Download Lecture 05 Gaus`s Law

Lecture 3
Gauss’Law
What’s Up???
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What’s that tingle? is due this week.
There will be a quiz on Friday covering
the Electric Field or Gauss’s Law.
This Week
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Some more problems on the Electric Field
We begin the study of Gauss’s Law which is
just an EASY way of calculating fields even
though it LOOKS HARD!
Let’s do a really HARD
problem
I had trouble with it too!
In the figure, particle 1 of charge +1.0 µC and particle 2 of
charge -2.0 µC, are held at separation L = 12.0 cm on an x
axis. If particle 3 of unknown charge q3 is to be located such
that the net electrostatic force on it from particles 1 and 2 is
zero, what must be the coordinates of particle 3? x=[-4.97]
First Region
+
1mC
-2mC
Region II
x
+
1mC
-2mC
F 1mC
2mC 
 2 
2
kq3  x
( x  L) 
NEVER ZERO
Region III (x<0)
x
+
1mC
-2mC
F  q3 (1mC ) q3 q2 



2
k  x
( x  L) 
Be careful not to use the
negative sign TWICE !!
q3 q 2 
F  q3 (1mC )



2
k  x
( x  L) 
Put some numbers in :
1
2

2
2
(
x

L
)
x
( x  L) 2
x
2
( x  L)
x
2
 2
L
12
x

 29.2
2  1 0.41
DIFFERENCE
The Graph
0
5
10
15
20
-1.0E-01
X
25
30
35
Graphical Solution-Better
Scale
DIFFERENCE
Solution.=29.3
0
5
10
15
20
-1.0E-03
X
25
30
35
NOTE
WebAssign answer is not
correct.
Gauss’s Law
http://en.wikipedia.org/wiki/Carl_Friedrich_Gauss
1777-1855
Quick Review – The Radian
definition
r
q
s
s
q
r
or
s  rq
full  circle : 2r  rq full
q full  2
Spherical Surface - New
Concept
dW  Solid Angle subtended
by the area on the
surface, da
In Three Dimensions:
The SOLID ANGLE
dA
dW  2
r
full  sphere
SOLID ANGLE
4r 2
W full  2  4
r
Unit = steradian
The solid angle is independent of the radius and
depends only on the area of the sphere that it
intersects.
Recall what we done so far?
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We were given Coulomb’s Law
We defined the electric field.
Calculated the Electric Field given a
distribution of charges using Coulomb’s Law.
qi
dq(r )
E
r

k
r

unit
unit
2

i 2
V
40
r1
r
1
(Units: N / C)
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??
Another QUESTION:
Solid Surface
Given the electric field at EVERY point
on a closed surface, can we determine
the charges that caused it??
The Answer …
The answer to the question is buried in something called
GAUSS’ LAW
or
Gauss’s Law
or
The Law Of Gauss
or
The Chapter that Everybody hates!
The “Area Vector”
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Consider a small area.
It’s orientation can be described by a
vector NORMAL to the surface.
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
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 a
vector
q
En  E n cos(q )  E  n
DEFINITION:
Element of Flux through a surface
DA
E
DF=ENORMAL x DA
(a scalar)
“Element” of Flux of a vector E
leaving a surface
DF  E  dA  E NORMAL  DA
also
DF  E  dA  E  ndA
n is a unit OUTWARD pointing vector.
This flux was LEAVING the
closed surface
q
Definition of TOTAL FLUX
through a surface
F   dF
surface
Total Flux of the Electric
Field LEAVING a surface is
F   E  n out dA
Total Flux
F leaving  
surface
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E  n out dA
F=Flux leaving the surface
It is a SCALAR
You can only do this integration
explicitly for “well behaved” surfaces
and fields.
Visualizing Flux
flux   E  ndA  F
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  F
E  n  E n cos(q )
n
q
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
F   E  ndA 
dA
2

40 r
Sphere
1
q
q

2
40 r
1
q
 dA  40 r 2 A
q
q
2

4r 
2
40 r
0
1
Gauss’ Law
Flux is total EXITING the
Surface.
n is the OUTWARD
pointing unit normal.
qenclosed
F   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
 E dA  
n
0
h
E  2rh 
0

E
20 r
From SYMMETRY E is
Radial and Outward
Infinite Sheet of Charge
s
h
E
cylinder
sA
EA  EA 
0
s
E
2 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 IN 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 
0
or
s
E
0
Charged Isolated Conductor
• The ELECTRIC FIELD is normal to the
surface outside of the conductor.
• The field is given by: E  s
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
dA

0

n
Because E=0 everywhere
inside the surface.
So Q=0 inside the hole
Including the surface.
A Spherical Shell with
A Charge Inside.
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
q
 E dA  
n
rV
4 3 1
E  4r 
 r r
0
3
0
E
rr
(Vectors)
E
3 0
2
O
r
0
Outside The Charge
 En dA 
R
q
0
r 4 3 Q
E  4r 
R 
0 3
0
2
r
or
O
E
1
Q
E
4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
+
+
+
+
+
+
+
+
Top
s
+
+
+
+
+
+
+
+
A
A
E
s
EA  0 A  EA  A
0
s
E
0
Bottom
s
EA  EA  2 EA  2 A
0
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
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 mC charge
In the center of the cube. Calculate the total flux exiting the cube.
6
1.8 10
5
2
F 

2
.
03

10
Nm
/C
12
 0 8.85 10
q
NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!
Easy, yes??
Problem #2
(15 from text)
Note: the problem is poorly stated in the text.
Consider an isolated conductor with an initial charge of 10 mC 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 mC initial
+3 mC added
Another Problem from the book
Gauss 
a
m,q both given as is a
s
Gaussian
Surface
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
 5.03 10 9 C 2
m
q
(all given)
A Hard One!
A uniformly charged VOLUME
With volume charge density=r
What is E at the end of the vector?
a
 En dA 
q
0
E  4r 2 
1
0
4
3
r r 3
or
E
ra
3 0
VECTOR!!!
(We already did this)
-2A cavity is now carved out in
The volume. Find the electric
field at the point indicated in the
Diagram at the end of the vector
c. Call the point b and let it be
Located at the end of the vector b
a
Vector sum: b=a+c
c
Solution:
We use the SUPERPOSITION of
the original volume and a new
negatively charged sphere (small)
as shown of same charge density
but opposite sign
-3
And the answer is…..
E(b)=E first sphere – E second sphere
rb rc r (b  c)
E(b) 


3 0 3 0
3 0
a
c
but
b ac
or
a  bc
ra
E(b) 
3 0
E is constant in
The cavity!
A Last Problem
A uniformly charged cylinder.
rR
s
R
s
r (r 2 h)
E (2rh) 
0
rr
E
2 0
rR
r (R 2 h)
E (2rh)  
0
0
q
rR 2
E
2 0 r