Download Motion of a charged particle in an electric field. Gauss`s Law

Lecture 3
Chapter 26
Physics II
Motion of a charged particle in
an electric field.
Gauss’s Law
95.144
Course website:
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Lecture Capture:
http://echo360.uml.edu/danylov201516/physicsII.html
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Electric field of a charged ring (Example 26.4)
A thin ring of radius R is uniformly charged with total charge Q.
Find the electric field at a point on the axis of the ring
(perpendicular to the ring)
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Electric field of a disc of charge
The ring results can be extended to calculate
the electric field of a uniformly charged disc.
Without derivation:
2
1
In the limit z<<R, it becomes
2
El. field created by an infinite charged plate
η - surface charge density
Interesting!!!
The field strength is independent of the distance z
95.144 Danylov Lecture 3
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The parallel-plate capacitor
Single plates
Let’s bring them very close to each other
z
It is called a
Parallel-plate capacitor
Outside the plates
=0
Inside the plates
=
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The field inside a parallel-plate
capacitor is uniform
Motion of a charged particle in
an electric field.
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Electron moving parallel to E
An electron is released from rest at the surface of the negative plate.
How long does it take the electron to cross to the positive plate? Example
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26.8
Chapter 27.
Gauss’s Law
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The idea behind Gauss’s law
It looks like number of lines passing through a closed surface
are related to the amount of charge inside (Gauss’s Law)
But, first, we need to learn how to count lines
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The Basic Definition of Flux
Imagine holding a rectangular wire loop of area A in front of a fan.
 The volume of air flowing through the loop each
second depends on the angle between the loop
and the direction of flow.
 The flow is maximum through a loop
that is perpendicular to the airflow.
 No air goes through the same loop if
it lies parallel to the flow.
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The Area Vector
 Let’s define an area vector
to be a vector in the
direction of , perpendicular to the surface, with a
magnitude A equal to the area of the surface.
 Vector
has units of m2.
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The Electric Flux
Consider 1) a uniform electric field E
2) a flat surface
The electric flux through a surface of area A
can be defined as the dot-product:
θ
95.144 Danylov Lecture 3
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Channel 61
ConcepTest 1 Electric Flux
The electric flux through the shaded surface is
A) 0
B) 400cos20 N m2/C
C) 400cos70 N m2/C
D) 400 N m2/C
E) Some other value
The Electric Flux (general case)
Consider 1) a non-uniform electric field E
2) area is not flat (curved)
Divide the surface into many small pieces of area A.
a) Each piece is small enough that it is essentially flat
b) The field is nearly uniform over each piece
c) Thus, the formula from the previous slide
can be used
The electric flux through each small piece is:
The electric flux through the whole surface is
the surface integral:
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The Electric Flux through a closed surface
Consider 1) a non-uniform electric field E
2) a closed surface
NOTE: For a closed surface, we use the convention that
the area vector dA is defined to always point toward the outside.
The electric flux through a closed surface:
closed surface
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Gauss’s Law
For any closed surface enclosing total charge Qin, the net electric flux
through the surface is:
Φ
∙
Gaussian surface
1) It works for any closed surface
q1
2) Qin is the net charge enclosed by the Gaussian surface
(charges outside must not be included)
3) Distribution of Qin doesn't matter
4) A Gaussian surface is an imaginary, mathematical surface
q4
q2
q3
q5
Gaussian surface
This result for the electric flux is known as Gauss’s Law.
Both, Gauss’s law and Coulomb’s law, help to find electric fields based on
distribution of charges.
•
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Gauss’s Law/Symmetry
Φ
∙
Gaussian surface
Evaluation of this surface integral is often difficult. However, when the
charge distribution has sufficient symmetry (spherical, cylindrical, planar),
evaluation of the integral becomes simple.
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Channel 61
ConcepTest 2 Electric Flux
Which spherical Gaussian
surface has
the larger electric flux?
A) Surface A
B) Surface B
C) They have the same flux
D) Not enough information to tell
Flux depends only on the enclosed
charge, not the radius.
Channel 61
Example
Electric Flux
Determine the electric flux through each surface
Φ
Flux depends only on the enclosed
charge, not the radius.
S1: Φ=(+Q-3Q)/ε0
S2: Φ=(+Q+2Q-3Q)/ε0
S3: Φ=(+2Q-3Q)/ε0
S4: Φ=0 (no charge inside)
S5: Φ=(+2Q)/ε0
∙
What you should read
Chapter 26 (Knight)
Sections:
 26.6
 26.7 (Skip it)
Chapter 27 (Knight)
Sections




27.1 (Not necessary, read if you want)
27.2
27.3
27.4
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Thank you
See you on Friday
95.144 Danylov Lecture 3
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Electron moving perpendicular to E
An electron with a horizontal initial
velocity v0 enters a parallel-plate capacitor
where the electric field is E. Write the
equation of its path.
Example
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