Transcript Document

General Physics (PHY 2140)
Lecture II
 Electrostatics
 Coulomb’s law
 Electric field
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 15
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Lightning Review
Last lecture:
1. Properties of electric charge
 two types: positive and negative
 always conserved and quantized
2. Insulators and conductors
 charges move freely in conductors; opposite
is true for insulators
 conductors can be charged by conduction and
induction; insulators can be polarized
Review Problem: Operating-room personnel must wear special conducting
shoes while working around oxygen. Why? What might
happen if personnel wore ordinary rubber shoes
(sneakers)?
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15.3 Coulomb’s Law - Observation
Charles Coulomb discovered in 1785 the fundamental law of
electrical force between two stationary charged particles.
An electric force has the following properties:



Inversely proportional to the square of the separation, r, between the
particles, and is along a line joining them.
Proportional to the product of the magnitudes of the charges |q1| and
|q2| on the two particles.
Attractive if the charges are of opposite sign and repulsive if the charges
have the same sign.
q1
q2
r
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15.3 Coulomb’s Law – Mathematical
Formulation
ke known as the Coulomb constant.
Value of ke depends on the choice of units.
SI units




Force: the Newton (N)
Distance: the meter (m).
Charge: the coulomb ( C).
Current: the ampere (A =1 C/s).
Experimentally measurement: ke = 8.9875109 Nm2/C2.
Reasonable approximate value: ke = 8.99109 Nm2/C2.
How do we know the units of ke?
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Example: Fun with units
Recall that units can be manipulated:
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Example
Charge and Mass of the Electron,
Proton and Neutron.
Particle
Charge ( C)
Mass (kg)
Electron
-1.60 10-19
9.11 10-31
Proton
+1.60 10-19
1.67 10-27
Neutron
0
1.67 10-27
1e = -1.60 10-19 c
Takes 1/e=6.6 1018 protons to create a total charge of 1C
Number of free electrons in 1 cm3 copper ~ 1023
Charge obtained in typical electrostatic experiments with
rubber or glass 10-6 C = 1 mc
A very small fraction of the total available charge
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15.3 Coulomb’s Law – Remarks
The electrostatic force is often called Coulomb force.
It is a force (thus, a vector):


a magnitude
a direction.
Second example of action at a distance.
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Mini-Quiz
Name the first action at a distance force you have
encountered in physics so far.
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Example: Electrical Force
Question:
The electron and proton of a hydrogen atom are separated (on the
average) by a distance of about 5.3x10-11 m. Find the magnitude of the
electric force that each particle exerts on the other.
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Question:
The electron and proton of a hydrogen atom are separated (on the average) by
a distance of about 5.3x10-11 m. Find the magnitude of the electric force that
each particle exerts on the other.
Observations:
We are interested in finding the magnitude of the force between two
particles of known charge, and a given distance of each other.
The magnitude is given by Coulomb’s law.
q1 =-1.60x10-19 C
q2 =1.60x10-19 C
r = 5.3x10-11 m
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Question:
The electron and proton of a hydrogen atom are separated (on the average) by
a distance of about 5.3x10-11 m. Find the magnitude of the electric force that
each particle exerts on the other.
Observations:
We are interested in finding the magnitude of the force between two
particles of known charge, and a given distance of each other.
The magnitude is given by Coulomb’s law.
q1 =-1.60x10-19 C
q2 =1.60x10-19 C
r = 5.3x10-11 m
Solution:
Fe  ke
e
r2
1.6 10
 5.3 10
19
2
 8.99 10
9 Nm 2
C2
11

