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Clicker Question
Two Finite Charged Planes are used to accelerate
charged particles.
A)
B)
C)
Infinite Plane with uniform charge distribution.
J.J. Thompson - 1897
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New concept: Electric Flux ΦE through a surface
A new technique for calculating E
- fields:
Gauss's Law
Surface with some area A
has some electric flux
(E
- field lines) through it.
Define surface vector
v
A = Anˆ
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Define surface vector
v
A = Anˆ
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v v
ΦE = E • A
(Vector Dot Product)
v v v v
Φ E = E • A =| E || A | cos θ
A = |A| = magnitude of the area of the surface [m2]
^n = direction of A = direction perpendicular to the surface
* In special case of a single flat surface.
E
A
A
θ
* Note that there are two possible
directions for A
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Clicker Question
A prism-shaped closed surface is in a constant, uniform electric
field E, filling all space, pointing right. The 3 rectangular faces of
the prism are labeled A, B, and C. Face A is perpendicular to the
E-field. The bottom face C is parallel to E. Face B is the leaning
face.
Which face has the largest
magnitude electric flux through it?
B
A) A
B) B
C) C
A
D) A and B have the same
C
magnitude flux
Think of water flowing through
a net.
Demonstration...
E
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Clicker Question
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Clicker Question
If I have a faucet tap turned on at the marked point, rank
order which line has the most to the least water flowing
through it per unit time. Assume the faucet has been on for a
long time.
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A) 1 > 2 > 3 > 4
B) 1 = 2 > 3 > 4
C) 1=3 > 2 > 4
D) 4 > 3 > 2 > 1
E) none of the above
1=2=3=4
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If I have a faucet tap turned on at the marked point, rank
order which line has the most to the least NET water flowing
through it per unit time. Assume the faucet has been on for a
long time.
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2
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A) 1 = 2 = 3 = 4
B) 1 > 2 > 3 > 4
C) 1 = 2 = 3 > 4
D) 1 = 3 > 2 > 4
E) none of the above
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One more complication...
What if our surface is not flat?
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v v
Φ E = ∫ E • da
3
E
da
Break up surface into many tiny segments of area
da, which must be flat in the infinitesimal limit.
Integral is over the surface.
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Clicker Question
Gauss's Law
v v Q
Φ E = ∫ E • da = inside
ε0
Circle = integral over a closed surface!
For a closed surface, da is always outward.
The electric flux thru any closed surface S is a constant
(1/ε0) times the net charge enclosed by S.
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Clicker Question
Which surface has the largest
magnitude electric flux?
-Q
+3Q
4
1
3
2
-2Q
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Gauss's Law can be derived from Coulomb's Law if charges
are stationary. Gauss's Law is actually more general.
A) 1 = 2 = 3 = 4
B) 1 > 2 = 4 > 3
C) 3 > 2 > 1 > 4
D) 3 > 2 > 1 = 4
E) None of the
above
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Example showing equivalence of two Laws.
Consider single charge +q
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Gauss's Law
Coulomb's Law
∫
v kq
1 q
| E |= 2 =
r
4πε 0 r 2
v v Q
E ⋅ da = inside
surf
ε0
1. Qinside = +q
2. E is always radially outward.
3. da is always radially outward.
4. Thus E.da = |E||da| everywhere!
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∫
v v Q
E ⋅ da = inside
surf
v
v
∫ | E || da |=
surf
v
|E|
∫
surf
How did that just work?
ε0
The key is symmetry !
+q
ε0
v +q
| da |=
ε0
v
q
| E | (4πr 2 ) =
ε0
v
| E |=
q
4πε 0 r 2
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Clicker Question
∫
v
v
Q
A)
E ⋅ da = inside
Consider a spherical Gaussian surface
ε0
surf
with a source charge +q at the center.
v v +q
If we move the charge slightly to the
B)
| E || da |=
ε0
right, which is the first step that is wrong?
surf
∫
v
C) | E |
∫
In situations with a high degree of symmetry, Gauss's Law can
allow for a quick calculation of the E-field.
It is only easy to compute
v
v if symmetric situation!
∫ E ⋅ da
surf
v +q
| da |=
ε0
surf
+q
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v
q
D) | E | (4πr 2 ) =
E) All correct.
ε0
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Example 1:
Find the E-field inside a uniform spherical shell of charge (η).
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+ +
+
Exam #1 Information:
By symmetry, E must be radial
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+
+
v v
E =| E | rˆ
You must show up on time to the exam. The exam will last 90
minutes.
Imagine a Gaussian surface as a
sphere of radius r.
In this case, E and da are parallel
everywhere. This is the symmetry!
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If you have permission/disability requiring a different
time/location for the exam, you will receive an email
by this Wednesday letting you know the time and location.
If you do not receive this email, let me know ASAP.
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The exam will consist of 20 multiple choice questions.
The exam will cover chapters 25, 26, 27.
Coulomb's Law, E
- field, E
- flux, Gauss's Law
This includes MP HW #1
- 4, Tutorials #1
- 3
Exam questions will look like clicker questions, parts
of MP homework problems, tutorial questions.
Formula sheet posted on the web will be included with
your exam. Exam is closed book, closed notes, no
calculators.
