Transcript Lesson 7 Gauss’s Law and Electric Fields

Lesson 7
Gauss’s Law and Electric Fields
Class 18
Today, we will:
• learn the definition of a Gaussian surface
• learn how to count the net number of field lines
passing into a Gaussian surface
• learn Gauss’s Law of Electricity
• learn about volume, surface, and linear charge
density
• learn Gauss’s Law of Magnetism
• show by Gauss’s law and symmetry that the
electric field inside a hollow sphere is zero
Section 1
Visualizing Gauss’s Law
Gaussian Surface
•A Gaussian surface is
–any closed surface
–surface that encloses a volume
•Gaussian surfaces include:
–balloons
–boxes
–tin cans
•Gaussian surfaces do not include:
–sheets of paper
–loops
Counting Field Lines
•To count field lines passing through
Gaussian surfaces:
–Count +1 for every line that passes
out of the surface.
–Count ─1 for every line that comes
into the surface.
+1
─1
Electric Field Lines
We have a +2 charge and a ─2 charge.
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
+8
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
+8
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
─8
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
─8
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
0
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
Electric Field Lines
What is the net number of field lines
passing through the Gaussian surface?
0
Electric Field Lines
From the field lines coming out of this
box, what can you tell about what’s
inside?
Electric Field Lines
The net charge inside must be +1 (if we
draw 4 lines per unit of charge).
Gauss’s Law of Electricity
The net number of electric field lines
passing through a Gaussian surface
is proportional to the charge
enclosed within the Gaussian
surface.
Section 2
Charge Density
Charge Density
Volume: ρ =
Charge
Volume
Surface: σ =
Charge
Area
Linear:
λ=
Charge
Length
Charge Density
In general, charge density can vary with position. In
this case, we can more carefully define density in
terms of the charge in a very small volume at each
point in space. The density then looks like a
derivative:

q
 r   lim
V 0 V

r
dq

dV
You need to understand what we mean by this
equation, but we won’t usually need to think of
density as a derivative.
Section 3
Gauss’s Law of Magnetism
Gauss’s Law and Magnetic Field Lines
If magnetic field lines came out from point
sources like electric field lines, then we would
have a law that said:
The net number of magnetic field lines
passing through a Gaussian surface is
proportional to the magnetic charge inside.
N
Gauss’s Law and Magnetic Field Lines
But we have never found a magnetic
monopole.
- The thread model suggests that there is no
reason we should expect to find a magnetic
monopole as the magnetic field as we know it
is only the result of moving electrical charges.
- The field line model suggests that there’s no
reason we shouldn’t find a magnetic
monopole as the electric and magnetic fields
are both equally fundamental.
Gauss’s Law and Magnetic Field Lines
What characteristic would a magnetic
monopole field have?
Gauss’s Law and Magnetic Field Lines
What characteristic would a magnetic
monopole field have?



F  qtest vtest  Bmonopole
Gauss’s Law and Magnetic Field Lines
All known magnetic fields have field lines that
form closed loops.
So what can we conclude about the number
of lines passing through a Gaussian surface?
Gauss’s Law of Magnetism
The net number of magnetic field lines passing
through any Gaussian surface is zero.
Section 4
Gauss’s Law and Spherical
Symmetry
Spherically Symmetric Charge Distribution
The charge density, ρ, can vary with r only.
Below, we assume that the charge density is
greatest near the center of a sphere.
Spherically Symmetric Charge Distribution
Outside the distribution, the field lines will go
radially outward and will be uniformly
distributed.
Spherically Symmetric Charge Distribution
The field is the same as if all the charge were
located at the center of the sphere!
Inside a Hollow Sphere
Now consider a hollow sphere of inside radius r with a
spherically symmetric charge distribution.
Inside a Hollow Sphere
There will be electric field lines outside the sphere and
within the charged region. The field lines will point
radially outward because of symmetry. But what about
inside?
Inside a Hollow Sphere
Draw a Gaussian surface inside the sphere. What is the
net number of electric field lines that pass through the
Gaussian surface?
Inside a Hollow Sphere
The total number of electric field lines from the hollow
sphere that pass through the Gaussian surface inside
the sphere is zero because there is no charge inside.
How can we get zero net field lines?
1. We could have some lines come in and go out again…
… but this violates symmetry!
How can we get zero net field lines?
2. We could have some radial lines come in and other
radial lines go out…
… but this violates symmetry, too!
How can we get zero net field lines?
3. Or we could just have no electric field at all inside the
hollow sphere.
How can we get zero net field lines?
3. Or we could just have no electric field at all inside the
hollow sphere.
This is the only way it can be done!
The Electric Field inside a Hollow Sphere
Conclusion: the static electric field inside a hollow
charged sphere with a spherically symmetric charge
distribution must be zero.

