Download Lect08

c
a
b
Gauss’ Law
•Choose either
E dS  E dS  EdS
or
E  dS  E dS  0
•And E constant over surface
•
is just the area of the Gaussian surface over which we
are integrating.
•Gauss’ Law
•This equation can now be solved for E (at the surface) if we
know qenclosed (or for qenclosed if we know E).
Gaussian Surfaces
Given an infinite sheet of charge as shown in the
figure. You need to use Gauss' Law to calculate the
electric field near the sheet of charge. Which of the
Gaussian surfaces are best suited for this purpose?
•
a cylinder with its axis along the plane?
–
•
No, E not parallel or perpendicular to the
cylinder
a cylinder with its axis perpendicular to the plane?
– Yes
•
a cube?
–
•
Yes
a sphere?
–
No
Infinite sheet of charge
• Symmetry:
+s
direction of E = x-axis
• Therefore, CHOOSE Gaussian
surface to be a cylinder whose
axis is aligned with the x-axis.
• Apply Gauss' Law:
• On the barrel,
• On the ends,
•
A
x
E
E
The charge enclosed = s A
Therefore, Gauss’ Law 
Conclusion: An infinite plane sheet of charge
creates a CONSTANT electric field .
Two Infinite Sheets
(into the screen)
• Field outside must be zero.
Two ways to see:
– Superposition
– Gaussian surface encloses
zero charge
• Field inside is NOT zero:
– Superposition
– Gaussian surface encloses
non-zero charge
0
-
+
E=0
s E=0
s
-
+
+
+
+
A
+
+
+
+
+
A
+
+
E
Question 1
The figure shows a Gaussian
surface enclosing charges 2q
and –q. The net flux through
the surface is:
a) q/0
b) 2q/0
-q
2q
c)-q/0
d)zero
Question 1
1.
2.
3.
4.
a
b
c
d
Question 1
The figure shows a Gaussian
surface enclosing charges 2q
and –q. The net flux through
the surface is:
a) q/0
b) 2q/0
-q
2q
c)-q/0
d)zero
•The net charge enclosed is +q. Gauss’ Law says the
flux through the surface is q/0
•q of the lines from the positive charge go to the
negative charge
•q go to infinity and thus pass through the surface
Gauss’ Law: Help for the Problems
• How to do practically all of the homework problems
•
•
Gauss’ Law is ALWAYS VALID!
What Can You Do With This?
If you have symmetry (a) spherical, (b) cylindrical,
or (c) planar AND:
• If you know the charge (RHS), you can calculate
the electric field (LHS)
• If you know the field (LHS, usually because
E=0 inside conductor), you can calculate the
charge (RHS).
Gauss’ Law: Help for the Problems
Spherical Symmetry: Gaussian surface = Sphere
of radius r
•
q = ALL charge inside radius r
Cylindrical symmetry: Gaussian surface = cylinder
of radius r
•
q = ALL charge inside radius r, length L
•
Planar Symmetry: Gaussian surface = Cylinder of area A
q = ALL charge inside cylinder =s A
Example 1: spheres
• A solid conducting sphere is concentric
with a thin conducting shell, as shown
• The inner sphere carries a charge Q1, and
the spherical shell carries a charge Q2,
such that Q2 = -3Q1.
Q2=-3Q1
Q1
R1
A
• How is the charge distributed on the
sphere?
B
• How is the charge distributed on the
spherical shell?
C
• What is the electric field at r < R1?
Between R1 and R2? At r > R2?
D
• What happens when you connect the two
spheres with a wire? (What are the
charges?)
R2
• How is the charge distributed on the sphere?
A
Q2=-3Q1
* The electric field inside a conductor is zero.
(A) By Gauss’s Law, there can be no net charge
inside the conductor, and the charge must
reside on the outside surface of the sphere
+
+
+
+
R1
R2
+
+
+
Q1
+
B
• How is the charge distributed on
the spherical shell?
Q2=-3Q1
* The electric field inside the conducting shell is zero.
(B) There can be no net charge inside the conductor,
therefore the inner surface of the shell must carry a
net charge of -Q1, to cancel the inner conductor
charge.
The outer surface must carry the charge +Q1 + Q2, so
that the net charge on the shell equals Q2.
Q1
The charges are distributed uniformly over the inner
and outer surfaces of the shell, hence
s inner
Q1

