#### Transcript Brief review: Force and Electric Field for point charges

```Phys1101 (P10D)
Electricity and Magnetism
1
Review
2
Review: Class Objectives



Briefly reintroduce the basic
electrostatic concepts for points
charges: electric force, field strength.
Give examples of the electric force and
field strength for simple configurations.
Give the example of the mechanics of
calculating the electric field for the
special case of the electric dipole.
3
Review: Student Objectives


Understand how to set up problems for
basic electrostatic concepts for points
charges: electric force, field strength.
You should be able to calculate the
electric force and field strength for
simple configurations.
4
Review: Student Objectives


You should be able to draw free body
diagrams for all problems.
You should be able to apply the
problem solving concepts to any given
problem (Understand that the same
problem solving technique is used for all
problems!).
5
Electric Charge
Discrete charge distributions
6
Introduction



Previous Knowledge:
Electric Force for point charges
Electric fields for point charges
7
Introduction




Previous Knowledge:
Electric Force for point charges
Electric fields for point charges
However we want to extend these
techniques for continuous distributions.
Eg. Discs and lines of charge.
8
Electric Charge




Review:
There are two types of charge –
positive and negative.
All objects have a charge!
If there is an equal amount of each
type, the object is neutral.
9
Electric Charge




Charge with the same sign repel.
Opposite sign charge attract.
Charge is quantised. That is, it comes in
discrete quantities.
The fundamental charge e = 1.6 1019 C
All other quantities are multiples of
q  0  e  2e...
10
Electric Charge

Charge is conserved.
11
Electric Charge
12
Electric Charge
Coulomb’s Law
13
Coulomb’s Law


The interaction between charged particles is
described by Coulomb’s Law.
For two charged particles with charge q1 and
q2 with separation r, the electrostatic force of
attraction or repulsion is directly proportional
to the product of their charges and inversely
proportional to the square of their separation.
1
k q1 q2
9
2
2
where
,
k


9

10
N

m
/
C
F
4 0
r2
14
Coulomb’s Law
 0 : permittivity of free space  8.851012 C 2 N 1m2
15
Coulomb’s Law

NB: Coulomb’s Law is of similar form to
the gravitational Law,
Gm1m2
F
r2

An indication of it’s validity.
16
Coulomb’s Law


The electrostatic force obeys the law of
superposition.
So that, the force on a charge due to a
group of charges is the vector sum due
to each charge separately.
17
Coulomb’s Law


The electrostatic force obeys the law of
superposition.
So that, the force on a charge due to a
group of charges is the vector sum due
to each charge separately.
 


q1 +
- q2
F3  F13  F23  F43
q3 +
+ q4
18
Coulomb’s Law
ia

So that in general, Fa   Fia
i
19
Coulomb’s Law
ia

So that in general, Fa   Fia
i


Recommended website:
http://hyperphysics.phy-astr.gsu.edu/
hbase/electric/elecforce.html
20
Coulomb’s Law
Solving problems with Coulomb’s
Law
21
Coulomb’s Law

1.
2.
Problem Solving Technique.
Draw a free body diagram.
Indicate only the forces acting on the
particular particle and their direction.
22
Coulomb’s Law

Example:
q1 +
+ q2
r

The figure shows two positive charges fixed in
place along the x-axis. q1 1.6 1019 C and
q2  3.2 1019 C. Their separation is 0.02 m . What
is the magnitude and direction of the electrostatic
force on particle1 from particle2?
23
Coulomb’s Law
q1 +
+ q2
r

Since particles have the same charge they
will repel.
F12
+
q1
24
Coulomb’s Law
q1 +
+ q2
r

Since particles have the same charge they
will repel.
F12
F12 
+
q1
k q1 q2
r2
25
Coulomb’s Law
q1 +
+ q2
r

Since particles have the same charge they
will repel.
+
F12
F12 
q1
k q1 q2
r2
9 109 1.6 1019 C  3.2 1019 C

0.02m2

F12  1.151024 N to theleft
 1.151024 N iˆ


26
Coulomb’s Law

In unit vector notation:

F12 
1
q1q2
rˆ12
2
4 0 r12
27
The Electric Field
28
The Electric Field


An electric field exists at a point if an
electric force F, is exerted on a test
charge q0 placed at that point.

The electric field E can be defined as:

 F
E
q0

Unit N C
The electric force per unit charge.
29
The Electric Field


The direction of E is the direction of the
force acting on the positive test charge.
E
F
+
30
The Electric Field


The direction of E is the direction of the
force acting on the positive test charge.
E
F
+

NB: the field is produced independent
of the test charge.
31
The Electric Field

The electric field due to a point charge
is given by
F
1 q
E

q0 4 0 r 2
or E 
q
4 0 r
2
rˆ
32
The Electric Field

The principle of superposition also holds
for electric fields. Therefore
  

E  E1  E2  ... En
33
The Electric Field

q1
Example: Three particles with charge
q1=+2Q, q2 =-2Q and q3 =-4Q each a
distances d from the origin. What is
the net electric field at the origin?
+
d
d
d
-
q3
- q2
34
The Electric Field

Solution: E1  1 2Q2
4 0 d
1
2Q
E2 
4 0 d 2
q1
+
d
30 
d
1
4Q
E3 
4 0 d 2
-
E3
q3
30 
30 
d
E2
-
q2
E1
35
The Electric Field

Solution: E1  1 2Q2
4 0 d
1
2Q
E2 
4 0 d 2
1
4Q
E3 
4 0 d 2
E3
E3
30 
30 
E1 + E2
30 
30 
E1 E2
36
The Electric Field

Solution: E1  E2  1 2Q2  1 2Q2
4 0 d
4 0 d
1 4Q

4 0 d 2
E3
E3
30 
30 
E1 + E2
30 
30 
E1 E2
37
The Electric Field



We decompose the vectors into
components.