E 0
Class 19
Today, we will:
• learn how to use Gauss’s law and symmetry to
find the electric field inside a spherical charge
distribution
• show that all the static charge on a conductor
must reside on its outside surface
• learn why cars are safe in lightning but cows
aren’t
Spherically Symmetric Charge Distribution
Electric field lines do not start or end outside charge
distributions, but that can start or end inside charge
distributions.
Spherically Symmetric Charge Distribution
What is the electric field inside a spherically symmetric
charge distribution?
Spherically Symmetric Charge Distribution
Inside the distribution, it is difficult to draw field lines, as
some field lines die out as we move inward. – We need
to draw many, many field lines to keep the distribution
uniform as we move inward.
Spherically Symmetric Charge Distribution
But we do know that if we drew enough lines, the
distribution would be radial and uniform in every
direction, even inside the sphere.
Spherically Symmetric Charge Distribution
Let’s draw a spherical Gaussian surface at radius r.
r
Spherically Symmetric Charge Distribution
Now we split the sphere into two parts – the part outside
the Gaussian surface and the part inside the Gaussian
surface.
r
r
Spherically Symmetric Charge Distribution
The total electric field at r will be the sum of the electric
fields from the two parts of the sphere.
r
r
Spherically Symmetric Charge Distribution
Since the electric field at r from the hollow sphere is
zero, the total electric field at r is that of the “core,” the
part of the sphere within the Gaussian surface.
r
r
Spherically Symmetric Charge Distribution
Outside the core, the electric field is the same as that of
a point charge that has the same charge as the total
charge inside the Gaussian surface.
r
Spherically Symmetric Charge Distribution
Inside a spherically symmetric charge distribution, the
static electric field is:
qenc
E (r ) 
2
4 0 r
1
r
Example: Uniform Distribution
A uniformly charged sphere of radius R has a total
charge Q. What is the electric field at r < R ?
Example: Uniform Distribution
A uniformly charged sphere of radius R has a total
charge Q. What is the electric field at r < R ?
qenc
E (r ) 
2
4 0 r
1
r
Since the charge density
is uniform:
qenc Venc

Q
V
Example: Uniform Distribution
1 qenc
E (r ) 
2
4 0 r
1 1 Venc

Q
2
4 0 r
V
3

r
1 1

Q
2
3
4 0 r
R
4
3
4
3
1 Qr

3
4 0 R
Section 5
Gauss’s Law and Conductors
Gauss’s Law and Conductors
Take an arbitrarily shaped conductor with charges on
the outside.
+
+
+
+
+
+
+
+
Gauss’s Law and Conductors
The static electric field inside the conductor must be
zero. – Draw a Gaussian surface inside the conductor.
+
+
+
+
+
+
+
+
Gauss’s Law and Conductors
No field lines go through the Gaussian surface because
E=0. Hence, the total enclosed charge must be zero.
+
+
+
+
+
+
+
+
Gauss’s Law and Conductors
The same must be true of all Gaussian surfaces inside
the conductor.
+
+
+
+
+
+
+
+
Surface Charge and Conductors
What if there are no charges on the outside and the net
charge of the conductor is zero?
-- The volume charge density inside the conductor
must be zero and the surface charge density on the
conductor must also be zero.
Surface Charge and Conductors
What if there are no charges on the outside and there
is net charge on the surface of a conductor?
+++ +
+
+
+
++
+
+
+
+
+
+
+
Surface Charge and Conductors
The charge distributes itself so the field inside is zero
and the surface is at the same electric potential
everywhere.
+ + +
+
+
+
+
+
+
+
+
+
+
+
+
+
Example: Surface Charge on a Spherical
Conductor
A spherical conductor of radius R has a voltage V. What
is the total charge? What is surface charge density?
Example: Surface Charge on a Spherical
Conductor
A spherical conductor of radius R has a voltage V. What
is the total charge? What is surface charge density?
1 Q
V (r ) 
, rR
4 0 r
1 Q
V ( R) 
4 0 R
Q  4 0VR
Q 4 0VR  0V
 