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∫
+
+
+
surf
+
+
+
+
+
+
+
+
+
+
+
+
+
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Clicker Question
v v Q
E ⋅ da = inside
ε0
∫
v v Q
E ⋅ da = inside
ε0
surf
v
| E | ∫ da = 0
v
| E | (4πr 2 ) = 0
v
Electric Field is
| E |= 0 zero everywhere
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+
+
+
+
+
+
+
+
+
+
inside!
+ + +
+ +
Using Gauss's Law we determined
that E=0 inside a sphere with a
uniform charge distribution.
What if we change the sphere
to a cube? Is E=0 inside?
A) Yes, E=0 everywhere inside.
B) No, E is not zero everywhere
inside.
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Example 2:
Find the E-field outside a uniform spherical shell of charge (η).
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+
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+ +
+
Clicker Question
∫
surf
By symmetry, E must be radial
+
+
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+
+
v v
E =| E | rˆ
+
+
+
ε0
+
+
In this case, E and da are parallel
everywhere. This is the symmetry!
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v
+
+
Imagine a Gaussian surface as a
sphere of radius r.
The total charge on the sphere is Q.
Using Gauss's Law, find the E-field
outside the sphere.
v v Q
E ⋅ da = inside
+
+
+
+
+ +
+
+
+
A) E =
Q
rˆ
4πε 0 r 2
B)
v
E = 0
C)
v
Q
E=
rˆ
4πε 0
D) None of
the above.
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E-Field near an infinite, uniform charged (η) plane of charge.
Think about Gauss's Law and gravity.
Previously we found
Newton's Problem?
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++
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++
++
++
++
++
+
v
η
| E |=
2ε 0
Now try using Gauss's Law.
Key Step - What shape should we use
for the imaginary Gaussian surface?
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Clicker Question
A uniform, infinite plane of positive charge creates a uniform Efield. An imaginary Gaussian surface in the shape of a cylinder
is shown. The flux through the surface is:
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∫
v v
E ⋅ da =
surf
v
v
v
v
v
v
∫ E ⋅ da + ∫ E ⋅ da + ∫ E ⋅ da
endcap1
endcap 2
sideroll
v v v v
v v
∫ E ⋅ da =| E | | Aendcap1 | + | E || Aendcap1 | +0
surf
v
v
v
∫ E ⋅ da = 2 | E | | πr
2
|
surf
E
v v
v
Qinside πr 2η
2
π
E
d
a
E
r
⋅
=
2
|
|
|
|
=
=
∫
E
surf
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ε0
ε0
v
η
| E |=
2ε 0
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Clicker Question
Conductors in Electrostatic Equilibrium
Consider a cube of side (s) of copper (conductor) in equilibrium.
What is the E-field magnitude at any point inside the cube?
E = 0 inside a conductor in equilibrium !
A) |E| = +infinite
B) |E| = k/s2
C) |E| = 0
D) None of the above
E) Cannot determine from information give.
Proof: If -E field were not zero, then conduction
electrons would feel a force F=qE and will move.
Moving charges = not electrostatic equilibrium
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Consider any Gaussian surface completely inside the cube.
The E-field is zero everywhere inside a conductor at
equilibrium, and thus the E-field is zero on all our surfaces!
∫
surf
∫
v v Q
E ⋅ da = inside
ε0
v v
Q
E ⋅ da = 0 = inside
ε0
surf
If we place a net charge Q on a conductor and let it settle to
equilibrium, all the excess charge Q must be on the surface !
+
+
+
+
+
+
+
+
+
Qinside = 0
This is a very powerful and
interesting feature of equilibrium
metals.
+
+
+
+
+
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This also tells us that the E-field at the surface of conductors
in equilibrium must be only perpendicular to the surface.
E
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Clicker Question
An excess charge +Q is put on a copper sphere. We then
place a point charge +q outside the sphere. We allow the
charges to settle into equilibrium.
Think about what an
- field parallel to the
E
surface would do.
+q
r
Van de Graff demonstration.
+Q
What is the E-field at the center
of the sphere?
A) |E| = k(q+Q)/r2
B) |E| = kq/r2
C) |E| = kQ/r2
D) |E| = 0
E) None of the above.
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Clicker Question
An excess charge +Q is put uniformly on a plastic (insulating)
sphere. We then place a point charge +q outside the sphere.
+q
r
+Q
What is the E-field at the center
of the sphere?
A) |E| = k(q+Q)/r2
B) |E| = kq/r2
C) |E| = kQ/r2
D) |E| = 0
E) None of the above.
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Instead of an infinite sheet of charge, what if we have an infinite
block of conductor.
All the charge will be on the surface and thus have density η.
The electric field is then calculated by Gauss's Law.
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Clicker Question
Imagine the Gaussian surface drawn below. Calculate the Efield magnitude a distance (s) away from the conductor
surface.
A) |E| = η/2ε0
B) |E| = η/ε0
C) |E| = 0
E) none of above
D) |E| = (1/4πε0)(η/s2)
Just a comment on divergence!
"Surface Integrals and the Divergence"
Chapter 2, Div, Grad, Curl and All That by H.M. Schey
∫
v v Q
E ⋅ da = inside
ε0
surf
v v ρ
∇⋅E =
ε0
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Equivalent forms of
the first of Maxwell's
equations.
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