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Lecture 3: Chapter 19
Cont
Electric Charges, Forces and
Electric Fields
Agenda






Compare the electric force to
Gravitational force
Superposition of forces
Electric Field
Superposition of Electric Field
Shielding and Charging by
Induction
Electric Flux and Gauss’s Law
What did we study last
lesson?
Electric Charges: positive and neg.
Electron (neg) and proton (pos)
 Charges are conserved
 Polarization
 Insulators & Conductors
 Coulomb’s Law

Recap

Two rods and a cat fur:

Glass Rod (positive charges)

Rubber Rod (negative)
Charging by Conduction
Recap


Coulomb’s Law
(electrostatic force between point
charges)
F  ke
q1 q2
r
2
Recap



7
Polarization is realignment of
charge within individual molecules.
Produces induced charge on the
surface of insulators.
how e.g. rubber or glass can be
used to supply electrons.
Student Discussion
A positively charged object hanging from a string is brought near
a non conducting object (ball). The ball is seen to be attracted to
the object.
1.Explain why it is not possible to determine whether the object is
negatively charged or neutral.
2.What additional experiment is needed to reveal the electrical
charge state of the object?
?
8
+
Cont


Two possibilities:
Attraction between objects of unlike
charges.
-
+
Attraction between a charged object and a
neutral object subject to polarization.
++ -++ -9
+
What additional experiment is needed to
reveal the electrical charge state of the
object?

Two Experiments:

Bring known neutral ball near the object and observe whether
there is an attraction.
?
0
Bring a known negatively charge object near the first one. If there is
an attraction, the object is neutral, and the attraction is achieved
by polarization.
+-+
+
+
+
-+-+
10
-
Question

Name the first action at a distance
force you have encountered in
physics so far.
Electric Force
11
Electrical Forces are Field
Forces

This is the second example of a field force



Gravity was the first
Remember, with a field force, the force is
exerted by one object on another object
even though there is no physical contact
between them
There are some important similarities and
differences between electrical and
gravitational forces
Electrical Force Compared
to Gravitational Force


Both are inverse square laws
The mathematical form of both laws is the
same




Masses replaced by charges
G replaced by ke
Electrical forces can be either attractive or
repulsive
Gravitational forces are always attractive
Example: Student
Participation
Consider a proton (mp=1.67x10-27 kg; qp=+1.60x10-19 C)
and an electron (me=9.11x10-31 kg; qe=-1.60x10-19 C)
separated by 5.29x10-11 m. The particles are attracted to
each other by both the force of gravity and by Coulomb’s
law force. Which of these has the larger magnitude?
A. Gravitational force
B. Coulomb’s law force
14
Which of these has the larger magnitude?
A. Gravitational force
B. Coulomb’s law force
Gravity : F 
Gmp me
Coulomb : F 
r
2
k e q p qe
r
2

6.67 10 11 1.67 10 27  9.1110 31

5.29 10 
11 2
8.99 109 1.60 10 19 1.60 10 19
5.29 10 Pg.
 659.
11 2
 3.6 10  47 N
 8.2 10 8 N
By what factor the electric
We will mostly ignore gravitational
effects
when we
force is greater
than the
gravitational force?
consider electrostatics.
15
Problem solving steps
1. Visualize problem – labeling variables
2. Determine which basic physical principle(s) apply
3. Write down the appropriate equations using the variables
defined in step 1.
4. Check whether you have the correct amount of
information to solve the problem (same number of
knowns and unknowns).
5. Solve the equations.
6. Check whether your answer makes sense (units, order of
magnitude, etc.).
16
Superposition Principle


From observations: one finds that
whenever multiple charges are present,
the net force on a given charge is the
vector sum of all forces exerted by other
charges.
Electric force obeys a superposition
principle.
17
The Superposition
Principle

How to work the problem?


