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1104/19/2304/19/23
Physics
Lecture 3Lecture 3 Electrostatics
Electric field (cont.) Conductors in electrostatic
equilibrium The oscilloscope Electric flux and Gauss’s law
Chapter 15
http://www.physics.wayne.edu/~apetrov/PHY2140/
2204/19/2304/19/23
Lightning ReviewLightning Review
Last lecture:
1. Coulomb’s law
the superposition principle
2. The electric field
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.
1 22e
q qF k
r
0
FE
q
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Example: Electric Field Due to Two Point ChargesExample: Electric Field Due to Two Point ChargesQuestion: Question:
Charges qCharges q11=4.9 =4.9 C and charge qC and charge q22=-2.8 =-2.8 C are placed as shown, 2 cm C are placed as shown, 2 cm and 4 cm from the origin. Find the electric field at point P, which has and 4 cm from the origin. Find the electric field at point P, which has coordinates (0, 0.015) m.coordinates (0, 0.015) m.
q1 (0, 0.02 m)
q2 (0, 0.04 m)
P (0, 0.015 m)
0
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2E��������������
1 2E
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Question: Question: Charges qCharges q11=4.9 =4.9 C and charge qC and charge q22=-2.8 =-2.8 C are placed as shown, 2 cm C are placed as shown, 2 cm
and 4 cm from the origin. Find the electric field at point P, which has and 4 cm from the origin. Find the electric field at point P, which has coordinates (0, 0.015) m.coordinates (0, 0.015) m.
Observations:Observations:
First find the field at point P due to charge qFirst find the field at point P due to charge q11 and q and q22..
Field EField E11 at P due to q at P due to q11 is directed away from q is directed away from q11..
Field EField E22 at due to q at due to q22 is directed towards q is directed towards q22(to the right).(to the right).
The net field at point P is the vector sum of EThe net field at point P is the vector sum of E11 and E and E22..
The magnitude is obtained withThe magnitude is obtained with
2e
qE k
r
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Question: Question: Charges qCharges q11=4.9 =4.9 C and charge qC and charge q22=-2.8 =-2.8 C C
are placed as shown, 2 cm and 4 cm from are placed as shown, 2 cm and 4 cm from the origin. Find the electric field at point P, the origin. Find the electric field at point P, which has coordinates (0, 0.015) m.which has coordinates (0, 0.015) m.
2
2
6
1 9 81 2 2 2 2 2
1 3
4.9.00 108.99 10 0.71 10 /
0.02 0.015Nm
e C
CqE k N C
d d m
Given:
qq11 = 4.9 = 4.9 C C
qq22 = -2.8 = -2.8 CC
dd11 = 0.02 m = 0.02 m
dd22 = 0.04 m = 0.04 m
dd33 = 0.015 m = 0.015 m
Find:Find:
EE1+21+2 = ? = ?
1d2d
3d
2
2
6
1 9 82 2 2 2
2 3
2.8.00 108.99 10 0.4 10 /
0.025Nm
e C
CqE k N C
md d
81 2, 1 2cos 0.83 10 /xE E E N C
81 2, 1 sin 0.57 10 /yE E N C
1
3
as arctan 53d
d
81 2 1.0 10 /
35
E N C
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15.5 Electric Field Lines15.5 Electric Field Lines
A convenient way to visualize field patterns is to draw A convenient way to visualize field patterns is to draw lines in the direction of the electric field.lines in the direction of the electric field.
Such lines are called Such lines are called field linesfield lines..
Remarks:Remarks:1.1. Electric field vector, E, is tangent to the electric field lines at Electric field vector, E, is tangent to the electric field lines at
each point in space.each point in space.
2.2. The number of lines per unit area through a surface The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the perpendicular to the lines is proportional to the strength of the electric field in a given region.electric field in a given region.
E is large when the field lines are close together and small E is large when the field lines are close together and small when far apart.when far apart.
