Download Chapter 22

Gauss’s Law
Young and Freedman
Chapter 22
Electric Flux, Surface Integrals and Gauss’s Law
• From what we learned in the last chapter, we can
calculate the Electric Field at any point in space:
r
1 % (r )
ˆ
E= $
r
dv
2
all 4 "# 0 r
!
• However this becomes very difficult for anything other
than the simplest geometries
• Gauss’s law is a clever way to avoid this problem and
relies on two important concepts
– Electric Flux
– Surface Integrals
Electric Field on the Surfaces of Boxes
Zero net charge means zero net flux
(a) No charge in the box, no flux
(b) Equal but opposite charges in the box, no flux
Flux Through a Surface
" E = EA cos #
r r
"E = E • A
"=
Direction of A is perpendicular to the
surface
Magnitude of A is equal to the area of
the surface
r
A = Anˆ
$ E cos #dA = $ E dA = $
%
r r
E • dA
Electric Flux through a Sphere Ex 22.3
PROBLEM: A positive point charge q = 3.0µC is
surrounded by a sphere with radius = 0.20m centered on
the charge. Find the electric flux through the sphere
Set-up the problem
"=
# d"
Required vector equations
surface
r r
d" = E • dA
r
dA = rˆdA
Substitute for dΦ
"=
#
r r
E • dA
r
E=
surface
!
Substitute vector quantities
& 1 q)
"= # (
rˆ • rˆdA
2+
!
surface ' 4 $% 0 r *
1 q
ˆ
r
2
4 "# 0 r
Electric Flux through a Sphere cont.
• At any point on the sphere the area vector and the
electric field vector point in the same direction (i.e.
outwards)
% 1 q(
"= + '
rˆ • rˆdA
2*
surface & 4 #$ 0 r )
% 1 q(
"='
dA
2* +
& 4 #$ 0 r )surface !
rˆ • rˆ = (1)(1)cos(0) = 1
surface area
of a sphere
% 1 q(
2
"='
4
#
r
2*
& 4 #$ 0 r )
q
"=
Note: results does NOT depend on r
$0
For completeness, let’s put in the numbers
$q'
"=& )
%#0 (
Solution for charge in a sphere
+6
q = 3.0µC = 3.0 *10 C
# 0 = 8.854...*10
+12
2
C / Nm
2
3.0 *10 +6
2
"=
Nm
/C
+12
8.9 *10
" = 3.4 *10 5 Nm 2 /C Check units, do they make sense?
Remember
r r
"= E•A
so units should be N/C times m2
Total Flux is Independent of Size of Sphere
Electric field decreases as 1/r2
but
Area increases as 1/r2
Charges outside the closed surface do
NOT contribute to the flux
Total Flux is Independent of Shape
As φ increases:
r
dA increases by cosφ
but
dot product decreases by cosφ
!
!
GAUSS’S LAW
r r Qenclosed
" = # E • dA =
$0
closed
surface
The total electric flux through a closed surface
is equal to 1/ε0 times the total net electric charge
inside the surface
Electric Field of a Charged Plane
Example 22.7
Find the electric field
caused by a thin, flat,
infinite sheet on which
there is a uniform positive
charge per unit area σ
!
Field of an infinite plane sheet of charge
"=
$
E# dA =
$
r r
E • dA
Qenclosed
"=
#0
• At the ends of the cylinder, E is parallel to the surface;
hence the flux through each end is +EA
• For the side walls of the cylinder, E is perpendicular to
the surface; hence the flux is 0 (we only have to consider
!
the end of the cylinder)
• In the problem we were given the charge per unit area.
The net charge enclosed within the Gaussian surface is
Qencl = "A hence:
"A
"
2EA =
;E =
#0
2# 0
Field of an infinite plane sheet of charge
• NOTE: this means that the field is independent of
distance from the sheet
• An “infinite” sheet is an idealization. However the
result holds as long as you are close compared to the
dimensions of the sheet
• A real, flat sheet will have a field which is close to
uniform and perpendicular to the sheet in such a
situation.
Example: Conducting Parallel Plates (See Y&F Example 22.8)
•
•
•
Two large parallel conducting plates are given charges of equal
magnitude and opposite sign; the charge per unit area is +σ for
one and –σ for the other.
Determine where the charges are located and what the field is
everywhere.
NOTE: It is important to note that the actual electric field will
have a complicated shape at the edges of the plates
We will generally disregard the fringe fields
when we do electric field calculations.
Sometimes we assume the plates are
“infinite” or we assume the size of the plate is
much, much bigger than the separation
Strategy for Solution of Gauss’s Law Problems:
Use a succession of Gaussian Surfaces. Start
with fields or charges you know. Use this information
to “bootstrap” your way along to learn about the
fields or charges you don’t know.
