Download Document 12306753

Physics 014
Chapter 22: Electric Fields
The Electric Field
?
Q: How does particle 1 “know” of the presence of particle 2?
That is, since the particles do not touch, how can particle 2 push
on particle 1—how can there be such an action at a distance?
Ans: particle 2 pushes on particle 1 not by touching it as you
would push on a coffee mug by making contact. Instead, particle
2 pushes by means of the electric field it has set up.
Electric Field
The electric field E at any point is defined in terms of the
electrostatic force F that would be exerted on a positive test
charge q0 placed there:
k |q|
| E | 2
R
Since E is the force per unit charge, it is measured in units of
N/C.
Electric Field Lines
• The field lines of a positive charge
radiate outward in all directions
– In three dimensions, the distribution is
spherical
• The field lines of a negative charge
radiate inward in all directions
Example 1. The isolated point charge of q=+15μC is in a
vacuum. A test charge qo=+15μC is 0.20m to the
right. Find the electric field at point p?
Fk
q qo
r2
8.99  10


E
9
N  m2 C 2 0.80  106 C 15  106 C
0.20m
2
F
2.7 N
6


3
.
4

10
NC
-6
qo 0.80 10 C

 2.7N
Electric Field Lines
• Field lines: useful way to Example: a negative point
visualize electric field E charge — note spherical
symmetry
• Field lines start at a
positive charge, end at
negative charge
• E at any point in space is
tangential to field line
• Field lines are closer
where E is stronger
Superposition
• Question: How do we figure out the field
due to several point charges?
• Answer: consider one charge at a time,
calculate the field (a vector!) produced by
each charge, and then add all the vectors!
(“superposition”)
qi
E  ke  2 rˆi
i ri
• Useful to look out for SYMMETRY to
simplify calculations!
Example 2
Total electric field
-2q
+q
• 4 charges are placed at the
corners of a square as shown.
• What is the direction of the
electric field at the center of
the square?
-q
+2q
y
(a) Field is ZERO!
(b) Along +y
(c) Along +x
x
Electric Field of a Dipole
• Electric dipole: two point
charges +q and –q separated
by a distance d
• Common arrangement in
Nature: molecules, antennae,
…
• Define “dipole moment”
vector p: from –q to +q, with
magnitude qd
Electric Field ON axis of dipole
-q
a
+q
P
x
Superposit ion : E  E  E
E 
kq
a

x 
2

2




1
1

E  kq 

2
2

a 
a 
 x    x   
2 
2 

E  
 kq
kq
a

x 
2

2 xa
2 2
 2 a
x  


4


2
Electric Field ON axis of dipole
E  kq
2 xa
2 2
 2 a
x  


4



p = qa
“dipole moment”
-- VECTOR
2kpx
2 2
 2 a
x  


4


-
+
What if x>> a? (i.e. very far away)
2kpx 2kp
E 4  3
x
x

E

p
r
3
E~p/r3 is actually true for ANY point far from a dipole
(not just on axis)
Continuous Charge Distribution
• Thus far, we have only dealt
with discrete, point charges.
• Imagine instead that a charge
Q is smeared out over a:
Q
Q
– LINE
– AREA
– VOLUME
• How to compute the electric
field E??
Q
Q
Electric Field – Continuous Charge
Distribution, equations
• For a single charge
k |q|
| E | 2
R
• For individual charge elements E  k q rˆ
e
2
r
• For continuous charge distribution
qi
dq
E  ke lim  2 rˆi  ke  2 rˆ
qi 0
ri
r
i
Charge Density
• Useful idea: charge density
• Line of charge:
charge per unit length = l
lQ/L
sQ/A
• Sheet of charge:
charge per unit area = s
• Volume of charge:
charge per unit volume = r
rQ/V
Charge Density
• Volume charge density: when a charge is distributed
evenly throughout a volume
ρ=Q/V
with units C/m3
• Surface charge density: when a charge is distributed
evenly over a surface area
σ =Q/A
with units C/m2
• Linear charge density: when a charge is distributed
along a line
λ = Q/ℓ
with units C/m
Computing electric field
of continuous charge distribution
• Approach: divide the continuous
charge distribution into
infinitesimally small elements
• Treat each element as a POINT
charge & compute its electric
field
• Sum (integrate) over all elements
• Always look for symmetry to
simplify life!
Example: Field on Bisector of Charged Rod
• Uniform line of charge +Q
spread over length L
• What is the direction of the
electric field at a point P on
the perpendicular bisector?
(a) Field is 0.
(b) Along +y
(c) Along +x
• Choose symmetrically
located elements of length dx
• x components of E cancel
P
y
a
x
dx
q
o
L
dx
Example 3: Find the electric field due to a
Line of Charge at point P?
• Uniform line of charge,
length L, total charge Q
• Compute explicitly the
magnitude of E at point P
on perpendicular bisector
• Showed earlier that the net
field at P is in the y
direction -- let’s now
compute this!
P
y
a
x
Q
o
L
- - - electric field due to a Line of Charge at P?
Distance
dE
d  a2  x2
Charge per unit length
P
k (dq)
dE  2
d
a
d
dx
Q
x o
L
q
l
L
k (l dx)a
dE y  dE cos  2
(a  x 2 )3 / 2
a
cos  2
2 1/ 2
(a  x )
- - - electric field due to a Line of Charge at P?
L/2
L/2
dx


x
E y  kl a 
2
2 3 / 2  kl a  2
2
2
(
a

x
)
 a x  a  L / 2
L / 2

2klL
a 4a  L
2
2
What is E very far away from the line (L<<a)?
What is E if the line is infinitely long (L >> a)?
2klL
2kl
Ey 