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15.5 Electric Field Lines (2)15.5 Electric Field Lines (2)
Electric field lines of single positive (a) and (b) negative Electric field lines of single positive (a) and (b) negative charges.charges.
+ q
a)
- q
b)
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15.5 Electric Field Lines (3)15.5 Electric Field Lines (3)
Rules for drawing electric field lines for any charge Rules for drawing electric field lines for any charge distribution.distribution.1.1. Lines must begin on positive charges (or at infinity) and must Lines must begin on positive charges (or at infinity) and must
terminate on negative charges or in the case of excess charge terminate on negative charges or in the case of excess charge at infinity.at infinity.
2.2. The number of lines drawn leaving a positive charge or The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude approaching a negative charge is proportional to the magnitude of the charge.of the charge.
3.3. No two field lines can cross each other.No two field lines can cross each other.
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15.5 Electric Field Lines (4)15.5 Electric Field Lines (4)
Electric field lines of a Electric field lines of a dipoledipole..
+ -
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Application: Measurement of the atmospheric electric fieldApplication: Measurement of the atmospheric electric field
The electric field near the surface of the Earth is about The electric field near the surface of the Earth is about 100 N/C downward. Under a thundercloud, the electric 100 N/C downward. Under a thundercloud, the electric field can be as large as 20000 N/C.field can be as large as 20000 N/C.
How can such a (large) field be measured?How can such a (large) field be measured?
A
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AAAAAAAAAAAA
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15.6 Conductors in Electrostatic Equilibrium15.6 Conductors in Electrostatic Equilibrium
Good conductors (e.g. copper, gold) contain charges Good conductors (e.g. copper, gold) contain charges (electron) that are not bound to a particular atom, and (electron) that are not bound to a particular atom, and are free to move within the material.are free to move within the material.
When no net motion of these electrons occur the When no net motion of these electrons occur the conductor is said to be in conductor is said to be in electro-static equilibriumelectro-static equilibrium..
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15.6 Conductors in Electrostatic Equilibrium15.6 Conductors in Electrostatic Equilibrium
Properties of an isolated conductor (insulated from the Properties of an isolated conductor (insulated from the ground).ground).1.1. Electric field is zero everywhere within the conductor.Electric field is zero everywhere within the conductor.
2.2. Any excess charge field on an isolated conductor resides Any excess charge field on an isolated conductor resides entirely on its surface. entirely on its surface.
3.3. The electric field just outside a charged conductor is The electric field just outside a charged conductor is perpendicular to the conductor’s surface. perpendicular to the conductor’s surface.
4.4. On an irregular shaped conductor, the charge tends to On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the accumulate at locations where the radius of curvature of the surface is smallest – at sharp points.surface is smallest – at sharp points.
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1.1. Electric field is zero everywhere within the conductor.Electric field is zero everywhere within the conductor.
If this was not trueIf this was not true, the field inside would be finite., the field inside would be finite.
Free charge there would move under the influence of the Free charge there would move under the influence of the field.field.
A current would be induced.A current would be induced.
The conductor would not be in an electrostatic state.The conductor would not be in an electrostatic state.
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2.2. Any excess charge field on an isolated conductor resides entirely Any excess charge field on an isolated conductor resides entirely on its surface.on its surface.
This property is a direct result of the 1/rThis property is a direct result of the 1/r22 repulsion repulsion between like charges.between like charges.
If an excess of charge is placed within the volume, the If an excess of charge is placed within the volume, the repulsive force pushes them as far apart as they can go.repulsive force pushes them as far apart as they can go.
They thus migrate to the surface.They thus migrate to the surface.
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3.3. The electric field just outside a charged conductor is The electric field just outside a charged conductor is perpendicular to the conductor’s surface. perpendicular to the conductor’s surface.
If not true, the field would have components parallel to If not true, the field would have components parallel to the surface of the conductor.the surface of the conductor.