m
C
2
2
 8.2 108 N
Attractive force with a magnitude of 8.2x10-8 N.
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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.
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Example: Using the Superposition Principle
Consider three point charges at the corners of a triangle, as shown
below. 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
y
q2
F31
-
3.00 m
q1 +
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4.00 m
F32
+
q3
37.0o
x
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Consider three point charges at the corners of a triangle, as shown
below. Find the resultant force on q3.
y
q2
F31
-
3.00 m
q1 +
4.00 m
F32
+
q3
37.0o
x
Observations:
The superposition principle tells us that the net force on q3 is the vector sum
of the forces F32 and F31.
The magnitude of the forces F32 and F31 can calculated using Coulomb’s
law.
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Consider three point charges at the corners of a triangle, as shown
below. Find the resultant force on q3.
y
q2
3.00 m
-
F32
4.00 m
F32  ke
q3 q2
F31  ke
q3 q1
r
r
37.0o
+
q3
5.00 m
q1 +
Solution:
F31
2
2
x
 5.00 10 C  2.00 10 C   5.62 10
9
 8.99 109
Nm
C2
2
9
 4.00m 
 5.00 10 C  6.00 10 C   1.08 10
9
 8.99 109
Nm
C2
2
 5.00m 
9
2
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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15.4 Electric Field - Discovery
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).
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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15.4 Electric Field – Quantitative Definition
A field : generally changes with position (location)
A vector quantity : magnitude and direction.
Magnitude at a given location

Expressed as a function of the force imparted by the field on
a given test charge.
E
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F
qo
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15.4 Electric Field – Quantitative Definition (2)
Direction defined as the direction of the electrical force
exerted on a small positive charge placed at that
location.
E
+
E
- -
-
-
-
-
- -
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+
+
+
-
-
+
+
+
+
+
+
+
+ +
+
+
+
+
+ + + +
+ + +
+ +
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15.4 Electric Field – Electric Field of a
Charge “q”
Given
One finds
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F  ke
E  ke
q qo
r
2
q
r
2
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• If q>0, field at a given point is radially outward from q.
• If q<0, field at a given point is radially inward from q.
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Problem-Solving Strategy
Electric Forces and Fields

Units:
For calculations that use the Coulomb constant, ke, charges must
be in coulombs, and distances in meters.
Conversion are required if quantities are provided in other units.

Applying Coulomb’s law to point charges.
It is important to use the superposition principle properly.
Determine the individual forces first.
Determine the vector sum.
Determine the magnitude and/or the direction as needed.
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Example:
An electron moving horizontally passes between two
horizontal planes, the upper plane charged negatively,
and the lower positively. 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.
-
-
- -
-
+
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+
+ +
-
- - - - - - - - - - - - - - - - -
vo
+
+ + + + + + + + + + + + + + + + +
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-
-
- -
-
+
+
+ +
-
- - - - - - - - - - - - - - - - -
vo
+
+ + + + + + + + + + + + + + + + +
Observations:
Horizontally:




No electric field
No force
No acceleration
Constant horizontal velocity
Ex  0
Fx  0
ax  0
v x  vo
x  vo t
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-
-
- -
-
+
+
+ +
-
- - - - - - - - - - - - - - - - -
vo
+
+ + + + + + + + + + + + + + + + +
Observations:
Vertically:




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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
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-
-
- -
-
- - - - - - - - - - - - - - - - -
-
+
+
+ +
+
+ + + + + + + + + + + + + + + + +
Conclusions:
The charge will follow a parabolic path downward.
Motion similar to motion under gravitational field only except the
downward acceleration is now larger.
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Example: Electric Field Due to Two Point Charges
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
P
E
0.400 m
E2
q1
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0.300 m
q2
x
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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 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
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q
r
2
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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:
E1  ke
E2  k e
q1
r12
q2
r22
 8.99  109

