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c
a
b
Today…
• Gauss’ Law: Motivation & Definition
• Coulomb’s Law as a consequence of Gauss’ Law
• Charges on Conductors:
– Where are they?
• Applications of Gauss’ Law
–
–
–
–
–
Uniform Charged Sphere
Infinite Line of Charge
Infinite Sheet of Charge
Two infinite sheets of charge
Shortcuts
Appendix: Three Gauss’ Laws examples
Fundamental Law
of Electrostatics
• Coulomb’s Law
Force between two point charges
OR
• Gauss’ Law
Relationship between Electric Fields
and charges
Preflight 4:
dS
dS
1
2) A positive charge is contained inside a spherical shell.
How does the electric flux dФE through the surface
element dS change when the charge is moved from position
1 to position 2?
a) dФE increases
b) dФE decreases
c) dФE doesn’t change
2
dS
1
3) A positive charge is contained inside a spherical shell.
How does the flux ФE through the entire surface change
when the charge is moved from position 1 to position 2?
a) ФE increases
ФE decreases
c) ФE doesn’t change
b)
dS
2
Gauss’ Law
• Gauss’ Law (a FUNDAMENTAL LAW):
The net electric flux through any closed surface is
proportional to the charge enclosed by that surface.
• How do we use this equation??
•The above equation is ALWAYS TRUE but it doesn’t
look easy to use.
•It is very useful in finding E when the physical
situation exhibits massive SYMMETRY.
Gauss’ Law…made easy
•To solve the above equation for E, you have to be able to CHOOSE A
CLOSED SURFACE such that the integral is TRIVIAL.
(1) Direction: surface must be chosen such that E is known to be
either parallel or perpendicular to each piece of the surface;
If
then
If
then
(2) Magnitude: surface must be chosen such that E has the same
value at all points on the surface when E is perpendicular to the
surface.
Gauss’ Law…made easy
•With these two conditions we can bring E outside of the
integral…and:
Note that
is just the area of the Gaussian surface over which
we are integrating. Gauss’ Law now takes the form:
This equation can now be solved for E (at the surface) if we know
qenclosed (or for qenclosed if we know E).
Geometry and Surface Integrals
• If E is constant over a surface, and normal to it everywhere, we
can take E outside the integral, leaving only a surface area
you may use different E’s
for different surfaces
of your “object”
z
c
a
y
b
z
x
R
R
L
Gauss  Coulomb
• We now illustrate this for the field of the
point charge and prove that Gauss’ Law
implies Coulomb’s Law.
• Symmetry E-field of point charge is radial and
spherically symmetric
• Draw a sphere of radius R centered on the charge.
• Why?
E normal to every point on the surface

E has same value at every point on the surface
 can take E outside of the integral!
• Therefore,
!
– Gauss’ Law
– We are free to choose the surface in such
problems… we call this a “Gaussian” surface
E
R
+Q
Uniform charged sphere
What is the magnitude of the
electric field due to a solid
sphere of radius a with uniform
charge density r(C/m3)?
r
a
r
• Outside sphere: (r>a)
– We have spherical symmetry centered on the center of
the sphere of charge
– Therefore, choose Gaussian surface = hollow sphere of
radius r