This field component would cause free charges of the This field component would cause free charges of the conductor to move.conductor to move.
A current would be created.A current would be created.
There would no longer be a electro-static equilibrium.There would no longer be a electro-static equilibrium.
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4.4. On an irregular shaped conductor, the charge tends to accumulate at On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest – at locations where the radius of curvature of the surface is smallest – at sharp points.sharp points.
Consider, for instance, a conductor fairly flat at one end and relatively pointed at the Consider, for instance, a conductor fairly flat at one end and relatively pointed at the other. other.
Excess of charge move to the surface.Excess of charge move to the surface.
Forces between charges on the flat surface, tend to be parallel 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.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.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 There is less of tendency for charges located at sharp edges to move away from one another. another.
Produces large fields (and force) near sharp edges.Produces large fields (and force) near sharp edges.
-
- --
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RemarksRemarks
Property 4 is the basis for the use of lightning rods near Property 4 is the basis for the use of lightning rods near houses and buildings. (Very important application)houses and buildings. (Very important application)
Most of any charge on the house will pass through the sharp Most of any charge on the house will pass through the sharp point of the lightning rod.point of the lightning rod.
First developed by B. Franklin.First developed by B. Franklin.
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Faraday’s ice-pail experimentFaraday’s ice-pail experiment
++++ ++
+++
+++
-- -
-
---
--
-
-
+
+
+
++
+
+
+
++
+
+
+
+
++
+
+
+
++
+
+
+
+
++
+
+
+
++
Demonstrates that the charge resides on the surface of a conductor.
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Mini-quizMini-quiz
Question:Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves, Suppose a point charge +Q is in empty space. Wearing rubber gloves, we sneak up and surround the charge with a spherical conducting we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge?shell. What effect does this have on the field lines of the charge?
+ q +
?
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Question:Question:Suppose a point charge +Q is in empty space. Wearing rubber gloves, we sneak up and surround the 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?charge with a spherical conducting shell. What effect does this have on the field lines of the charge?
Answer:Answer:Negative charge will build up on the inside of the shell.Negative charge will build up on the inside of the shell.Positive charge will build up on the outside 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.There will be no field lines inside the conductor but the field lines will remain outside the shell.
+ q+
--
-
-
--
- -
-
-
-
-
++
+
+
+
+
++
+
+
+
+
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Mini-QuizMini-Quiz
Question:Question:
Is it safe to stay inside an automobile during a lightning Is it safe to stay inside an automobile during a lightning storm? Why?storm? Why?
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Question:Question:
Is it safe to stay inside an automobile during a lightning storm? Why?Is it safe to stay inside an automobile during a lightning storm? Why?
Answer:Answer:
Yes. It is. The metal body of the car carries the excess charges on its 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 external surface. Occupants touching the inner surface are in no danger.danger.
SAFE
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15.8 The Van De Graaff Generator15.8 The Van De Graaff Generator
Read Textbook + Discuss in Lab.Read Textbook + Discuss in Lab.
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15.9 The oscilloscope15.9 The oscilloscope
Changing E field applied on the deflection plate (electrodes) moves the electron beam.
V1
d
V2
L
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Oscilloscope: deflection angle (additional)Oscilloscope: deflection angle (additional)
1
2
2
2 2 22
1
2electron gun
between plates
tan2 2
xe
ye e
y y x
ye x
y e
x e x e
eVv
m
eVeEa
m m d
v a t L v t
eV Lv
m d v
v LmeV eV V LL
v m d v m d eV d eV
V1
d
V2
L
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15.10 Electric Flux and Gauss’s Law15.10 Electric Flux and Gauss’s Law
Discuss a technique introduced by Karl F. Gauss (1777-Discuss a technique introduced by Karl F. Gauss (1777-1855) to calculate electric fields.1855) to calculate electric fields.
Requires symmetric charge distributions.Requires symmetric charge distributions.
