Transcript Mr. Gauss`s Law

Flux and
Gauss’s Law
Spring 2008
Calendar for the Week
Today (Wednesday)
– Comments on the rest of chapter 23
– Introduction to the concept of FLUX
Friday
– 7:30 review session
– No Quiz on Friday (sorry)
– More on Gauss’s Law
EXAM APPROACHING (Date soon but
start getting ready).
Notes about email & stuff
I need to know which class you are
in (9:30 or 11:30)
It is better if you use WebAssign for
your questions. It allows me to
make changes faster.
If a problem topic is not covered in
class you have my permission to
read the text to learn the material!
Last Time: Definition – Sort of –
Electric Field Lines
Field Lines  Electric Field
Last time we showed that
Ignore the Dashed Line …
Remember last time .. the big plane?
s/2e0
s/2e0
E=0
s/2e0
s/2e0
s/2e0
s/2e0
E=s/e0
We will use this a lot!
E=0
NEW RULES (Bill Maher)
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Imagine a region of space where the ELECTRIC
FIELD LINES HAVE BEEN DRAWN.
The electric field at a point in this region is
TANGENT to the Electric Field lines that have been
drawn.
If you construct a small rectangle normal to the field
lines, the Electric Field is proportional to the
number of field lines that cross the small area.


The DENSITY of the lines.
We won’t use this much
So far …
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The electric field exiting a closed surface
seems to be related to the charge inside.
But … what does “exiting a closed surface
mean”?
How do we really talk about “the electric field
exiting” a surface?


How do we define such a concept?
CAN we define such a concept?
Mr. Gauss
answered the
question with..
Another QUESTION:
Solid Surface
Given the electric field at EVERY point
on a closed surface, can we determine
the charges that caused it??
A Question:

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
Given the magnitude and direction of the
Electric Field at a point, can we
determine the charge distribution that
created the field?
Is it Unique?
Question … given the Electric Field at a
number of points, can we determine the
charge distribution that caused it?

How many points must we know??
Still another question
Given a small area, how can
you describe both the area
itself and its orientation with a
single stroke!
The “Area Vector”


Consider a small area.
It’s orientation can be described by a
vector NORMAL to the surface.



We usually define the unit normal vector n.
If the area is FLAT, the area vector is given by
An, where A is the area.
A is usually a differential area of a small part
of a general surface that is small enough to be
considered flat.
The “normal component” of the
ELECTRIC FIELD
E
n
En = E cos( )n
E n = E cos( ) n = E  n
En
DEFINITION FLUX
E
n
En
Flux = E n A = (E  n)A
 = E A cos( )
We will be considering CLOSED
surfaces

En = E n cos( ) = E  n
The normal vector to a closed surface is DEFINED as positive
if it points OUT of the surface. Remember this definition!
“Element” of Flux of a vector E
leaving a surface
d = E  dA = E NORMAL  A
also
d = E  dA = E  ndA
For a CLOSED surface:
n is a unit OUTWARD pointing vector.
This flux is LEAVING the closed
surface.

Definition of TOTAL FLUX through
a surface
 =  d
surface
Total Flux of the Electric
Field LEAVING a surface is
 =  E  n outdA
Visualizing Flux
flux =  E  ndA = 
n is the OUTWARD
pointing unit normal.
Definition: A Gaussian Surface
Any closed surface that
is near some distribution
of charge
Remember
flux =  E  ndA = 
E  n = E n cos( )
n

E
A
Component of E
perpendicular to
surface.
This is the flux
passing through
the surface and
n is the OUTWARD
pointing unit normal
vector!
Example
Cube in a UNIFORM Electric Field
Flux is EL2
E
Flux is -EL2
L
area
Note sign
E is parallel to four of the surfaces of the cube so the flux is zero across these
because E is perpendicular to A and the dot product is zero.
Total Flux leaving the cube is zero
Simple Example
1
r
q
 =  E  ndA =
dA
2

4e0 r
Sphere
1
q
q
=
2
4e0 r
1
q
 dA = 4e0 r 2 A
q
q
2
=
4r =
2
4e0 r
e0
1
Gauss’ Law
Flux is total EXITING the
Surface.
n is the OUTWARD
pointing unit normal.
qenclosed
 =  E  ndA =
e0
q is the total charge ENCLOSED
by the Gaussian Surface.
qenclosed
=
dA
E
n

e0
Simple Example
UNIFORM FIELD LIKE BEFORE
E
A
E
E
A
 EA  EA =
q
e0
=0
Line of Charge
Q
L
charge Q
=
=
length
L
Line of Charge
q
 E dA = e
n
0
h
E  2rh =
e0

2
2k
E=
=
=
2e0 r 4e0 r
r
From SYMMETRY E is
Radial and Outward
What is a Cylindrical
Surface??
Ponder
Looking at A Cylinder from its END
Circular
Rectangular
Drunk
Infinite Sheet of Charge
s
h
E
cylinder
sA
EA  EA =
e0
s
E=
2e 0
We got this same
result from that
ugly integration!
Materials