2
4R2
and
s outer
Q2 + Q1  2Q1


2
2
4R2
4R2
R1
R2
C
• What is the Electric Field at r < R1?
Between R1 and R2? At r > R2?
* The electric field inside a conductor is zero.
(C) r < R1:
Inside the conducting sphere
(C) Between R1 and R2 : R1 < r < R2
Charge enclosed = Q1
(C) r > R2
Charge enclosed = Q1 + Q2
Q2=-3Q1
Q1
R1

E  0.
R2
1 Q1
E
rˆ
2
4 0 r
1 Q1 + Q 2
1 2Q1
E
rˆ  
rˆ
2
2
4 0 r
4 0 r
• What happens when you connect the two
spheres with a wire? (What are the charges?)
D
Q2=-3Q1
Q1
R1
After electrostatic equilibrium is
reached, there is no charge on the
inner sphere, and none on the inner
surface of the shell. The charge Q1
+ Q2 on the outer surface remains.
-
-
-
-
- - - -
R2
-
- - -
Also, for r < R2
and for r > R2

E  0.
2Q1
E 
rˆ
2
4 0 r
1
Are Gauss’ and Coulomb’s Laws Correct?
The table shows the results of such
experiments looking for a deviation from an
inverse-square law:
Are Gauss’ and Coulomb’s Laws Correct?
•One problem with the above experiments is that they
have all been done at short range, 1 meter or so.
•Other experiments, more sensitive to cosmic-scale
distances, have been done, testing whether Coulomb’s
law has the form:
•No evidence for a nonzero μ has been found.
Example 2: Cylinders
An infinite line of charge passes directly through the
middle of a hollow, charged, CONDUCTING infinite
cylindrical shell of radius R. We will focus on a segment
of the cylindrical shell of length h. The line charge has a
linear charge density l, and the cylindrical shell has a net
surface charge density of stotal.
stotal
l
R
sinner
souter
h
souter
l
R
sinner
stotal
A
h
•How is the charge distributed on
the cylindrical shell?
•What is sinner?
•What is souter?
B
•What is the electric field at r<R?
C
•What is the electric field for r>R?
What is sinner?
A1
souter
l
R
sinner
stotal
h
•The electric field inside the cylindrical shell is zero.
•Choose as our Gaussian surface a cylinder, which
lies inside the cylindrical shell.
•Then the net charge enclosed is zero.
•Therefore, there will be a surface charge density on
the inside wall of the cylinder to balance out the
charge along the line.
What is sinner?
A1
souter
l
R
sinner
stotal
h
•The total charge on the enclosed portion (of length h) of the line charge
is l h
•Therefore, the charge on the inner surface of the conducting cylindrical
shell is
Qinner  l h
•The total charge is evenly distributed along the inside surface of the
cylinder.
•Therefore, the inner surface charge density sinner is just Qinner divided by
total area of the cylinder:
•Notice that the result is independent of h.
s inner
l h
l


2 Rh 2 R
What is souter?
A2
souter
l
R
stotal
sinner
h
•We know that the net charge density on the cylinder is stotal.
• The charge densities on the inner and outer surfaces of the
cylindrical shell have to add up to stotal.
•Therefore,
souter =stotal – sinner = stotal +l /(2 R).
B
What is the Electric Field at r<R?
h
Gaussian surface
l
r
R
h
•Whenever we are dealing with electric fields created by symmetric
charged surfaces, we must always first chose an appropriate
Gaussian surface.
•In this case, for r <R, the appropriate surface surrounding the line
charge is a cylinder of radius r.
B
What is the Electric Field at r<R?
h
Gaussian surface
l
r
R
h
Using Gauss’ Law, the E field is given by
qenclosed
0

lh
 AE r  2 rhEr
0
qenclosed is the charge on the enclosed line charge lh,
(2rh) is the area of the barrel of the Gaussian surface.
The Field at radius r is:
C
What is the Electric field for r>R?
Gaussian surface
R
stotal
•
•
•
•
•
r
h
Chose a Gaussian surface as indicated above.
Charge enclosed in our Gaussian surface
•Net charge enclosed on the line: lh
•Net charge on the cylindrical shell: = 2Rh stotal
Therefore, net charge enclosed is 2 Rhs total + l h
The surface area of the Gaussian surface is 2rh
Gauss’ Law:
2 Rhs total + l h

AEr  2 rhEr 
0
•Solve for Er to find
l
Summary
• Practice Gauss’ Law problems
• Remember SYMMETRY, CONSTANT FIELD
• Choose appropriate Gaussian surface
EA 
qenclosed
0
• Next week Electric Potential. Read Chapter 24
• Try Chapter 23 problems 25, 33, 47, 51, 56
• Next week: Quiz on Thursday and Friday will be on
chapters 21,22,23