2
A 4 R
R
On the outside, the
potential is that of a
point charge.
On the surface, the
voltage is V(R).
Take Two Conducting Spheres
with the Same Voltage
The smaller sphere has a larger charge density.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Now Connect the Two Spheres
The charge density is greater near the “pointy” end.
The electric field is also greater near the “pointy” end.
+
+
+
+
+
+
+
+
+ +
+
+
+
+
Edges on Conductors
Charge moves to sharp points on conductors.
Electric field is large near sharp points.
Smooth, gently curved surfaces are the best for holding
static charge.
Lightning rods are pointed.
A Hollow Conductor
What if there’s a hole in the conductor?
+
+
+
+
+
+
+
+
A Hollow Conductor
Draw a Gaussian surface around the hole.
+
+
+
+
+
+
+
+
A Hollow Conductor
There is no net charge inside the Gaussian surface.
+
+
+
+
+
+
+
+
A Hollow Conductor
Is there surface charge on the surface of the hole?
+
+
+
+
++
+
+
+
+
A Hollow Conductor
There is no field surrounding the charge to hold the
charges fixed, so the charges migrate and cancel each
other out.
+
+
+
+
+
+
+
+
Charge on a Conductor
Static charge moves to the outside surface of a
conductor.
+
+
+
+
+
+
+
+
Lightning and Cars
Why is a car a safe place to be when lightning strikes?
Note: Any car will do – it doesn’t need to be a Cord….
Lightning and Cars
Is it the insulating tires?
Lightning and Cars
Is it the insulating tires?
If lightning can travel 1000
ft through the air to get to
your car, it can go another
few inches to go from your
car to the ground!
Lightning and Cars
A car is essentially a hollow conductor.
Charge goes to the outside.
The electric field inside is small.
Lightning and Cars
A car is essentially a hollow conductor.
Charge goes to the outside.
The electric field inside is small.
How should a cow stand to avoid injury
when lightning strikes nearby?
Physicist’s Cow
Cow
Earth
d
I
Physicist’s Cow
When d is bigger, the resistance along the ground
between the cow’s feet is bigger, the voltage across
the cow is bigger, and the current flowing through the
Cow
cow is bigger.
V2
P
R
Earth
d
I
How should a cow stand to avoid injury
when lightning strikes nearby?
So the cow should keep her feet
close together!
Class 20
Today, we will:
• learn how integrate over linear, surface, and
volume charge densities to find the total charge on
an object
• learn that flux is the mathematical quantity that
tells us how many field lines pass through a
surface
Section 6
Integration
Gauss’s Law of Electricity
The net number of electric field lines
passing through a Gaussian surface
is proportional to the enclosed
charge.
But, how do we find the enclosed
charge?
Charge and Density
q  V is valid when?
Charge and Density
q  V when ρ is uniform.
If ρ is not uniform over the whole volume,
we find some small volume dV where it is
uniform. Then:
dq   dV
If we add up all the little bits of dq, we get
the entire charge, q.
q   dq    dV
Integration
The best way to review integration is to work
through some practical integration problems.
Integration
The best way to review integration is to work
through some practical integration problems.
Our goal is to turn two- and three- dimensional
integrals into one-dimensional integrals.
Fundamental Rule of Integration
Identify the spatial variables on which the
integrand depends.
You must slice the volume (length or surface)
into slices on which these variables are
constant.
Fundamental Rule of Integration
