Transcript Electric Fields

Electric field
Chapter 22
Week-2
Electric Fields
(22-1)
In this chapter we will introduce the concept of an electric field.
As
long as charges are stationary Coulomb’s law described adequately the
forces among charges. If the charges are not stationary we must use an
alternative approach by introducing the electric field (symbolE ). In
connection with the electric field, the following topics will be covered:
-Calculate the electric field generated by a point charge.
-Using the principle of superposition determine the electric field created
by a collection of point charges as well as continuous charge
distributions.
-Once the electric field at a point P is known we will be calculate the
electric force on any charge placed at P
-Define the notion of an “electric dipole”. Determine the net force , the
net torque, exerted on an electric dipole by a uniform electric field, as
well as the dipole potential energy.
Fields
 Imagine an object is placed at a particular point in





space.
When placed there, the object experiences a force F.
We may not know WHY there is a force on the object,
although we usually will.
Suppose further that if we double some property of
the object (mass, charge, …) then the force is found
to double as well.
Then the object is said to be in a force field.
The strength of the field (field strength) is defined as
the ratio of the force to the property that we are
dealing with.
Electric Field
 If a charge Q is in an electric field E then it
will experience a force F.
 The Electric Field is defined as the force per
unit charge at the point.
 Electric fields are caused by charges and
consequently we can use Coulombs law to
calculate it.
 For multiple charges, add the fields as
VECTORS.
Two Charges
F  1  qq0
q
E     k 2 runit  k 2 runit
q0  q0  r
r
Doing it
Q
A Charge
r
F
q
The spot where we want
to know the Electric Field
qQ
F k
runit
2
r
F
Q
E
 k 2 runit
q
r
GeneralqQ
F  k 2 runit
r
Q
F
 k 2 runit
E
r
q
General
E  E j  
Fj
q
 k
Qj
r
2
j
r j ,unit
Force  Field
E
qoP
r
q
Electric field generated by a point charge
Consider the positive charge q shown in
the figure. At point P a distance r from
q we place the test charge qo . The force
exerted on qo by qo is equal to:
q qo
F
4 o r 2
1
E
F
1 q qo

qo 4 o qo r 2

q
4 o r 2
1
The magnitude of E is a positive number
E
(22-4)
1
q
4 o r 2
In terms of direction, E points radially
outwards as shown in then figure.
If q were a negative charge the magnitude
of E remains the same. The direction
of E points radially inwards instead
O
O
Electric field generated by a group of point charges. Superposition
The net electric electric field E generated by a group of point charges is equal
to the vector sum of the electric field vectors generated by each charge.
In the example shown in the figure E  E1  E2  E3
Here E1 , E2 , and E3 are the electric field vectors
generated by q1 , q1 , and q3 , respectively
Note: E1 , E2 , and E3 must be added as vectors
Ex  E1x  E2 x  E3 x , E y  E1 y  E2 y  E3 y , Ez  E1z  E2 z  E3 z
(22-5)
What is the electric field at the
center of the square array?
Electric Dipole
A system of two equal charges of opposite sign
(22-6)
+q
 q 
placed at a distance d is known as an "electric
dipole". For every electric dipole we associate
d
-q
a vector known as "the electric dipole moment"
(symbol p )defined as follows:
+q/2 The magnitude p  qd
+q/2
The direction of p is along the line that connects
the two charges and points from - q to  q.
Many molecules have a built-in electric dipole
-q
moment. An example is the water molecule (H 2 O)
The bonding between the O atom and the two H
atoms involves the sharing of 10 valence electrons
(8 from O and 1 from each H atom)
Electric field generated by an electric dipole
We will determine the electric field E generated by the
electric dipole shown in the figure using the principle of
superposition. The positive charge generates at P an electric
1 q
field whose magnitude E(  ) 
The negative charge
2
4 o r
creates an electric field with magnitude E(  )
1
q

4 o r2
The net electric field at P is: E  E(  )  E(  )

1 q
q
1 
q
q
E



 2  2
2
2

4 o  r
r  4 o   z  d / 2 
 z  d / 2  
2
2

d 
d  
d

E
1


1

We
assume:
<< 1


 
2 
4 o z  2 z 
2z
 2 z  
q  d   d  
qd
1 p
E
1    1    =

2 
3
4 o z  z   z   2 o z
2 o z 3
q
1  x 
2
(22-7)
= 1 2x
Kinds of continuously distributed
charges
 Line of charge
 l = the charge per unit length.
 dq= lds (ds= differential of length along the line)
 Area
 s = charge per unit area
 dq=sdA
 dA = dxdy (rectangular coordinates)
 dA= 2rdr for elemental ring of charge
 Volume
 r=charge per unit volume
 dq=rdV
 dV=dxdydz or 4r2dr or some other expressions we will
look at later.
Continuous Charge Distribution
Symmetry
Let’s Do it Real Time
Concept – Charge per
unit length l
dq= l ds
The math
ds  rd
Ey  0
Why?
0
dq
E x  (2)  k 2 cos( )
r
0
0
E x  (2)  k
0
0
lrd
r
2
cos( )
2k
2k
Ex 
l  cos( )d  l sin(  0 )
r 0
r
Example : Detemine the electric field E generated at point P
by a uniformly charged ring of radius R and total charge q.
Point P lies on the normal to the ring plane that passes through
the ring center C, at a distance z. Consider the charge element
of length dS and charge dq shown in the firgure. The distance
between the element and point P is r  z 2  R 2
The charge dq generates at P an electric field of magnitude
dE and which points outwards along the line AP.
dq
dE 
The z-component of dE is given by:
2
4 o r
Adq
dq
(22-9)
C
dEz  dE cos  From triangle PACwe have: cos   z / r
 dEz 
Ez 
zdq
zdq