Conductors

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



Electrons are free to move.
In equilibrium, all charges are a rest.
If they are at rest, they aren’t moving!
If they aren’t moving, there is no net force on them.
If there is no net force on them, the electric field must be
zero.
THE ELECTRIC FIELD INSIDE A
CONDUCTOR IS ZERO!
More on Conductors


Charge cannot reside in the volume of a
conductor because it would repel other
charges in the volume which would move and
constitute a current. This is not allowed.
Charge can’t “fall out” of a conductor.
Isolated Conductor
Electric Field is ZERO in
the interior of a conductor.
Gauss’ law on surface shown
Also says that the enclosed
Charge must be ZERO.
Again, all charge on a
Conductor must reside on
The SURFACE.
Charged Conductors
Charge Must reside on
the SURFACE
-
E=0
-
-
s
Very SMALL Gaussian Surface
E
sA
EA =
e0
or
s
E=
e0
Charged Isolated Conductor

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The ELECTRIC FIELD is normal to the
surface outside of the conductor.
s
The field is given by:
E=
e0
Inside of the isolated conductor, the Electric
field is ZERO.
If the electric field had a component parallel
to the surface, there would be a current flow!
Isolated (Charged) Conductor with
a HOLE in it.
E
n
dA = 0 =
Q
e0
Because E=0 everywhere
inside the conductor.
So Q (total) =0 inside the hole
Including the surface.
A Spherical Conducting Shell with
A Charge Inside.
Insulators

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
In an insulator all of the charge is bound.
None of the charge can move.
We can therefore have charge anywhere in
the volume and it can’t “flow” anywhere so it
stays there.
You can therefore have a charge density
inside an insulator.
You can also have an ELECTRIC FIELD in
an insulator as well.
Example – A Spatial Distribution of
charge.
Uniform charge density = r = charge per unit volume
q
 E dA = e
n
rV
4 3 1
E  4r =
= r r
e0
3
e0
E
rr
(Vectors)
E=
3e 0
2
O
r
A Solid SPHERE
0
Outside The Charge
 En dA =
R
q
e0
r 4 3 Q
E  4r =
R =
e0 3
e0
2
r
or
O
E
1
Q
E=
4e0 r 2
Old Coulomb Law!
Graph
E
R
r
Charged Metal Plate
s
E
+
+
+
+
+
+
+
+
s
+
+
+
+
+
+
+
+
A
A
E
E is the same in magnitude EVERYWHERE. The direction is
different on each side.
Apply Gauss’ Law
s
+
+
+
+
+
+
+
+
Top
s
+
+
+
+
+
+
+
+
A
A
E
s
EA  0 A = EA = A
e0
s
E=
e0
Bottom
s
EA  EA = 2 EA = 2 A
e0
Same result!
Negatively Charged
ISOLATED Metal Plate
s
s
--
E is in opposite direction but
Same absolute value as before
Bring the two plates together
s1
s1
A
e
s1
s1
B
e
As the plates come together, all charge on B is attracted
To the inside surface while the negative charge pushes the
Electrons in A to the outside surface.
This leaves each inner surface charged and the outer surface
Uncharged. The charge density is DOUBLED.
Result is …..
E=0
2s1
2s1
s
s
E
A
E=0
B
e
e
sA
e0
s 2s 1
E=
=
e0 e0
EA =
VERY POWERFULL IDEA

Superposition

The field obtained at a point is equal to the
superposition of the fields caused by each of
the charged objects creating the field
INDEPENDENTLY.
Problem #1
Trick Question
Consider a cube with each edge = 55cm. There is a 1.8 C charge
In the center of the cube. Calculate the total flux exiting the cube.
6
1.8 10
5
2
= =
=
2
.
03

10
Nm
/C
12
e 0 8.85 10
q
NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!
Easy, yes??
Note: the problem is poorly stated in the text.
Consider an isolated conductor with an initial charge of 10 C on the
Exterior. A charge of +3mC is then added to the center of a cavity.
Inside the conductor.
(a) What is the charge on the inside surface of the cavity?
(b) What is the final charge on the exterior of the cavity?
+10 C initial
+3 C added
Another Problem
Charged Sheet
a
m,q both given as is a
s
Gaussian
Surface
Gauss 
sA
2 EA =
e0
s
E=
2e 0
-2
a
m,q both given as is a
a
T
qE
s
T cos(a ) = mg
mg
qs
T sin( a ) = qE =
2e 0
Free body diagram
-3
Divide
qs
Tan(a ) =
2e 0 mg
and
2e 0 mg tan(a )
s=
= 5.03 10 9 C 2
m
q
(all given)
A Last Problem
A uniformly charged cylinder.
rR
s
R
s
r (r 2 h)
E (2rh ) =
e0
rr
E=
2e 0
rR
r (R 2 h)
E (2rh ) = =
e0
e0
q
rR 2
E=
2e 0 r