When integrating densities to find the total charge, the
density must be a constant on the slice or we cannot write
dq  dV
Fundamental Rule of Integration
Examples
Square in x-y plane
Cylinder
Sphere
  x
  y
  r
  z
  r
Fundamental Rule of Integration
Examples
   x, dq  dA
Consider a very thin slice.
Is
dq constant on this slice?
Fundamental Rule of Integration
Examples
   x, dq  dA
Consider a very thin slice.
Is
dq constant on this slice?
Fundamental Rule of Integration
Examples
Square in x-y plane
Cylinder
Sphere
  x
  y
  r
  z
  r
Fundamental Rule of Integration
Examples
Square in x-y plane
Cylinder
Sphere
  x
  y
  r
  z
  r
Fundamental Rule of Integration
Examples
Square in x-y plane
Cylinder
Sphere
  x
  y
  r
  z
  r
Fundamental Rule of Integration
Examples
Square in x-y plane
Cylinder
Sphere
  x
  y
  r
  z
  r
Fundamental Rule of Integration
Examples
Square in x-y plane
Cylinder
Sphere
  x
  y
  r
  z
  r
Rules for Areas and Volumes of Slices
Memorize These!!!
Square in x-y plane
dA  L dx
Disk
dA  2 r dr
Cylinder
dV  2 r L dr
dV   r 2 dz
Sphere
dV  4 r 2 dr
Rules for Areas and Volumes of Slices
Memorize These!!!
Square in x-y plane
dA  L dx
Disk
dA  2 r dr
Cylinder
dV  2 r L dr
dV   r 2 dz
Sphere
dV  4 r 2 dr
Rules for Areas and Volumes of Slices
Memorize These!!!
Square in x-y plane
dA  L dx
Disk
dA  2 r dr
Cylinder
dV  2 r L dr
dV   r 2 dz
Sphere
dV  4 r 2 dr
Rules for Areas and Volumes of Slices
Memorize These!!!
Square in x-y plane
dA  L dx
Disk
dA  2 r dr
Cylinder
dV  2 r L dr
dV   r 2 dz
Sphere
dV  4 r 2 dr
Rules for Areas and Volumes of Slices
Memorize These!!!
Square in x-y plane
dA  L dx
Disk
dA  2 r dr
Cylinder
dV  2 r L dr
dV   r 2 dz
Sphere
dV  4 r 2 dr
Rules for Areas and Volumes of Slices
Memorize These!!!
Square in x-y plane
dA  L dx
Disk
dA  2 r dr
Cylinder
dV  2 r L dr
dV   r 2 dz
Sphere
dV  4 r 2 dr
Let’s Do Some Integrals
Charge on a Cylinder
A cylinder of length L and radius R has a charge
density    z 4 where  is a constant and z is
the distance from one end of the cylinder. Find the
total charge on the cylinder.
How do you slice the cylinder?
What is the volume of each slice?
Charge on a Cylinder
dq   dV   z  R dz
4
q   dq  R
L
2
 z dz
4
0
5
L
q   R
5
2
2
Charge on a Sphere
A sphere of radius R has a charge density    r
where  is a constant. Find the total charge on the
sphere.
How do you slice the sphere?
What is the volume of each slice?
Charge on a Sphere
dq   dV   r 4 r dr
2
R
q   dq  4  r dr
3
0
4
R
4
q  4
  R
4
Section 7
Gauss’s Law and Flux
Field Lines and Electric Field
N
Ek
A
1
 N  EA
k
This is valid when
1) .A is the area of a section of a perpendicular surface.
2) The electric field is constant on A.
Field Lines and Electric Field
N
Ek
A
1
 N  EA
k
This is valid when
1) A is the area of a section of a perpendicular surface.
2) The electric field is constant on A.
-- But E is a constant on A only in a few cases of high
symmetry: spheres, cylinders, and planes.
Electric Flux
Gauss’s Law states that:
N  qenc
1
 N  EA  qenc
k
 EA  qenc
EA is called the electric flux. We write it as  E or just
.
Electric Flux
Gauss’s Law states that:
N  qenc
1
 N  EA  qenc
k
 EA  qenc
EA is called the electric flux. We write it as  E or just
Flux is a mathematical expression for number of field
lines passing through a surface!
.
Electric Flux and a Point Charge
Lets calculate the electric flux from a point charge
passing through a sphere of radius r.
  EA
1
q
2