Technique based on the notion of Technique based on the notion of electrical fluxelectrical flux..
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15.10 Electric Flux15.10 Electric Flux
To introduce the notion of flux, consider To introduce the notion of flux, consider a situation where the electric field is a situation where the electric field is uniform in magnitude and direction. uniform in magnitude and direction.
Consider also that the field lines cross a Consider also that the field lines cross a surface of area A which is surface of area A which is perpendicular to the field.perpendicular to the field.
The number of field lines per unit of The number of field lines per unit of area is constant.area is constant.
The flux, The flux, , is defined as the product of , is defined as the product of the field magnitude by the area crossed the field magnitude by the area crossed by the field lines.by the field lines.
EA
Area=A
E
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15.10 Electric Flux15.10 Electric Flux
Units: NmUnits: Nm22/C in SI units./C in SI units.
Find the electric flux through the area A = 2 mFind the electric flux through the area A = 2 m22, which is , which is perpendicular to an electric field E=22 N/Cperpendicular to an electric field E=22 N/C
EA Answer: Answer: = 44 Nm2/C. = 44 Nm2/C.
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15.10 Electric Flux15.10 Electric Flux
If the surface is not perpendicular to the field, the If the surface is not perpendicular to the field, the expression of the field becomes:expression of the field becomes:
Where Where is the angle between the field and a normal to is the angle between the field and a normal to the surface.the surface.
cosEA
N
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15.10 Electric Flux15.10 Electric Flux
Remark: Remark:
When an area is constructed such that a closed surface When an area is constructed such that a closed surface is formed, we shall adopt the convention that the flux is formed, we shall adopt the convention that the flux lines passing lines passing intointo the interior of the volume are the interior of the volume are negativenegative and those passing and those passing out ofout of the interior of the volume are the interior of the volume are positivepositive..
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Example:Example:
Question:Question:
Calculate the flux of a constant E field (along x) through a Calculate the flux of a constant E field (along x) through a cube of side “L”.cube of side “L”.
x
y
z
E1 2
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Question:Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:Reasoning: Dealing with a composite, closed surface.Dealing with a composite, closed surface. Sum of the fluxes through all surfaces.Sum of the fluxes through all surfaces. Flux of field going in is negativeFlux of field going in is negative Flux of field going out is positive.Flux of field going out is positive. E is parallel to all surfaces except surfaces labeled 1 and 2.E is parallel to all surfaces except surfaces labeled 1 and 2. So only those surface contribute to the flux.So only those surface contribute to the flux.
x
y
z
E1 2
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Question:Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:Reasoning: Dealing with a composite, closed surface.Dealing with a composite, closed surface. Sum of the fluxes through all surfaces.Sum of the fluxes through all surfaces. Flux of field going in is negativeFlux of field going in is negative Flux of field going out is positive.Flux of field going out is positive. E is parallel to all surfaces except surfaces labeled 1 and 2.E is parallel to all surfaces except surfaces labeled 1 and 2. So only those surface contribute to the flux.So only those surface contribute to the flux.
Solution:Solution:
x
y
z
E1 2
21 1 1
22 2 2
2 2
cos
cos
0net
EA EL
EA EL
EL EL
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15.10 Gauss’s Law15.10 Gauss’s Law
The net flux passing through a closed surface The net flux passing through a closed surface surrounding a charge Q is proportional to the magnitude surrounding a charge Q is proportional to the magnitude of Q:of Q:
In free space, the constant of proportionality is 1/In free space, the constant of proportionality is 1/oo
where where oo is called the permittivity of of free space. is called the permittivity of of free space.
cosnet EA Q
9 2 2
1 1
4 4 8.99 10 /o
ek Nm C
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15.10 Gauss’s Law15.10 Gauss’s Law
The net flux passing through any closed surface is equal The net flux passing through any closed surface is equal to the net charge inside the surface divided by to the net charge inside the surface divided by oo..
cosneto
QEA
