Download E - Del Barco Lab

Gauss and All That Jazz
????
Bindell
Del Barco
The Geometry – Past Chapter??
Define surface charge density
s=charge/unit-area
(z2+r2)1/2
dq=sdA
dA=2prdr
dq=s dA = 2psrdr
dq cos( ) k 2prsdr
z
dE z  k

2
2
z r
z2  r2 z2  r2




R
E z  2pksz 
(z2+r2)1/2

0
z
rdr
2
r

2 3/ 2
 s 
z

Ez  
 2 
1 
z 2  R2
0 

When R  ,
s
Ez 
2 0





1/ 2
s
s
2 0
s
2 0
s
2 0
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Infinite sheet
Of charge
s
2 0
s
2 0
s
2 0
s
2 0
s
2 0
s
E0
s
2 0
s
2 0
s
2 0
s
2 0
s
2 0
s
2 0
+
+
+
+
+
+
+
+
+
+
+
+
+
+
s
s
s
E


2 0 2 0  0
Infinite sheet
Of charge
s
-
Infinite sheet
Of charge
E0
Field Lines  Electric Field
know
If we
where
the charges are, we
can calculate the
Electric Field.
QUESTION:
If we know the electric
field can we determine
the charges?
Summary
Consider 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
Charge
Positive or Negative?
Charge
Positive or Negative?
We just showed that
So far …
The electric field exiting or entering a 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?
Conclusion from Demo
An area can be represented by a VECTOR.
The Vector is perpendicular to the AREA that it represents.
The area itself must be FLAT
If not, use a teeny tiny differential area that is small enough that we can consider it flat. The
magnitude of this area is dA.
The area VECTOR is the unit vector dotted into the area vector:
dA  ( dA)n
where
n is a unit vector perpendicular to
the surface.
Check this out!
For this example, since everything is perpendicular to the
spherical surface, consider the following calculation:
q
1 q
1 q
q
2
EdA

k
dA

dA

4
p
r

2
2
2

 r

4p 0 r
4p 0 r
0
or
E

n
dA


qinside
0
But you
didn’t
derive it!
This is
E

n
dA


qinside
Inside what??
0
FLUX
What the flux?
FLUX     E  ndA 
qinside
0
That’s MY
LAW !!
Another QUESTION:
Solid Surface
Given the electric field at EVERY point
on a closed surface, can we determine
the charges that caused it??
Types of G-surfaces we can actually use
Sphere
Charge distribution
Must be
Centro-symmetric

Cylinder
Charge must be axially
Symmetric.
PLANE OF CHARGE
Line of
Charge
Point Charge
GENERALLY OF INFINITE EXTENT
The “normal component” of the ELECTRIC FIELD
E
n
En  E cos( )n
En  E cos( ) n  E  n
En
DEFINITION: FLUX
E
n
En
Flux  E n A  (E  n)A or E  dA
  E A cos( )
Types of Surfaces
1. Open Surfaces
2. Closed Surfaces
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.

Flux is a scalar! How can we ascribe a DIRECTION to a scalar???
Flux leaving is q/0
Flux leaving is zero
FLUX is like the “amount” of
Electric Field (Light) passing
through a surface.
This isn’t really a correct statement but it should give you an idea and I will deny I ever said this!
“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.
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 and is used for
Gauss’s Law Calculations
flux   E  ndA  
Remember
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 so there is NO charge in the cube!
Simple Example (We already did this!)
1
r
q
   E  ndA 
dA
2

4p0 r
Sphere
1
q
q

2
4p0 r
1
q
 dA  4p0 r 2 A
q
q
2

4pr 
2
4p0 r
0
1
Gauss’ Law
Flux is total EXITING the
Surface.
n is the OUTWARD
pointing unit normal.
qenclosed
   E  ndA 
0
q is the total charge ENCLOSED
by the Gaussian Surface.
qenclosed

dA
E
n

0
Simple Example
UNIFORM FIELD LIKE BEFORE
E
A
E
E
A
 EA  EA 
q
0
0
Line of Charge
Q
L
charge Q


length
L
Let’s review a little
y=L/2
dy
Total charge = Q
Total Length=L
=Q/L
dq=dy
y
O
x-axis 
dq
1
dq
1
r
dEx  (2)
cos( )  (2)
2
2
4p 0 (r  y )
4p 0 (r 2  y 2 ) (r 2  y 2 )1/2
2
r
Q
dEx 
dy
2
2 3/2
4p 0 (r  y ) L
Doing the math (Yuck!)
L /2
2
r
Q
dEx 
dy
2
2 3/2
4p 0 (r  y ) L
2r Q
dy
dEx 
4p 0 L (r 2  y 2 )3/2
2r Q
y
Ex 
4p 0 L r 2 r 2  y 2
2r Q
Ex 
4p 0 L
L/2
r2
L /2
2r Q
Ex 
4p 0 L

dy
(r 2  y 2 )3/2
0
From tables of integrals …
Ex 
Q
4p 0
0
2
L
r2 
4
1
2
L
r r2 
4
What happens when L gets really big
compared to r? (Infinite line of charge)
Ex 
Q
1
4p 0
L2
r r 
4
Q