4 r
2
4 0 r
Electric Flux and a Point Charge
Gauss’s law says this is proportional to the charge
enclosed in the sphere!
  EA   q
1 q
2
4 r   q
2
4 0 r

1
0
Electric Flux and Gauss’s Law
This means that we can write Gauss’s Law of Electricity as

1
0
qenc
A Few Facts about Flux
For our purposes, we will (almost) always calculate flux
through a section of perpendicular surface where the
field is constant. So we will evaluate flux simply as:
  EA
A Few Facts about Flux
But we do need to find a more general expression for
flux so you’ll know what it really means…
An Area Vector
We wish to define a vector area. To do this
1) we need a flat surface.
2) the direction is perpendicular to the plane of the area.
(Don’t worry about the fact there are two choices of direction that
are both perpendicular to the area – up and down in the figure
below.)

A
3) the magnitude of vector is the area.
A Few Facts about Flux
First, Let’s consider the flux passing through a frame
oriented perpendicular to the field.
A
 0  EA
A Few Facts about Flux
If we tip the frame by an angle θ, the angle between the
field and the normal to the frame, there are fewer field
lines passing through the frame.
A
A

 0  EA   EA cos 
A Few Facts about Flux
Or, using the vector area of the loop, we may write:
 
  E A
A
A

 0  EA   EA cos 
A Few Facts about Flux
 
  E  A only holds when the frame is flat and the
field is uniform.
What if the surface (frame) isn’t flat, or the electric field
isn’t uniform?
Area Vectors on a Gaussian Surface
1) We must take a small region of the surface dA that is
essentially flat.
2) We choose a unit vector perpendicular to the plane of
dA going in an outward direction.

dA
A Few Facts about Flux
The flux through this small region is:
 
d  E  dA
A Few Facts about Flux
To find the total flux, we simply add up all the
contributions from every little piece of the surface.
 
   d   E  dA
Recall that the normal to each small area is taken to be
in the outward direction.
A Few Facts about Flux
Thus, the most general equation for flux through a
surface is:
 
   E  dA
If we take the flux through a Gaussian surface, we
usually write the integral sign with a circle through it to
emphasize the fact that the integral is over a closed
surface:
 
   E  dA
Class 21
Today, we will:
• learn how to use Gauss’s law to find the electric
fields in cases of high symmetry
• insdide and outside spheres
• inside and outside cylinders
• outside planes
Section 7
Gauss’s Laws in Integral Form
Gauss’s Law of Electricity
Integral Form
The number of electric field lines passing through a
Gaussian surface is proportional to the charge enclosed
by the surface.

1
0
qenc
We can make this simple expression look much more
impressive by replacing the flux and enclosed charge
with integrals:
  1
 E  dA 
S
0
  dV
S
Gauss’s Law of Magnetism
Integral Form
The number of magnetic field lines passing through a
Gaussian surface is zero
 B  0
With the integral for magnetic flux, this is:
 
 B  dA  0
S
Gauss’s Law of Electricity
Tee-Shirt Form
  1
 E  dA 
S
0
  dV
S
This can be written in many different ways. A popular
form seen on many tee-shirts is:
  q
 E  dA 
0
Gauss’s Law of Electricity
Tee-Shirt Form
  1
 E  dA 
S
0
  dV
S
This can be written in many different ways. A popular
form seen on many tee-shirts is:
  q
 E  dA 
0
This is a good form of Gauss’s law to use if
you want to impress someone with how
smart you are.
Gauss’s Law of Electricity
Practical Form
  1
 E  dA 
S
 EA 
0
1
0
  dV
S
  dV
This is the form of Gauss’s law you will use
when you actually work problems.
Gauss’s Law of Electricity
Practical Form
EA 
1
0
  dV
Now let’s think about what this
equation really means!
Gauss’s Law of Electricity
Practical Form
EA 
Electric field on
Gaussian surface
-- Must be the same
everywhere on the
surface!
1
0
  dV
Gauss’s Law of Electricity
Practical Form
EA 
Electric field on
Gaussian surface
-- Must be the same
everywhere on the
surface!
1
0
  dV
Area of the
entire
Gaussian
surface – Must
be a perpendicular
surface (an element
of a field contour)!
Gauss’s Law of Electricity
Practical Form
EA 
Electric field on
Gaussian surface
-- Must be the same
everywhere on the
surface!
1
0
  dV
Area of the
entire
Gaussian
surface – Must
be a perpendicular
surface (an element
of a field contour)!
Integral of
the charge
density over
the volume
enclosed by the
Gaussian
surface!
Section 9
Using Gauss’s Law to Find
Fields
Problem 1: Spherical Charge Distribution
Outside
Basic Plan:
1) Choose a spherical Gaussian
surface of radius r outside the
charge distribution.
2) EA  4 r E 
2
qtotal
0
3) Integrate the charge over the
entire charge distribution.
Problem 1: Spherical Charge Distribution
Outside
 r
R
EA  4 r E 
2
E
R
1
4 0 r
2
qtotal
0
  dV
0
r
Problem 1: Spherical Charge Distribution
Outside
E
E
E
R
1
4 0 r