Gauss’
Law

1
q
4  0 r 2
same as point charge!
Uniform charged sphere
r
• Outside sphere: (r > a)
a
• Inside sphere: (r < a)
r
– We still have spherical symmetry centered on the center of
the sphere of charge.
– Therefore, choose Gaussian surface = sphere of radius r
Gauss’
Law
But,
E
Thus:
a
r
Gauss’ Law and Conductors
• We know that E=0 inside a conductor (otherwise
the charges would move).
• But since
 E  dS  0 
Qinside  0 .
1
Charges on a conductor only
reside on the surface(s)!
+
+
+
+
+
+
+
+
Conducting
sphere
Preflight 4:
A
B
A blue sphere A is contained within a
red spherical shell B. There is a charge
QA on the blue sphere and charge QB
on the red spherical shell.
7) The electric field in the region between the
spheres is completely independent of QB the charge
on the red spherical shell.
True
False
Lecture 4, ACT 1
Consider the following two topologies:
A)
s2
A solid non-conducting sphere
carries a total charge Q = -3 mC
distributed evenly throughout. It
is surrounded by an uncharged
conducting spherical shell.
s1
-|Q|
E
B)
Same as (A) but conducting shell removed
1A
•Compare the electric field at point X in cases A and B:
(a) EA < EB
1B
(b) EA = EB
(c) EA > EB
•What is the surface charge density s1 on the inner
surface of the conducting shell in case A?
(b) s1 = 0
(c) s1 > 0
(a) s1 < 0
Lecture 4, ACT 1
Consider the following two topologies:
A) A solid non-conducting sphere
carries a total charge Q = -3 mC
distributed evenly throughout. It
is surrounded by an uncharged
conducting spherical shell.
1A
s2
s1
-|Q|
E
•Compare the electric field at point X in cases A and B:
(a) EA < EB
(b) EA = EB
(c) EA > EB
• Select a sphere passing through the point X as the Gaussian surface.
•How much charge does it enclose?
•Answer: -|Q|, whether or not the uncharged shell is present.
(The field at point X is determined only by the objects with NET CHARGE.)
Lecture 4, ACT 1
s2
Consider the following two topologies:
A solid non-conducting sphere carries a
total charge Q = -3 mC and is surrounded
by an uncharged conducting spherical
shell.
s1
-|Q|
E
B) Same as (A) but conducting shell removed
1B
•What is the surface charge density s1 on the inner
surface of the conducting shell in case A?
(a) s1 < 0
•
•
•
(b) s1 = 0
(c) s1 > 0
Inside the conductor, we know the field E = 0
Select a Gaussian surface inside the conductor
• Since E = 0 on this surface, the total enclosed charge must be 0
• Therefore, s1 must be positive, to cancel the charge -|Q|
By the way, to calculate the actual value: s1 = -Q / (4  r12)
Infinite Line of Charge
• Symmetry  E-field
must be ^ to line and
can only depend on
distance from line
2
y
Er
Er
• Therefore, CHOOSE
Gaussian surface to be a
+ + +++++++ + +++++++++++++ + + + + + +
cylinder of radius r and
x
length h aligned with the
x-axis.
h
•Apply Gauss’ Law:
• On the ends,
• On the barrel,
AND

NOTE: we have obtained here the same result as we did last lecture using
Coulomb’s Law. The symmetry makes today’s derivation easier.
Lecture 4, ACT 2
• A line charge l (C/m) is placed along
the axis of an uncharged conducting
cylinder of inner radius ri = a, and
outer radius ro = b as shown.
– What is the value of the charge density so
(C/m2) on the outer surface of the
cylinder?
(a)
(b)
(c)
b
a
l
Lecture 4, ACT 2
• A line charge l (C/m) is placed along
the axis of an uncharged conducting
cylinder of inner radius ri = a, and
outer radius ro = b as shown.
– What is the value of the charge density so
(C/m2) on the outer surface of the
cylinder?
a
l
(c)
(b)
(a)
b
View end on:
Draw Gaussian tube which surrounds only the outer edge
so
b
0
Preflight 4:
5) 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 following Gaussian
surfaces are best suited for this purpose?
Note: you may choose more than one answer
a) a cylinder with its axis along the plane
b) a cylinder with its axis perpendicular to the plane
c) a cube
d) a sphere
Infinite sheet of charge,
surface charge density s
• 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:
A
x
E
• On the barrel,
E
• On the ends,
• The charge enclosed =
Therefore, Gauss’ Law
sA

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
Gauss’ Law: Help for the Homework Problems
(See also: mandatory on-line “tutorial exercises”)
• Gauss’ Law is ALWAYS VALID!
• What Can You Do With This?
If you have (a) spherical, (b) cylindrical, or (c) planar symmetry
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).
• Spherical Symmetry: Gaussian surface = sphere of radius r
LHS:
RHS: q = ALL charge inside radius r
• Cylindrical symmetry: Gaussian surface = cylinder of radius r
LHS:
RHS: q = ALL charge inside radius r, length L
• Planar Symmetry: Gaussian surface = cylinder of area A
LHS:
RHS: q = ALL charge inside cylinder =sA
Sheets of Charge
σ1
A
σL
B
σR
C
D
E1 + EL - ER = 0
σ1 + σ L - σ R = 0
σ L + σ R = 0 (uncharged conductor)
σ1 + 2σ L = 0
σ L=
σ1
2
σ R=
+σ1
2
Uncharged Conductor
σ1
EA 
xˆ
2E 0
+σ1
EB 
xˆ
2E 0
EC  0
+σ1
ED 
xˆ
2E 0
Hints:
1.
2.
3.
4.
Assume σ is positive. If it’s negative, the answer will still work.
Assume +xˆ to the right.
Use superposition, but keep signs straight
Think about which way a test charge would move.
Summary
• Gauss’ Law: Electric field flux through a
closed surface is proportional to the net
charge enclosed
– Gauss’ Law is exact and always true….
• Gauss’ Law makes solving for E-field easy
when the symmetry is sufficient
– spherical, cylindrical, planar
• Gauss’ Law proves that electric fields
vanish in conductor
– extra charges reside on surface
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
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
Q1
* The electric field inside a conductor is zero.
R1
(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
+
+
+
+
+
+
+
+
R2
B
• How is the charge distributed on
the spherical shell?
Q2
Q1
* 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, and the outer surface must carry the charge
+Q1 + Q2, so that the net charge on the shell equals Q2.
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?
Q2
Q1
R1
* 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

