Chapter 22: Gauss`s Law
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Transcript 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
4
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, technicall y,
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
Fq
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 ) Ar 5
3
3
2
qenclosed
2
E
d
A
E
(
4
r
)
4
E (4r )
Ar 5
3 0
Outside the sphere :
qtotal V
4 3
4
( Ab )( b ) Ab 5
3
3
2
qenclosed
2
E dA E (4r )
E (4r 2 )
2
E
A 3
r
3 0
4
Ab 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 (2rL)
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 (2rL)
What if you had an
infinitely long line of q
enclosed lL
charge with a linear
l
L
charge density, l?
E (2rL )
0
l
l
E
or E
rˆ
2r 0
2r 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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