 0r
2
0
R
2
3
r
 dr
0
 R
 0r
 r 4r dr
2
2
4
4
R
r
Problem 2: Spherical Charge Distribution
Inside
Basic Plan:
1) Choose a spherical Gaussian
surface of radius r inside the
charge distribution.
2) EA  4 r E 
2
qtotal
0
3) Integrate the charge over the
inside of the Gaussian surface
only.
Problem 2: Spherical Charge Distribution
Inside
 r
r
EA  4 r E 
2
E
r
1
4 0 r
2
qtotal
0
  dV
0
R
Problem 2: Spherical Charge Distribution
Inside
E
E
r
1
4 0 r

 0r
 r 4r dr
2
2
0
r
 r dr
3
2
0
 r
r
E

2
 0 r 4 4 0
4
2
r
R
Problem 3: Cylindrical Charge Distribution
Outside
Basic Plan:
1) Choose a cylindrical Gaussian
surface of radius r and length L
outside the charge distribution.
2) EA  2 rLE 
qtotal
0
3) Integrate the charge over the
entire charge distribution.
Problem 3: Cylindrical Charge Distribution
Outside
Basic Plan:
4) Note that there are no field lines
coming out the ends of the
cylinder, so there is no flux
through the ends!
Problem 3: Cylindrical Charge Distribution
Outside
   r7
R
EA  2 rLE 
E
1
2 0 rL
R
qtotal
0

dV

0
r
Problem 3: Cylindrical Charge Distribution
Outside
E
R
1
r

2 rL
0
0

8
E
r dr

 0r 0
R
 R
E
 0r 9
9
7
L 2 rdr
R
r
Problem 4: Cylindrical Charge Distribution
Inside
Basic Plan:
1) Choose a cylindrical Gaussian surface
of radius r and length L inside the
charge distribution.
2)
EA  2 rLE 
qtotal
0
3) Integrate the charge over the inside of
the Gaussian surface only.
Problem 4: Cylindrical Charge Distribution
Inside
 r
r
7
EA  2 rLE 
E
1
2 0 rL
r
qtotal
0

dV

0
R
Problem 4: Cylindrical Charge Distribution
Inside
E
r
1
r

2 rL
0
7
0

8
E
r dr

 0r 0
r
 r
r
E

 0 r 9 9 0
9
8
2rLdr
r
R
Infinite Sheets of Charge
Basic Plan:
1) Choose a box with faces parallel to
the plane as a Gaussian surface.
Let A be the area of each face.
2) Find the charge inside the box. No
integration is needed.
Problem 5: Infinite Sheet of Charge
(Insulator with σ given)
Note there is flux out both
sides of the box!
A
2 EA 
0

E
2 0
A
A
Problem 6: Infinite Sheet of Charge
(Conductor with σ on each surface)
Note there is flux out both
sides of the box, and the
total charge density is 2σ!
2 EA 

E
0
2A
0
A
A
Problem 6: A second way…
Now there is flux out only
one side of the box, but the
total charge density inside is
just σ!
A
EA 
0

E
0
A
Ein  0
Problem 7: A Capacitor
The area of the plate is A and the area of
the box is A.
There is flux out only one
side of the box!
A
EA 
0

Q
E

0 0 A
A
A
A Word to the Wise!
If you can do these seven examples, you can do
every Gauss’s law problem I can give you! Know
them well!