Nm 2
C2
7.00  106 C
 0.400m 
2
  3.93 10 N / C
5
10.00 10 C   3.60 10 N / C
6
 8.99  109
Nm
C2
2
 0.500m 
5
2
Ex  53 E2  2.16  105 N / C
E y  E1  E2 sin   E1  54 E2  1.05 105 N / C
E  Ex2  E y2  2.4  105 N / C
  arctan( E y / Ex )  25.9o
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15.5 Electric Field Lines
A convenient way to visualize field patterns is to draw
lines in the direction of the electric field.
Such lines are called field lines.
Remarks:
1.
2.
Electric field vector, E, is tangent to the electric field lines at
each point in space.
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.
E is large when the field lines are close together and small
when far apart.
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15.5 Electric Field Lines (2)
Electric field lines of single positive (a) and (b) negative
charges.
a)
b)
+ q
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- q
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15.5 Electric Field Lines (3)
Rules for drawing electric field lines for any charge
distribution.
1.
2.
3.
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Lines must begin on positive charges (or at infinity) and must
terminate on negative charges or in the case of excess charge
at infinity.
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 each other.
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15.5 Electric Field Lines (4)
Electric field lines of a dipole.
+
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-
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Application: Measurement of the atmospheric electric field
The electric field near the surface of the Earth is about
100 N/C downward. Under a thundercloud, the electric
field can be as large as 20000 N/C.
How can such a (large) field be measured?
A
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A
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15.6 Conductors in Electrostatic Equilibrium
Good conductors (e.g. copper, gold) contain charges
(electron) that are not bound to a particular atom, and
are free to move within the material.
When no net motion of these electrons occur the
conductor is said to be in electro-static equilibrium.
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15.6 Conductors in Electrostatic Equilibrium
Properties of an isolated conductor (insulated from the
ground).
1.
2.
3.
4.
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Electric field is zero everywhere within the conductor.
Any excess charge field on an isolated conductor resides
entirely on its surface.
The electric field just outside a charged conductor is
perpendicular to the conductor’s surface.
On an irregular shaped conductor, the charge tends to
accumulate at locations where the radius of curvature of the
surface is smallest – at sharp points.
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1.
Electric field is zero everywhere within the conductor.
If this was not true, the field inside would be finite.
Free charge there would move under the influence of the
field.
A current would be induced.
The conductor would not be in an electrostatic state.
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2.
Any excess charge field on an isolated conductor resides entirely
on its surface.
This property is a direct result of the 1/r2 repulsion
between like charges.
If an excess of charge is placed within the volume, the
repulsive force pushes them as far apart as they can go.
They thus migrate to the surface.
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3.
The electric field just outside a charged conductor is
perpendicular to the conductor’s surface.
If not true, the field would have components parallel to
the surface of the conductor.
This field component would cause free charges of the
conductor to move.
A current would be created.
There would no longer be a electro-static equilibrium.
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4.
On an irregular shaped conductor, the charge tends to accumulate at
locations where the radius of curvature of the surface is smallest – at
sharp points.
Consider, for instance, a conductor fairly flat at one end and relatively pointed at the
other.
Excess of charge move to the surface.
Forces between charges on the flat surface, tend to be parallel to the surface.
Those charges move apart until repulsion from other charges creates an equilibrium.
At the sharp ends, the forces are predominantly directed away from the surface.
There is less of tendency for charges located at sharp edges to move away from one
another.
Produces large fields (and force) near sharp edges.
-
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-
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Remarks
Property 4 is the basis for the use of lightning rods near
houses and buildings. (Very important application)


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Most of any charge on the house will pass through the sharp
point of the lightning rod.
First developed by B. Franklin.
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Faraday’s ice-pail experiment
+++++
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+++++
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+
Demonstrates that the charge resides on the surface of a conductor.
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Mini-quiz
Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves,
we sneak up and surround the charge with a spherical conducting shell.
What effect does this have on the field lines of the charge?
?
+ q
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+
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Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves, we sneak up and surround the
charge with a spherical conducting shell. What effect does this have on the field lines of the
charge?
Answer:
Negative charge will build up on the inside of the shell.
Positive charge will build up on the outside of the shell.
There will be no field lines inside the conductor but the field lines will remain outside the shell.
+
+
+
+
+
-
-
+ q
-
+
-
-
-
+
-
-
+
+
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+
-
-
+
+
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Mini-Quiz
Question:
Is it safe to stay inside an automobile during a lightning
storm? Why?
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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
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