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General Physics (PHY 2140)
Lecture 3
¾ Electrostatics
9 Electric field (cont.)
9 Conductors in electrostatic
equilibrium
9 The oscilloscope
9 Electric flux and Gauss’s law
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 15
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1
Lightning Review
Last lecture:
F = ke
1. Coulomb’s law
9
q1 q2
r2
the superposition principle
2. The electric field
E=
JG
F
q0
Review Problem: A “free” electron and “free” proton are placed in an
identical electric field. Compare the electric force on each particle.
Compare their accelerations.
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2
Example: Electric Field Due to Two Point Charges
Question:
Charges q1=4.9 µC and charge q2=-2.8 µC are placed as shown, 2 cm
and 4 cm from the origin. Find the electric field at point P, which has
coordinates (0, 0.015) m.
q1 (0, 0.02 m)
JG
E2
0
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P (0, 0.015 m)
JG
E1
JG
E1+ 2
q2 (0, 0.04 m)
3
Question:
Charges q1=4.9 µC and charge q2=-2.8 µC are placed as shown, 2 cm
and 4 cm from the origin. Find the electric field at point P, which has
coordinates (0, 0.015) m.
Observations:
First find the field at point P due to charge q1 and q2.
Field E1 at P due to q1 is directed away from q1.
Field E2 at due to q2 is directed towards q2(to the right).
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
4
Question:
Charges q1=4.9 µC and charge q2=-2.8 µC
are placed as shown, 2 cm and 4 cm from
the origin. Find the electric field at point P,
which has coordinates (0, 0.015) m.
Given:
q1 = 4.9 µC
q2 = -2.8 µC
d1 = 0.02 m
d2 = 0.04 m
d3 = 0.015 m
E1 = ke
d2
d3
( 4.9.00 ×10 C ) = 0.71×10 N / C
( 0.02 + 0.015 ) m
( 2.8.00 ×10 C ) = 0.4 ×10 N / C
( 0.025 ) m
−6
q1
d12 + d32
E2 = ke
d1
= 8.99 × 109
8
2
2
2
= 8.99 × 109
Nm
C2
2
Find:
E1+ 2, x = E1 cos α + E2 = 0.83 × 108 N / C
E1+2 = ?
E1+ 2, y = − E1 sin α = −0.57 × 10 N / C
8

d1
°
=
=
as
α
arctan
53


d3


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2
−6
q1
( d 2 − d3 )
Nm
C2
2
8
2
2
E1+ 2 = 1.0 × 108 N / C
φ = −35D
5
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
7
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.
8
15.5 Electric Field Lines (4)
Electric field lines of a dipole.
+
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-
9
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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10
A
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11
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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12
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.
13
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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15
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.
18
Faraday’s ice-pail experiment
+++++
+
+
+
+
+
-
-
-
+
+
++++
+
+
+
+
-
+
+
+
+
+
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
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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+
20
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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+
-
+
-
-
+
+
21
Mini-Quiz
Question:
Is it safe to stay inside an automobile during a lightning
storm? Why?
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22
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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23
15.8 The Van De Graaff Generator
Read Textbook + Discuss in Lab.
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15.9 The oscilloscope
Changing E field applied on the deflection plate (electrodes)
moves the electron beam.
V2
θ
d
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V1
L
25
Oscilloscope: deflection angle (additional)
V2
θ
d
L
V1
vx =
electron gun
between plates
vy = a yt
vy =
eV 2
eE
=
me
me d
L = vxt
eV 2 L
me d v x
tan θ =
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ay =
2 eV1
me
vy
vx
=
eV 2 L
eV 2 Lm e
V2 L
=
=
m e d v x 2 m e d 2 eV1 d 2 eV
26
15.10 Electric Flux and Gauss’s Law
Discuss a technique introduced by Karl F. Gauss (17771855) to calculate electric fields.
Requires symmetric charge distributions.
Technique based on the notion of electrical flux.
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27
15.10 Electric Flux
To introduce the notion of flux, consider
a situation where the electric field is
uniform in magnitude and direction.
Consider also that the field lines cross a
surface of area A which is
perpendicular to the field.
The number of field lines per unit of
area is constant.
The flux, Φ, is defined as the product of
the field magnitude by the area crossed
by the field lines.
Φ = EA
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Area=A
E
28
15.10 Electric Flux
Units: Nm2/C in SI units.
Find the electric flux through the area A = 2 m2, which is
perpendicular to an electric field E=22 N/C
Φ = EA
Answer: Φ = 44 Nm2/C.
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15.10 Electric Flux
If the surface is not perpendicular to the field, the
expression of the field becomes:
Φ = EA cos θ
Where θ is the angle between the field and a normal to
the surface.
N
θ
θ
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30
15.10 Electric Flux
Remark:
When an area is constructed such that a closed surface
is formed, we shall adopt the convention that the flux
lines passing into the interior of the volume are negative
and those passing out of the interior of the volume are
positive.
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Example:
Question:
Calculate the flux of a constant E field (along x) through a
cube of side “L”.
y
1
2
E
x
z
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32
Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:
z Dealing with a composite, closed surface.
z Sum of the fluxes through all surfaces.
z Flux of field going in is negative
z Flux of field going out is positive.
z E is parallel to all surfaces except surfaces labeled 1 and 2.
z So only those surface contribute to the flux.
y
1
2
E
x
z
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33
Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:
z Dealing with a composite, closed surface.
z Sum of the fluxes through all surfaces.
z Flux of field going in is negative
z Flux of field going out is positive.
z E is parallel to all surfaces except surfaces labeled 1 and 2.
z So only those surface contribute to the flux.
2
Solution:
1
1
1
Φ = − EA cos θ = − EL
Φ 2 = EA2 cos θ 2 = EL2
Φ net = − EL2 + EL2 = 0
y
1
2
E
x
z
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34
15.10 Gauss’s Law
The net flux passing through a closed surface
surrounding a charge Q is proportional to the magnitude
of Q:
Φ net = ∑ EA cos θ ∝ Q
In free space, the constant of proportionality is 1/εo
where εo is called the permittivity of of free space.
1
1
εo =
=
4π ke 4π 8.99 ×109 Nm 2 / C 2
(
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)
35
15.10 Gauss’s Law
The net flux passing through any closed surface is equal
to the net charge inside the surface divided by εo.
Φ net = ∑ EA cos θ =
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Q
εo
36