3
E ~ 1/z3
 E =>0 as d =>0
 Valid for “far field”

September 18, 2007
Electric Field of a Continuous Charge
Distribution

E 
q
rˆ
2
40 r
1

Find an expression for dq:




1
E
40
qi
i r 2 rˆi
i

q
1
1
dq
E
lim  2 i rˆi 
rˆ
2


q

0
40
40 r
ri
i

dq = λdl for a line distribution
dq = σdA for a surface distribution
dq = ρdV for a volume distribution
Represent field contributions at P
due to point charges dq located in
the distribution. Use symmetry,

dE 

dq
rˆ
2
40 r
Add up (integrate the contributions)
over the whole distribution, varying
the displacement as needed,

E   dE
September 18, 2007
Example: Electric Field Due to a
Charged Rod

A rod of length l has a uniform positive charge per unit length λ and a total
charge Q. Calculate the electric field at a point P that is located along the
long axis of the rod and a distance a from one end.

Start with

dq  dx
1 dq
1 dx
dE 

40 x 2 40 x 2
then,
l a
E

So

a
 dx
 l  a dx



40 x 2 40 a x 2 40
l a
 1
 x 
a

E
1 Q1
1 
Q
 

40 l  a l  a  40 a(l  a)
Finalize


l => 0 ?
a >> l ?
September 18, 2007
Electric Field Lines

The electric field vector is tangent to
the electric field line at each point. The
line has a direction, indicated by an
arrowhead, that is the same as that of
the electric field vector. The direction of
the line is that of the force on a
positive test charge placed in the field.

The number of lines per unit area
through a surface perpendicular to the
lines is proportional to the magnitude
of the electric field in that region. Thus,
the field lines are close together where
the electric field is strong and far apart
where the field is weak.
September 18, 2007
Electric Field Lines

The lines must begin on a positive
charge and terminate on a negative
charge. In the case of an excess of
one type of charge, some lines will
begin or end infinitely far away.

The number of lines drawn leaving a
positive charge or approaching a
negative charge is proportional to
the magnitude of the charge.

No two field lines can cross.
September 18, 2007
Electric Field
.B
3. Rank the magnitudes E of
the electric field at points
A, B, and C shown in the
figure.
A) EC>EB>EA
B) EB>EC>EA
C) EA>EC>EB
D) EB>EA>EC
E) EA>EB>EC
.C
.A
September 18, 2007
Motion of a Charged Particle in a
Uniform Electric Field


F  qE



F  qE  ma

 qE
a
m

If the electric field E is uniform
(magnitude and direction), the electric
force F on the particle is constant.

If the particle has a positive charge, its
acceleration a and electric force F are in
the direction of the electric field E.

If the particle has a negative charge, its
acceleration a and electric force F are in
the direction opposite the electric field E.
September 18, 2007
A Dipole in an Electric Field

Start with
  Fx sin   F (d  x) sin   Fd sin 

Then
F  qE and

So
p  qd
|  | pE sin 
  p E
September 18, 2007
A Dipole in an Electric Field

Start with

Since
dW  d
f
f
f
i
i
i
U f  U i   d 
 pE sin d  pE  sin d

 pE[ cos  ]if  pE (cos  i  cos  f )


Choose
So
U i  0 at  i  90
U   pE cos 
U  pE
September 18, 2007
4. In which configuration, the potential energy of the
dipole is the greatest?
a
c
b
E
d
e
September 18, 2007
Summary

Electric field E at any point is defined in terms of the electric force F that acts on a small positive test

charge placed at that point divided by the magnitude q0 of the test charge: 
F
E
q0

Electric field lines provide a means for visualizing the direction and magnitude of electric fields. The
electric field vector at any point is tangent to a field line through that point. The density of field lines in
any region is proportional to the magnitude of the electric field in that region.
Field lines originate on positive charge and terminate on negative charge.

 F
Field due to a point charge:
1 Q



E
q0

40 r 2
rˆ
The direction is away from the point charge if the charge is positive and toward it if the charge is negative.
Field due to an electric dipole:
1 p
E
20 z 3

Field due to a continuous charge distribution: treat charge elements as point charges and then summing
via inegration, the electric field vectors produced by all the charge elements.


Force on a point charge in an electric field:

Dipole in an electric field:

F  qE

The field exerts a torque on the dipole

The dipole has a potential energy U associated with its orientation in the field
  p E
U  pE
September 18, 2007