In this case the components in the ydirection cancel.
The only components are in x-direction
(both in the positive x-direction).
E3x
E1x+ E2x
38
The Electric Field

1
4
Q
1
4
Q
Solution: E 
cos30 
cos30
2
2
4 0 d
4 0 d
1 6.93Q

4 0 d 2
E3x
E1x+ E2x
39
Electric Dipole

The configuration of two oppositely
charged particles with magnitude q and
separation d is called a dipole.
+ +q
d
- -q
40
Electric Dipole: Interlude


The same technique used for the
example on slide 34 is used to calculate
the electric field of the dipole.
The net electric field is the vector sum
of the electric field due to each charge.
41
Electric Dipole

r(+)
r(-)
P

The configuration of two oppositely
charged particles with magnitude q and
separation d is called a dipole.
Let’s calculate the electric field at a
point p.
+ +q
d
- -q
42
Electric Dipole

The particles are chosen along the z
axis.
P
r(+)
z
r(-)
+ +q
d
- -q
43
Electric Dipole

The particles are chosen along the z
axis. Because of symmetry the electric
field for each charge is along the z axis.
E(+)
r(+)
P
E(-)
z
r(-)
+ +q
d
- -q
44
Electric Dipole

The resultant electric field E is:
E  E( )  E( )
E(+)
r(+)
P
E(-)
z
r(-)
+ +q
d
- -q
45
Electric Dipole

The resultant electric field E is:
E  E(  )  E(  )
E(+)
r(+)
1
q
1 q


2
2
4 0 r(  ) 4 0 r(  )
P
E(-)
z
r(-)
+ +q
d
- -q
46
Electric Dipole

1
r

z

Note: (  )
2 d
r(  )  z  12 d
E(+)
r(+)
P
E(-)
z
r(-)
+ +q
d
- -q
47
Electric Dipole

The resultant electric field E is:
E  E(  )  E(  )
E(+)
r(+)
P
E(-)
1
q
1 q


2
4 0 r(  ) 4 0 r(2 )
z
r(-)
+ +q
q
q


2
2
1
1
4 0 ( z  2 d ) 4 0 ( z  2 d )
d
- -q
48
Electric Dipole

Taking out the common term (z) we
get:
2
2

q
d 
d  


E
1
  1 
 
2 
4 0 z 
2z 
2 z  

E(+)
r(+)
P
E(-)
z
r(-)
+ +q
d
- -q
49
Electric Dipole

Taking out the common term (z) we
get:
2
2

q
d 
d  


E
1
  1 
 
2 
4 0 z 
2z 
2 z  

E(+)
r(+)
P
E(-)
z
r(-)
+ +q
Considering distances z  d
We can use a binomial expansion.
d
- -q
50
Electric Dipole

The binomial expansion is:
a  x 
n

 a  na
n
n 1
nn  1 n  2 2
x
a x  ...  x n
2
http://hyperphysics.phyastr.gsu.edu/hbase/alg3.html
51
Electric Dipole

So that:
 d
  d

E
1   ...   1   ... 
2 
z
z
4 0 z 
 

q
52
Electric Dipole

So that:
 d
  d

E
1   ...   1   ... 
2 
z
z
4 0 z 
 

q

Simplifying we get:
E
E
q
2d
4 0 z 2 z
2qd
4 0 z 3
E 
qd
2 0 z 3
53
Electric Dipole

So that:
 d
  d

E
1   ...   1   ... 
2 
z
z
4 0 z 
 

q

Simplifying we get:
E
E
q
2d
4 0 z 2 z
2qd
4 0 z 3
E 
Dipole Moment
qd
2 0 z 3
54
Electric Dipole

E(+)
r(+)
The product qd is called the dipole
moment p. The direction of p is from
the -ve charge to the +ve charge.
P
E(-)
z
r(-)
+
+ +q
d

p
- -q
55
Continuous Charge
56
Class Objectives
57
Class Objectives
58
Continuous Charge
Distributions

When we look at charge distributions
consisting of many closely spaced point
charges, these distributions are
considered continuous rather than
discrete.
59
Continuous Charge
Distributions


When we look at charge distributions
consisting of many closely spaced point
charges, these distributions are
considered continuous rather than
discrete.
This charge may be distributed over a
line, surface or volume.
60
Continuous Charge
Distributions


When we look at charge distributions
consisting of many closely spaced point
charges, these distributions are
considered continuous rather than
discrete.
This charge may be distributed over a
line, surface or volume. Ie. A rod (line),
a disc (surface or area) and a sphere
(volume)
61
Continuous Charge
Distributions


When dealing with continuous charge
distributions we express the charge as a
charge density instead of the total
charge.
That is:
62
Continuous Charge Distributions
Charge is distributed over the surface of
a rod. The charge density is the total
charge divided by the total length.
Charge is distributed over the surface of
a disc (top or bottom). The charge density
is the total charge divided by the total
surface area.
Charge is distributed in the volume of
the sphere. The charge density is the
total charge divided by the total volume.
63
Continuous Charge Distributions

The following table shows the common
charge densities:
64
Continuous Charge Distributions

Table.
Name
Charge
Linear charge
density
Surface charge
density
Volume Charge
density
Symbol
q
SI Unit
C

C/m

C/m2

C/m3
dq

dx
dq

dA
dq

dV
65
Electric field for a charge
distribution

We consider the distribution to be made
up of differential charges dq.
dx
dq dq dq dq

dq dq
To find E at a point we sum the electric
field due to each dq (at that point).
66
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