L
1
k
4p 0
2
L  r
Ex 
Q
1
4p 0
L2
r
4

2Q 1
2 Q1

4p 0 rL
4p 0 L r
2 k
E
r
Line of Charge – The Easy Way! GAUSS
q
 E dA  
n
0
h
E  2prh 
0

2
2k
E


2p0 r 4p0 r
r
From SYMMETRY E is
Radial and Outward
By the way ..
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 
0
sA
2 EA 
0
We got this same
s
E
result from that
2 0
ugly integration!
Gauss – Part 2
WELCOME BACK TO THE TWILIGHT ZONE
CORRECTION
Gauss was NOT wealthy, as I said last week. I
lied. He was born into a poor family but was
very bright and went into politics where he
passed Gauss’s Law. It only works in Germany
where Gauss was born.
Gauss is now dead.
LAST TIME:
We defined FLUX
Flux     E  dA  EdA cos
We “derived” Gauss’s Law
Flux     E  dA 
q
0
Gauss’s LAW
Flux     E  dA 
q
0
 Only certain problems can be solved in THIS class and
only for certain shapes.
 The electric field E must be either perpendicular to the
surface or parallel to it.
 E needs to be uniform over the surface that you are
using.
SHAPES
Small
area
NOPE!
Only for small areas
W
Flux=-EA
E
A’
L’

L
Flux=0
Flux=EA’cos

Flux  EA' cos  EL'W cos  E
A'  WL'
L  L' cos
L
L' 
cos
LW
cos   EA
cos
PROBLEM
dA  2prdr
r  R sin  dr  Rd
d  E 2pR 2 sin d 

Q
4p 0 R

FROM THE TOP
r

2
2
p
R
2
 sin d
0

Q
Q
Q

sin d 
cos | 
(1  cos )

2 0 0
2 0
2 0
0
Materials
Conductors
◦
◦
◦
◦
◦
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.
Charge Must reside on
the SURFACE
of a conductor in
Equilibrium
Charged Conductors
sA
EA 
0
-
E=0
-
-
or
E
s
Very SMALL Gaussian Surface (cylinder)
s
E
0
Charged Isolated Conductor
The ELECTRIC FIELD is normal to the surface outside of the conductor.
The field is given by:
s
E
0
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
0
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.
An interesting example
+s
Infinite
sheet of
positive
charge
s
E
2 0
s
- E
+
2 0
+
+
+
+
- Metal +
s +
E
2 0 +
- s
+
E
- 
+
2 0
+
E
s
2 0
Hey- Even an
idiot knows the
electric field in
the metal is
zero!
Insulators
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  
n
0
rV
4 3 1
E  4pr 
 r pr
0
3
0
E
rr
(Vectors)
E
3 0
2
O
r
A Solid INSULATING SPHERE
Outside The Charge
 En dA 
R
q
0
r 4 3 Q
E  4pr 
pR 
0 3
0
2
r
E
O
or
1
Q
E
4p0 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
s
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Top
A
E
A
s
EA  0 A  EA  A
0
s
E
0
Bottom
s
EA  EA  2 EA  2 A
0
Same result!
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
0
s 2s
E  1
0 0
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
 0 8.85 10
q
NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!
Easy, yes??
Note: the problem is poorly stated in the text it was stolen from.
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 
0
s
E
2 0
-2
a
m,q both given as is a
a
T
qE
s
T cos(a )  mg
mg
qs
T sin( a )  qE 
2 0
Free body diagram
-3
Divide
qs
Tan(a ) 
2 0 mg
and
2 0 mg tan(a )
s
q
The charge density r for this solid sphere is non uniform and is given by rAr2 . What is the
Electric field as a function of distance from the center for r>R?
R
The TOTAL charge on the sphere is given by:
rR
R
R
R
5
R
2
2
4
Q   rdV   ( Ar )( 4pr dr )   ( 4 A)(pr dr ) 4 Ap
5
0
0
0
1
5
5
5
5
1
R
1 1
R
1 1
R
AR
E
4pAR

AR

AR

2
2
2
4p 0 r
5 0 r
5 0 r
5 5 0 r 2
Inside the slab of stuff :
q rV r (2 x) A
2AE 


d
r
0
0
r ( x)
E
0
E(x)
E(x)
A
x
Outside the slab of stuff :
q rAd
2 AE 

0
E
Uniformly
charged
0
rd
2 0
0
STOLEN SLIDE!
Oh, Oh!
ENOUGH!