Find the Physics
SP212
Ch. 23 - Gauss’ Law
Maj Jeremy Best USMC
Physics Department, U.S. Naval Academy
January 20, 2016
Figure:
newswatch.nationalgeographic.com
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
Figure: Stormtrack.org
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The Electric Field
The Electric Field is everywhere, and we cannot escape
it, not here, nor in the fleet. We have to work smarter,
not harder. We’ll still use calculus, but now let’s try to
simplify our lives with Gauss’s Law.
Figure:
http://www.physicsclassroom.com/Class/estatics/u8l4e.cfm
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Flux
Defining our Terms
~ to represent
We define an area vector, A
our area.
~ has magnitude equal to the area in
A
question
~ has direction perpendicular to the
A
surface, pointing out
The flux Φ through the surface is
Gauss’s Law depends on the concept of electric flux
through a surface, which can be thought of as the
amount of electric field piercing the surface.
~
Φ = vA cos θ = ~v · A
That’s for any vector field ~v. It therefore applies to an
electric field!
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SP212
January 20, 2016
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Demonstration
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Flux of an Electric Field
Demonstrations using mathematica are a fun way to get
an intuitive grasp of many different concepts. http:
//demonstrations.wolfram.com/ElectricFlux/
Now that we know how to calculate the flux through a
single area (like a window) it’s easy to define the flux
through an object with more than one side: Just add
the flux through each side
X
~E · ∆A
~
Φ=
We can extend that to complicated, curved surfaces by
using calculus in the usual way:
I
~
Φ = ~E · d A
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SP212
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Gauss’ Law
Gauss law relates the net flux of an electric field
through a closed surface (a Gaussian surface) to the net
charge qenc that is enclosed by that surface.
The net charge qenc is the algebraic sum of
all the enclosed positive and negative
charges.
Net can be positive, negative, or zero.
If qenc is positive then net flux is
outward .
If qenc is negative then net flux is
inward .
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SP212
January 20, 2016
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Flux φ
Problem: Flux φ
The square surface shown in the figure measures 3.2 mm
on each side. It is immersed in a uniform electric field
with magnitude E = 1800 N/C and with field lines at an
angle of 35◦ with a normal to the surface, as shown.
Calculate the electric flux through the surface.
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SP212
January 20, 2016
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The Second Example
Problem: Flux 2
A cylinder is immersed in a uniform electric field ~E as
shown. The axis of the cylinder is parallel to the field.
What is the total flux Φnet through the entire closed
surface?
Solution:
Apply our definition of flux:
~
Φ = ~E · A
= EA cos θ
= (1800 N/C)(0.0032 m)2 cos(145◦ )
Φ = −1.5 × 10−2 Nm2 /C
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SP212
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Do it side by side!
Side a
Solution:
On this side, the electric field is anti-parallel to the area
vector, so θ = 180◦
Z
E cos(180◦ ) dA
a
Z
−E
dA
Solution:
I
~E · d A
~
Φ=
Z
Z
=
E cos θ dA +
a
Z
E cos θ dA +
b
E cos θ dA
c
− EA
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Side b
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Side c
Solution:
On this side, the electric field is parallel to the area
vector, so θ = 0◦
Z
E cos(0◦ ) dA
a
Z
E
dA
Solution:
On this side, the electric field is perpendicular to the
area vector, so θ = 90◦
Z
E cos(90◦ ) dA
a
0
EA
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Add ’em up!
Gauss’s Law
Solution:
Now add our results from all three sides to get the total
flux:
Gauss’s law states:
I
Φ = −EA + 0 + EA = 0
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Gauss’s Law and Coulomb’s Law
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Gauss’s Law and Coulomb’s Law
Let’s look at a single positive point
charge q, surrounded by a spherical
gaussian surface. We know that all the
~ must be
differential area vectors d A
pointed radially outward. We also know
the electric field must be radially
directed outward everywhere, so
everywhere they are parallel, and θ = 0.
Gauss’s Law then becomes
I
I
q
enc
~E · d A
~ =
= E dA
0
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
0 Φ = qenc
~E · d A
~ = qenc
0
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E is constant all over the surface of the sphere, so it
comes out of the integral. The integral is then just the
surface area of the sphere:
Solution:
Z
q
0
q
E (4πr 2 ) =
0
1 q
E=
4π0 r 2
E
dA =
This is the same expression we obtained fromJanuary
Coulomb’s
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Law!