2
a
a L
Problem-Solving Strategy
• Conceptualize
– Establish a mental representation of the
problem
– Image the electric field produced by the charges
or charge distribution
• Categorize
– Individual charge?
– Group of individual charges?
– Continuous distribution of charges?
Problem-Solving Strategy, cont
• Analyze
– Units: when using the Coulomb constant, ke, the charges
must be in C and the distances in m
– Analyzing a group of individual charges:
• Use the superposition principle, find the fields due
to the individual charges at the point of interest and
then add them as vectors to find the resultant field
• Be careful with the manipulation of vector quantities
– Analyzing a continuous charge distribution:
• The vector sums for evaluating the total electric field at
some point must be replaced with vector integrals
• Divide the charge distribution into infinitesimal pieces,
calculate the vector sum by integrating over the entire
charge distribution
Problem Solving Hints, final
• Analyze, cont.
– Symmetry:
• Take advantage of any symmetry to simplify calculations
• Finalize
– Check to see if the electric field expression is consistent with
your mental representation
– Check to see if the solution reflects any symmetry present
– Image varying parameters to see if the mathematical result
changes in a reasonable way
Example
– Find
theDue
electric
fieldofof
a uniformly
22-4 The 4
Electric
Field
to a Line
Charge
Charged Ring at point P?
• A differential element of charge
occupies a length ds.
•This element sets up an electric field
dE at point P.
• The components perpendicular to the
z axis cancel.
• The parallel components add.
© 2014 John Wiley & Sons, Inc.
All rights reserved.
The Electric
Due to a charged
Line of Charge
- - -22-4
electric
field ofField
a uniformly
ring at P?
• Adding Components & replace
cosθ .
• Integrating from a starting point
(call it s=0) through the full
circumference (s=2πR).
Example
– FindField
the Due
electric
field of Disk
a uniformly
22-5 The5Electric
to a Charged
Charged Disc at point P?
• We superimpose a ring on the disk as shown
• The ring has radius r, radial width dr, and a charge
element dq.
• It sets up a differential electric field dE at point P
on its central axis.
• Then, we can sum all the dE contributions with
• We find
© 2014 John Wiley & Sons, Inc.
All rights reserved.
A
Point
Charge
ininan
Field
22-6
A Point
Charge
anElectric
Electric Field
If a particle with charge q is placed in an external electric field E,
an electrostatic force F acts on the particle:
Ink-jet printer. Drops shot from
generator G receive a charge in
charging unit C. An input signal
from a computer controls the
charge and thus the effect of field
E on where the drop lands on the
paper.
© 2014 John Wiley & Sons, Inc.
All rights reserved.
A Dipole
in an
Field
22-7
A Dipole
in anElectric
Electric Field
The torque on an electric dipole of dipole
moment p when placed in an external
electric field E is given by a cross product:
A potential energy U is associated with the
orientation of the dipole moment in the field,
as given by a dot product:
(a) An electric dipole in a uniform external electric field E. Two centers of equal but opposite charge
are separated by distance d. The line between them represents their rigid connection.
(b) Field E: causes a torque τ on the dipole. The direction of τ is into the page, as represented by the
symbol (x-in a circle) .
© 2014 John Wiley & Sons, Inc.
All rights reserved.
Summary
22 Summary
Definition of Electric Field
• The electric field at any
point
Eq. 22-1
Electric Field Lines
• provide a means for
visualizing the directions and
the magnitudes of electric
fields
Field due to a Point Charge
• The magnitude of the electric
field E set up by a point
charge q at a distance r from
the charge is
Field due to an Electric Dipole
• The magnitude of the
electric field set up by the
dipole at a distant point on
the dipole axis is
Eq. 22-9
Field due to a Charged Disk
• The electric field
magnitude at a point on
the central axis through a
uniformly charged disk is
given by
Eq. 22-3
© 2014 John Wiley & Sons, Inc.
All rights reserved.
Eq. 22-26
22
Summary
Summary
Force on a Point Charge in an
Electric Field
• When a point charge q is
placed in an external electric
field E
Dipole in an Electric Field
• The electric field exerts a
torque on a dipole
Eq. 22-34
Eq. 22-28
• The dipole has a potential
energy U associated with its
orientation in the field
Eq. 22-38
© 2014 John Wiley & Sons, Inc.
All rights reserved.
Adding vectors: Review
Graphical method (triangle method):
• Draw vector A.
• Draw vector B starting at the tip
of vector A.
• The resultant vector R = A + B is
drawn from the tail of A to the tip
of B.
Components of a vector
The x- and y-components of a
vector:
Ax  A cos 
Ay  A sin 
The magnitude of a vector:
A  A
Ax  Ay
2
2
The angle  between vector and x-axis:
 Ay 
  tan  
 Ax 
1
Vector addition using
unit vectors:
Only add components!!!!!
We want to calculate: R = A + B
From diagram:
R = (Ax + Bx)i + (Ay + By)j
The components of R:
Rx = Ax + Bx
Ry = Ay + By
Vector addition using
unit vectors:
The magnitude of a R:
R  Rx  Ry  ( Ax  Bx ) 2  ( Ay  By ) 2
2
2
 Ry   Ay  By 

The angle  between vector R and x-axis: tan      
 Rx   Ax  Bx 