Download Chapter 22: Gauss`s Law

Chapter 22
1
Flux

Number of objects passing through a
surface
2
Electric Flux, F

is proportional to the number of electric field
lines passing through a surface
 Assumes that the surface is perpendicular
to the lines


If not, then we use a cosine of the angle
between them to get the components that
are parallel
Mathematically:
 
F   E A cos   E  A
3
Simple Cases
A
A
A

F=EA
F=0
E
E
F=EAcos
E cos
E
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From S to 
A S represents a sum over a large a
collection of objects
 Integration is also a sum over a collection of
infinitesimally small objects, in our case,
small areas, dA
 
 So
F  E  dA



Since dA represents dxdy then, technically,
 
F   E  dA
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Gauss’s Law
The field lines emitted by a charge are
proportional to the size of the charge.
 Therefore, the electric field must be
proportional to the size of the charge
 In order to count the field lines, we must
enclose the charges in some geometrical
surface (one that we choose)

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Mathematically
Fq
F
Charge enclosed
within bounding limits
of this closed surface
integral
q
0
  qenclosed
E

d
A


0
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Fluxes, Fluxes, Fluxes
8
3 Shapes
Sphere
 Cylinder
 Pillbox

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Sphere


When to use: around
spherical objects (duh!) and
point charges
 Hey! What if an object is
not one of these objects?
Closed surface integral
yields:
 
2
E

d
A

E
(
4

r
)


r is the radius of the
geometrical object that you
are creating
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Sphere Example
What if you had a sphere of radius, b, which contained a material whose
charge density depend on the radius, for example, =Ar2 where A is a
constant with appropriate units?
Inside the sphere :
qenclosed  V
4 3
4
 ( Ar )( r )  Ar 5
3
3
2
qenclosed
 
2
E

d
A

E
(
4

r
)

4
E (4r ) 
Ar 5
3 0
Outside the sphere :
qtotal  V
4 3
4
 ( Ab )( b )  Ab 5
3
3
2
qenclosed
 
2
 E  dA  E (4r )
E (4r 2 ) 
2
E
A 3
r
3 0
4
Ab 5
3 0
A 5
b
3 0
E
r2
At r=b, both of these expressions should be equal
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Cylinder


When to use: around
cylindrical objects and
line charges
Closed surface integral
yields:
 
 E  dA  E (2rL)

L
r is the radius of the
geometrical object that
you are creating and L is
the length of the cylinder
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Cylinder Example
 
 E  dA  E (2rL)
What if you had an
infinitely long line of q
enclosed  lL
charge with a linear
lL
charge density, l?
E (2rL ) 
0

l
l
E
or E 
rˆ
2r 0
2r 0
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Pillbox

When to use: around
flat surfaces and
sheets of charge
 Closed surface
integral yields:
 
 E  dA  EA

A is the area of the
pillbox
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Charge Isolated Conductor in
Electrostatic Equilibrium

If excess charge is placed on an isolated conductor,
the charge resides on the surface. Why?


If there is an E-field inside the conductor then it would exert
forces on the free electrons which would then be in motion.
This is NOT electrostatic.
Therefore, if there is no E-field inside, then, by
Gauss’s Law, the charge enclosed inside must be zero


If the charges are not on the outside, you are only left with the
surface
A caveat to this is that E-field lines must be perpendicular to
the surface else free charges would move.
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Electric field on an infinitely large
sheet of charge
Let  
q
A
A
E
+++++++++++++++++++++++++++++++++++++++++++++
E
 
 E  dA  EA  ( E )( A)  2 EA
qenclosed  A
So
2 EA 
A
0

E
2 0
or
 
E
nˆ
2 0
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Electric field on a conducting sheet
Let  
q
A
A
E
+++++++++++++++++++++++++++++++++++++++++++++
 
 E  dA  EA  0  EA
qenclosed  A
So
So a conductor has 2x the
electric field strength as
the infinite sheet of charge
A
0
 

E
or E  nˆ
0
0
EA 
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A differential view of Gauss’s Law

Recall the
Divergence of a field
of vectors


Div (v )    v
How much the vector
diverges around a given
point
Div=+large
Div=0
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Divergence Theorem (aka Gauss’s
Thm or Green’s Thm)

 




v
d


v

d
A


Suspiciously like LHS
of Gauss’s Law
A place of high divergence is like a faucet
Bounded surface
of some region
Sum of the faucets in a volume = Sum of the water
going thru the
surface
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Div(E)
  qenclosed
 E  dA 
0
 


qenclosed
  E d 
0
and
qenclosed    d
so
 
E 
0
So how the E-field spreads out from a point depends on the amount of
charge density at that point
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