Find the electrical forces between pairs
of charges separately
Then add the vectors

Remember to add the forces as vectors
Superposition Principle
Example: 3 charges in a
line
F12
F13
q1 = 6.00 uC
q2 = 1.50 uC
q3 = -2.00 uC
d1 = 3.00 cm
d2 = 2.00 cm
K = 8.99 x 109 N.m2 /C2
19
Example: Superposition
Principle (Not along a line)
Consider three point charges at the corners of a
triangle, as shown in the next slide. Find the
resultant force on q3 if
q1 = 6.00 x 10-9 C
q2 = -2.00 x 10-9 C
q3 = 5.00 x 10-9 C
20
Superposition Principle
Example
The force exerted
by q1 on q3 is F13
 The force exerted
by q2 on q3 is F23
 The total force
exerted on q3 is
the vector sum of
F13and F23

Consider three point charges at the corners of a triangle, as shown
below. Find the resultant force on q3.
Solution:
F32  ke
q3 q2
F31  ke
q3 q1
r
r
2
2
5.00 10 C  2.00 10 C   5.62 10
9
 8.99 109
Nm
C2
2
9
 4.00m 
2
5.00 10 C  6.00 10 C   1.08 10
9
 8.99 109
Nm2
C2
 5.00m 
9
N
9
2
8
N
Fx  F32  F31 cos 37.0o  3.01109 N
Fy  F31 sin 37.0o  6.50 109 N
F  Fx2  Fy2  7.16 109 N
  65.2o
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

Spherical Charge Distribution
Students need to read and work on
pg 663/664
Electric Field - Discovery


24
Electric forces act through space
even in the absence of physical
contact.
Suggests the notion of electrical
field (first introduced by Michael
Faraday (1791-1867).
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Electric Field


25
An electric field is said to exist in a
region of space surrounding a
charged object.
If another charged object enters a
region where an electrical field is
present, it will be subject to an
electrical force.
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Electric Field



A charged particle,
with charge Q,
produces an electric
field in the region of
space around it
A small test charge,
qo, placed in the
field, will experience
a force.
“The electric field is
the force per charge
at a given location”
Electric Field


Mathematically,
E
F
qo
SI units are N / C
Given:
F  ke
One finds:
q qo
r
2
E  ke
q
r
2
Electric Field

Use this for the magnitude of the field

The electric field is a vector quantity

The direction of the field is defined to
be the direction of the electric force that
would be exerted on a small positive
test charge placed at that point
Direction of Electric Field

The electric field
produced by a
negative charge is
directed toward
the charge

A positive test
charge would be
attracted to the
negative source
charge
Direction of Electric Field,
cont

The electric field
produced by a
positive charge is
directed away
from the charge

A positive test
charge would be
repelled from the
positive source
charge
More About a Test Charge
and The Electric Field

The test charge is required to be a
small charge



It can cause no rearrangement of the
charges on the source charge
The electric field exists whether or not
there is a test charge present
The Superposition Principle can be
applied to the electric field if a group of
charges is present
Electric Fields and
Superposition Principle

The superposition principle holds
when calculating the electric field
due to a group of charges



Find the fields due to the individual
charges
Add them as vectors
Use symmetry whenever possible to
simplify the problem
Electric Field, final
Problem Solving Strategy


Draw a diagram of the charges in
the problem
Identify the charge of interest


You may want to circle it
Units – Convert all units to SI

Need to be consistent with ke
Problem Solving Strategy,
cont

Apply Coulomb’s Law



Sum all the x- and y- components


For each charge, find the force on the
charge of interest
Determine the direction of the force
This gives the x- and y-components of the
resultant force
Find the resultant force by using the
Pythagorean theorem and trig
Problem Solving Strategy,
Electric Fields

Calculate Electric Fields of point
charges



Use the equation to find the electric
field due to the individual charges
The direction is given by the direction
of the force on a positive test charge
The Superposition Principle can be
applied if more than one charge is
present
Examples of Electric Field
a)
b)
c)
E will be constant (a charged plane)
E will decrease with distance as 1
(a point charge)
r2
E in a line charge, decrease with a
distance 1/r
Example:

-
An electron moving horizontally
passes between two horizontal
planes, the upper plane charged
negatively, and the lower positively.
-
- -
-
- - - - - - - - - - - - - - - - vo
-
+
38
+
+ +
+
+ + + + + + + + + + + + + + + + +
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Example:

-
A uniform, upward-directed electric field
exists in this region. This field exerts a
force on the electron. Describe the motion
of the electron in this region.
-
- -
-
- - - - - - - - - - - - - - - - vo
-
+
39
+
+ +
+
+ + + + + + + + + + + + + + + + +
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-
-
- -
-
- - - - - - - - - - - - - - - - vo
-
+
+
+ +
+
+ + + + + + + + + + + + + + + + +
Observations:

Horizontally:




No electric field
No force
No acceleration
Constant horizontal
velocity
Ex  0
Fx  0
ax  0
vx  vo
x  vo t
40
-
-
- -
-
- - - - - - - - - - - - - - - vo
-
+
+
+ +
+
+ + + + + + + + + + + + + + + + +
Observations:
Vertically:




41
Constant electric field
Constant force
Constant acceleration
Vertical velocity
increase linearly with
time.
E y  Eo
Fy  qo Eo
a y  qo Eo / mo
v y  qo Eot / mo
1
y  qo Eot 2 / mo
2
-
-
- -
-
- - - - - - - - - - - - - - - -
+
+
+ +
+
+ + + + + + + + + + + + + + + + +
Conclusions:
The charge will follow a parabolic path downward.
Motion similar to motion under gravitational field only
except the downward acceleration is now larger.
42
Example: Electric Field Due to Two Point
Charges (Superposition of Fields)
Question:
Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC
is on the x axis, 0.300 m from the origin. Find the electric field
at point P, which has coordinates (0,0.400) m.
y
E1
0.400 m
P
q1
43
E
E2
0.300 m
q2
x
Question:
Charge q1=7.00 mC is at the origin, and charge q2=-10.00
mC is on the x axis, 0.300 m from the origin. Find the
electric field at point P, which has coordinates (0,0.400) m.
Observations:





First find the Electric field at point P due to
charge q1 and q2.
Field E1 at P due to q1 is vertically upward.
Field E2 at due to q2 is directed towards q2.
The net field at point P is the vector sum of E1
and E2.
The magnitude is obtained with
E  ke
44
q
r
2
Question:
Charge q1=7.00 mC is at the origin, and charge q2=-10.00
mC is on the x axis, 0.300 m from the origin. Find the
electric field at point P, which has coordinates (0,0.400) m.
Set up the problem:
q1=7.00 mC
q2=-10.00 mC
K = 8.99 x 109 N. m2 /C2
r1 = 0.400m
r2= ? What do we do?
45
Question:
Charge q1=7.00 mC is at the origin, and charge q2=-10.00
mC is on the x axis, 0.300 m from the origin. Find the
electric field at point P, which has coordinates (0,0.400) m.
1)
Calculate r2
2)
Calculate the electric field at P
3)
46
I will work set it up, but you will need to finish (it
looks like example in pg. 669)
Question:
Charge q1=7.00 mC is at the origin, and charge q2=-10.00
mC is on the x axis, 0.300 m from the origin. Find the
electric field at point P, which has coordinates (0,0.400) m.
Solution:
47
Electric Field Lines


A convenient aid for visualizing
electric field patterns is to draw
lines pointing in the direction of
the field vector at any point
These are called electric field
lines and were introduced by
Michael Faraday
Electric Field Lines, cont.

The field lines are related to the
field in the following manners:
 The electric field vector, E , is tangent

to the electric field lines at each point
The number of lines per unit area
through a surface perpendicular to
the lines is proportional to the
strength of the electric field in a given
region
Electric Field Line Patterns



Point charge
The lines radiate
equally in all
directions
For a positive
source charge,
the lines will
radiate outward
Electric Field Line Patterns

For a negative
source charge,
the lines will point
inward
Electric Field Line Patterns


An electric dipole
consists of two
equal and
opposite charges
The high density
of lines between
the charges
indicates the
strong electric
field in this region
Electric Field Line Patterns




Two equal but like point
charges
At a great distance from
the charges, the field
would be approximately
that of a single charge of
2q
The bulging out of the
field lines between the
charges indicates the
repulsion between the
charges
The low field lines
between the charges
indicates a weak field in
this region
Electric Field Patterns