Important Note: The symmetry of the problem
implies that the component of the electric field
perpendicular to the plates must be zero.
r r Qenclosed
E " dA =
#
!0
closed
surface
+σ
-σ
Question #1: What is the Electric Field Outside the Plates?
Observation: Because the charge densities are equal and opposite, the total
charge will be zero if the surface encloses equal area of both plates
Pick Surface #1 to enclose both plates
Total charge inside
is zero because of
cancellation of
+ and – charges.
Contribution from “c” is
zero because field must
be parallel to surface
and dot product is zero
r r
0
Ab + 0 =
# E " dA = Eleftend Aa + Eright
$0
end
surface#1
Eleft + Eright = 0
end
Area a equals area b a
end
Eleft = "Eright
!
end
end
Symmetry of system tells us
the magnitudes of the the
fields outside are equal.
!
Eoutside = Eoutside = 0
left
right
+σ
-σ
c
Surface #1
b
Question #2: What is the Charge on the Outside Surface of the Plates?
Observation: We know that the field outside the plates is zero from previous step.
We know the field inside the conductor is zero.
Pick Surface #2 to pass through regions where we know E=0
#
+σ
r r
E " dA = Eoutside Aa + Einside
surface#2
left
Eoutside + Einside
left
conductor
Qenclosed
Ab + 0 =
$0
conductor
= 0 = Qenclosed
Area a equals area b
Qenclosed = 0
c
a
b
Surface #2
Conclusion: No charge on outside surfaces of Conductors
ALL CHARGE IS ON INSIDE SURFACE OF CONDUCTORS!
-σ
Question #3: What is the Electric Field Between the Plates?
Pick Surface #3 to have surface b in the region between the
plates and surface a in a region of zero field
#
r r
E " dA = Einside
conductor
surface# 3
Aa + Ebetween Ab =
Total charge is equal to
area times charge
Qenclosed density
plates
" # Ab
Einside Aa + Ebetween Ab =
!0
conductor
plates
=0
Ebetween
plates
"
=
!0
$0
+σ
+
+
+
+
+
+
+
a
b
+
+
Surface
#3
+
+
+
+
-σ
-
Electric field for oppositely charged parallel plates
+σ
E inside conductors =0
-σ
Electric field for oppositely charged parallel plates
+σ
E perpendicular =0
-σ
!
Electric field for oppositely charged parallel plates
+σ
-σ
S1 only need to
evaluate end between
plates
"A
EA =
#0
"
E=
#0
S4 the same (- signs
cancel)
Field of a charged conducting sphere
• PROBLEM: We place a positive charge q on a
conducting sphere with radius R. Find E at any point
inside or outside the sphere.
• Consider symmetry:
– no preferred region of the surface of the sphere and as it’s
conducting, the charge is free move and there is nothing to
make it concentrate more in some region than others.
– Rotation of sphere cannot change the field pattern
– Charge is uniform, field is radial
• Inside the sphere, there is no charge Qencl=0, hence
E=0
Field of a charged conducting sphere
Outside the sphere
# E dA = E(4 $r
"
2
)
q
E(4 $r ) =
%0
1 q
E=
2
4 $% 0 r
2
Same as for a point charge
How to build up a BIG charge
1) Rub comb against hair to get
a charge
2) Touch comb against inside
of metal sphere - All charge
goes to outside
3) REPEAT
Van de Graaff generator
2) Electrons drawn away from
conducting shell, leave net positive
charge on shell
1) Electron sink removes electrons,
leaving positive charge on belt
Field of a line charge
• PROBLEM: Find the electric field for an infinitely
long, thin wire with charge per unit length λ
• Consider symmetry:
– Cylindrical symmetry. Rotation around axis cannot change
the E field
– No component parallel to wire, why would one direction be
preferred?
– Field lines must be radial, lying in planes exactly
perpendicular to the wire
• Use a cylinder for the Gaussian surface
!
Field of a line charge
No flux through ends of
surface, as E" = 0
Flux through side walls is
perpendicular, hence
E" = E
!
Surface area of side walls is:
A = 2 "rl
Flux through side walls:
Qencl %l
" = E2 #rl =
=
$0 $0
Hence field is:
1 $
E=
2"# 0 r
Chapter 22 Summary
Chapter 22 Summary cont.
End of Chapter 22
You are expected to:
•
Understand the following:
Electric Flux, Vector Area, Gauss’s Law
•
Be able to perform surface integrals in simple geometries
(planes, spheres, cylinders, lines, etc..)
•
Be able to apply Gauss’s Law for simple geometries using
surfaces with appropriate symmetry.
•
Be able to apply Gauss’s Law in cases where charged and
uncharged conductors and insulators are involved.
•
Be able to reconstruct the reasoning used in examples 22.3
through 22.10