4 o r 3 4  z 2  R 2 3/ 2
o
z
4 o  z  R
2

2 3/ 2
 dq 
Ez   dEz
zq
4 o  z  R
2

2 3/ 2
The Geometry (the disk)
Define surface charge density
s=charge/unit-area
(z2+r2)1/2
dq=sdA
dA=2rdr
dq=s dA = 2srdr
(z2+r2)1/2
dq cos( ) k 2rsdr
z
dE z  k 2 2  2 2
z r
z  r z2  r2


 

R
E z  2ksz 
0
z
rdr
2
r

2 3/ 2

1/ 2
Final Result
(z2+r2)1/2
 s 
z
1 
E z  
2
2
2

z R
 0 
When R  ,
s
Ez 
2 0




3. Electric field lines extend away from positive charges (where they originate)
and towards negative charges (where they terminate)
Example 1 : Electric field lines of a negative point charge - q
E
1
q
4 o r 2
q
-The electric field lines point towards the point charge
-The direction of the lines gives the direction of E
-The density of the lines/unit area increases as the
distance from  q decreases.
Note : In the case of a positive point charge
the electric field lines have the same
form but they point outwards
q
(22-11)
Example 3.
Electric field lines generated by
an electric dipole (a positive
and a negative point charge of
the same size but of opposite
sign)
Example 4.
Electric field lines generated by
two equal positive point charges
(22-13)
Look at the “Field Lines”
F+
Forces and torques exerted on electric dipoles
by a uniform electric field
Consider the electric dipole shown in the figure in
the presence of a uniform (constant magnitude and
direction) electric field E along the x-axis
The electric field exerts a force F  qE on the
F-
x-axis
positive charge and a force F  qE on the
negatice charge. The net force on the dipole
Fnet  qE  qE  0
The net torque generated by F and F about the dipole center is:
d
d
sin   F sin    qEd sin    pE sin 
2
2
In vector form:   p  E
The electric dipole in a uniform electric field does not move
         F
but can rotate about its center F  0
net
  p E
(22-14)
Potential energy of an electric dipole
in a uniform electric field


90
90
U     d     pE sin  d 

U   pE  sin  d    pE cos    p  E
U   pE cos 
90
U  pE
p
At point A (  0) U has a minimum
value U min   pE
B
U

180˚
A
(22-15)
E
It is a position of stable equilibrium
At point B (  180) U has a maximum
value U max   pE
It is a position of unstable equilibrium
p
E
p
i
Fig.a
Work done by an external agent to rotate an electric
E
dipole in a uniform electric field
Consider the electric dipole in Fig.a. It has an electric
dipole moment p and is positioned so that p is at an angle
i with respect to a uniform electric field E
p
f
Fig.b
An external agent rotates the electric dipole and brings
it in its final position shown in Fig.b. In this position
E
p is at an angle f with respect to E
The work W done by the external agent on the dipole
is equal to the difference between the initial and
final potential energy of the dipole
W  U f  U i   pE cos  f    pE cos i 
W  pE  cos i  cos  f
(22-16)

What did we learn in this chapter??
 We introduced the concept of the Electric
FIELD.




We may not know what causes the field. (The
evil Emperor Ming)
If we know where all the charges are we can
CALCULATE E.
E is a VECTOR.
The equation for E is the same as for the force
on a charge from Coulomb’s Law but divided
by the “q of the test charge”.
What else did we learn in this
chapter?
 We introduced continuous distributions of
charge rather than individual discrete
charges.
 Instead of adding the individual charges we
must INTEGRATE the (dq)s.
 There are three kinds of continuously
distributed charges.
Kinds of continuously distributed
charges
 Line of charge
 l = the charge per unit length.
 dq= lds (ds= differential of length along the line)
 Area
 s = charge per unit area
 dq=sdA
 dA = dxdy (rectangular coordinates)
 dA= 2rdr for elemental ring of charge
 Volume
 r=charge per unit volume
 dq=rdV
 dV=dxdydz or 4r2dr or some other expressions we will
look at later.
Summary
qQ
F  k 2 runit
r
F
Q
E
 k 2 runit
q
r
General
Fj
Qj
E  E j  
  k 2 r j ,unit
q
rj
E  k
rdV (r )
r
2
 k
sdA(r )
r
2
 k
(Note: I left off the unit vectors in the last
equation set, but be aware that they should
be there.)
ds (r )
r2
To be remembered …
 If the ELECTRIC FIELD at a point is E, then
 E=F/q (This is the definition!)
 Using some
advanced mathematics we can derive
from this equation, the fact that:
F  qE