Unequal and
unlike charges
Note that two
lines leave the
+2q charge for
each line that
terminates on -q
Rules for Drawing Electric
Field Lines

The lines for a group of charges must
begin on positive charges and end on
negative charges



In the case of an excess of charge, some
lines will begin or end infinitely far away
The number of lines drawn leaving a
positive charge or ending on a negative
charge is proportional to the magnitude
of the charge
No two field lines can cross each other
Electric Field Lines, final



The electric field lines are not
material objects
They are used only as a pictorial
representation of the electric field
at various locations
They generally do not represent
the path of a charged particle
released in the electric field
Mini-Discussion
Question:
Is it safe to stay inside an
automobile during a lightning storm?
Why?
57
Question:
Is it safe to stay inside an automobile during a lightning storm? Why?
Answer:
Yes. It is. The metal body of the car carries the excess charges on its
external surface. Occupants touching the inner surface are in no
danger.
SAFE
58
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Shielding and Charging by
Induction


When no net motion of charge occurs
within a conductor, the conductor is
said to be in electrostatic equilibrium
An isolated conductor has the following
properties:


The electric field is zero everywhere inside the
conducting material
Any excess charge on an isolated conductor
resides entirely on its surface
Properties, cont


The electric field just outside a
charged conductor is perpendicular to
the conductor’s surface
On an irregularly shaped conductor,
the charge accumulates at locations
where the radius of curvature of the
surface is smallest (that is, at sharp
points)
Property 1

The electric field is zero
everywhere inside the conducting
material

Consider if this were not true


If there were an electric field inside the
conductor, the free charge there would
move and there would be a flow of
charge
If there were a movement of charge, the
conductor would not be in equilibrium
Property 2

Any excess charge on an isolated
conductor resides entirely on its surface


A direct result of the 1/r2 repulsion between
like charges in Coulomb’s Law
If some excess of charge could be placed
inside the conductor, the repulsive forces
would push them as far apart as possible,
causing them to migrate to the surface
Property 3, cont.





Any excess charge moves to its surface
The charges move apart until an equilibrium is achieved
The amount of charge per unit area is greater at the flat
end
The forces from the charges at the sharp end produce a
larger resultant force away from the surface
Why a lightning rod works
Van de Graaff
Generator



An electrostatic generator
designed and built by
Robert J. Van de Graaff in
1929
Charge is transferred to
the dome by means of a
rotating belt
Eventually an
electrostatic discharge
takes place
Electric Flux



Field lines
penetrating an
area A
perpendicular to
the field
The product of EA
is the flux, Φ
In general:

ΦE = E A cos θ
Electric Flux, cont.

ΦE = E A cos θ


The perpendicular to the area A is at
an angle θ to the field
When the area is constructed such
that a closed surface is formed, use
the convention that flux lines passing
into the interior of the volume are
negative and those passing out of the
interior of the volume are positive
Gauss’ Law

Gauss’ Law states that the electric flux
through any closed surface is equal to the
net charge Q inside the surface divided by
εo
E 


Qinside
o
εo is the permittivity of free space and equals
8.85 x 10-12 C2/Nm2
The area in Φ is an imaginary surface, a
Gaussian surface, it does not have to coincide
with the surface of a physical object
Electric Field of a Charged
Thin Spherical Shell

The calculation of the field outside the shell is
identical to that of a point charge
Q
Q
E
 ke 2
2
4r  o
r

The electric field inside the shell is zero
Electric Field of a Nonconducting
Plane Sheet of Charge



Use a cylindrical
Gaussian surface
The flux through the
ends is EA, there is
no field through the
curved part of the
surface
The total charge is Q
= σA

E
2 o

Note, the field is
uniform
Electric Field of a Nonconducting
Plane Sheet of Charge, cont.


The field must be
perpendicular to
the sheet
The field is
directed either
toward or away
from the sheet
Parallel Plate Capacitor


The device consists of
plates of positive and
negative charge
The total electric field
between the plates is
given by

E 
o

The field outside the
plates is zero