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
Sample Problem
Solution
The dashed circle represents the cross
section of a closed spherical Gaussian
surface. There are +7.0 C and -4.0 C
charges inside the Gaussian surface and
+2.0 C and -5.0 C outside the surface.
Find the net electric flux through the
Gaussian surface.
Solution:
The two charges +7.0 C and -4.0 C
have a net positive charge of +3.0 C,
so using what we know about The flux
):
φ through the gaussian surface ( qencl
0
the net electric flux through the
Gaussian surface is 3.0C
0 .
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SP212
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Sample Problem
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Conductors
A cube that is 0.20 m on a side is oriented with its edges
along the axes of a Cartesian coordinate system. The
cube is in an electric field ~E = (1.0 ı̂ − 2.0 ̂ + 3.0 k̂k ) NC
Find the electric flux through the square surface at the
top of the cube.
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SP212
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Gauss’s Law let’s us analyze how charges and conductors
interact. The excess charge placed on a conductor
distributes itself around the outside of the conductor in
such a way that it completely cancels the electric field
inside the conductor.
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Gauss’s Law and Conductors
Using Gauss’s Law - Thick Conductor
Think of an isolated, charged
conductor. Draw a Gaussian surface
just inside the surface.
There must be no ~E within the
conductor
Gauss’s Law then tells us qenc = 0
throughout the conductor
Therefore all the charge resides on
the surface
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Using Gauss’s Law - Thick Conductor
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Cylindrical Symmetry
Gauss’s Law therefore tells us:
qenc
Φ=
0
σA
EA =
0
σ
E=
0
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
So what’s the electric field just outside
the conductor? Construct a small
Gaussian surface right on the outside of
the conductor.
The field is parallel to the sides
The flux through the end is the
field times the area Φ = EA
The enclosed charge is qenc = σA
We see a portion of an infinitely long
charge plastic rod, of positive linear
charge density λ. Let us find the
electric field ~E a distance r from the
rod.
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SP212
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Cylindrical Symmetry
Planar Symmetry
By symmetry, ~E is perpendicular to the
outside of our Gaussian surface
everywhere: no angles!
Φ = EA = E (2πrh)
Let’s find the electric field a height r
above an infinite sheet of uniform
charge density σ.
0 Φ = qenc
Φ = EA + EA
qenc = Aσ
σ
E=
20
So Gauss’s Law is easy:
Φ = qenc /0
λh
E (2πrh) =
0
λ
E=
2π0 r
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Spherical Symmetry
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Spherical Symmetry
Gravity and electrostatics have very similar mathematical
underpinnings. Recall the shell theorem from Chapter 13
last semester. The electrostatic version is The electric
field from a shell of uniform charge is the same as if all
the charge were concentrated at the center.
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SP212
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And also a charged particle inside a uniform shell of
charge does not a feel a force from the shell.
Well, using Gauss’s Law, we can prove these easily!
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SP212
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Spherical Symmetry
Spherical Symmetry
And s2 is equally easy:
First, consider Gaussian surface S1 .
0 EA = qenc
0 EA = qenc
qenc = 0
E =0
qenc = q
0 E (4πr 2 ) = q
E=
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Spherical Symmetry
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SP212
1 q
4π0 r 2
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Spherical Symmetry
Now we have a charge distribution
through a spherical volume. Let’s
look at the electric field inside.
As we saw two slides ago, the
charge outside our Gaussian
Surface does not contribute to
the interior field.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
A = 4πr 2
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q
q
= 4 3
V
3 πR
0 EA = qenc = ρv
ρ=
0 E (4πr 2 ) = ρ(4/3)πr 3
q
0 E (4πr 2 ) =
(4/3)πr 3
3
(4/3)πR
q
E=
r
4π0 R 3
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SP212
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Wiley Plus Homework
Chapter 23: Questions 1, 2, 4, 8. Problems: 2, 5, 9, 14,
20, 23, 32, 33, 36, 52, 75, and 78.
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