E  0.
R2
Q1
E  k 2 rˆ
r
1 Q1 + Q 2
2Q1
E
rˆ  k 2 rˆ
2
4 0 r
r
D
-
-
- - - -
• What happens when you connect the two
spheres with a wire? (What are the charges?)
-
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.
- - -
Also, for r < R2
and for r > R2

E  0.

2Q1
E   k 2 r̂
r
Let’s try some numbers
Q1 = 10mC
Q2 = -30mC
B
 sinner = -162 mC/m2
 souter = -325 mC/m2
C
 Electric field
R1 = 5cm
R2 = 7cm
 r < 5cm: Er(r = 4cm) = 0 N/C
 5cm < r < 7cm: Er(r = 6cm) = 2.5 x 107 N/C
 r > 7cm: Er(r = 8cm) = -2.81 x 107 N/C
D
 Electric field
 r > 7cm: Er(r = 9cm) = -2.22 x 107 N/C
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
h
A
•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. Hence, if we choose as our
Gaussian surface a cylinder, which lies inside the cylindrical shell, we know that
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.
•The total charge on the enclosed portion (of length h) of the line charge is:
Total line charge enclosed = 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.
sinner is just Qinner divided by total
area of the cylinder: sinner = -lh / 2Rh = -l / 2R
Therefore, the inner surface charge density
•Notice that the result is independent of h.
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
l
•Whenever we are dealing with electric
E  fields created by symmetric
charged surfaces, we must always first chose an appropriate Gaussian
surface. In this case, for r <R, the surface surrounding the line charge is
actually a cylinder of radius r.
•Using Gauss’ Law, the following equation determines the E-field:
2rhEr = qenclosed / o
qenclosed is the charge on the enclosed line charge,
which is lh, and (2rh) is the area of the barrel of the
Gaussian surface.
r
The result is:
C
What is the Electric field for r>R?
Gaussian surface
l
R
stotal
r
h
• As usual, we must first chose a Gaussian surface as indicated above.
We also need to know the net charge enclosed in our Gaussian
surface. The net charge is a sum of the following:
•Net charge enclosed on the line: lh
•Net charge enclosed within Gaussian surface, residing on the
cylindrical shell: Q= 2Rh stotal
• Therefore, net charge enclosed is Q + lh
• The surface area of the barrel of the Gaussian surface is 2rh
• Now we can use Gauss’ Law: 2rh E = (Q + lh) / o
•You have all you need to find the Electric field now.
Solve for Er to find
Let’s try some numbers
R = 13 cm
h = 168 cm
A1
sinner = -61.21 mC/m2
A2
souter = 589.2 mC/m2
B
•Electric Field (r<R);
stotal = 528 mC/m2
Er(r = 5cm) = 1.798 x 107 N/C
C
•Electric Field (r>R);
Er(r = 20cm) = 4.328 x 107 N/C
l = 50mC/m
Example 3: planes
Suppose there are infinite planes positioned at x1 and x2. The plane
at x1 has a positive surface charge density of s1 while the plane at x2
has negative surface charge density of s2. Find:
A
the x-component of the electric field at a point x>x2
B
the x-component of the electric field at x1<x<x2
C
the x-component of the electric field at a point x<x1
s1
x1
y
s2
x2
x
Solutions
A
Ex(x>x2)
We can use superposition to find
.
The E-field desired is to the right of both sheets. Therefore;
y
s1
x1
s2
x2
E2
x
E1
B
Ex(x1<x<x2)
This time the point is located to the left of s2 and to the right
of s1, therefore;
y
s1
E2
s2
E1
x1
x2
x
C
Ex(x<x1)
When the point is located to the left of both sheets;
y
s1
E1
s2
E2
x1
x2
x
Let’s add some numbers...
x1= -2m
A
x2= 2m
s1 = +2mC/m2
s2 = -3mC/m2
E1x= 1.130 x 105 N/C
Ex= -0.565 x 105 N/C
E2x= -1.695 x 105 N/C
B
E1x= 1.130 x 105 N/C
E2x= 1.695 x 105 N/C
C
E1x= -1.130 x 105 N/C
E2x= 1.695 x 105 N/C
Ex= 2.825 x 105 N/C
Ex= 0.565 x 105 N/C