Transcript Document

c
a
b
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
•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).
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
y
c
a
b
z
x
R
R
L
Gauss  Coulomb
• We now illustrate this for the field of a
point charge and prove that Gauss’
Law implies Coulomb’s Law.
• Symmetry E-field of point charge is
radially and spherically symmetric
• Draw a sphere of radius R centered on
the charge.
E
R
+Q
• 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!
Gauss  Coulomb
• Therefore,
E
2
E

dS

EdS

E
dS

4

R
E



– Gauss’ Law
4 R E 
2
Q
0
– We are free to choose the surface in
such problems… we call this a
“Gaussian” surface
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
E
1
q
4  0 r 2
same as point charge!
Uniform charged sphere
• Outside sphere: (r > a)
a
• Inside sphere: (r < a)
r
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 until net field=0).
• But since
 E  dS  0

Qinside  0 .
s
Charges on a conductor only
reside on the surface(s)!
+
+
+
+
+
+
+
+
Conducting
sphere
Gauss’ Law and Conductors
• The electric field immediately
outside a conductor must be
perpendicular to the conductor
surface.
dS

– Otherwise the charges would move
along the surface until the field was
perpendicular everywhere.
• Applying Gauss’ law to a small
Gaussian cylinder perpendicular
to the surface
• The field just outside a
conductor is perpendicular to the
surface and proportional to the
surface charge density
A
 E  dS  EA   0   0

E
0
Q
Question 1
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.
The electric field in the region between the spheres is
completely independent of QB the charge on the red
spherical shell.
True
False
Question 1
1. a
2. b
0%
b
0%
a
0
of
5
10
Question 1
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.
The electric field in the region between the spheres is
completely independent of QB the charge on the red
spherical shell.
True
False
Infinite Line of Charge, charge/length=
• Symmetry  E-field must be 
to line and can only depend on
distance from line
y
Er
Er
• Therefore, CHOOSE Gaussian
surface to be a cylinder of
radius r and length h aligned
with the x-axis.
+ + +++++++ + +++++++++++++ + + + + + +
•Apply Gauss’ Law:
x
h
• On the ends,
• On the barrel,

NOTE: we have obtained here the same result as we did previously
using Coulomb’s Law. The symmetry makes today’s derivation easier.
Charge density on a conducting cylinder
• A line charge  (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 o
(C/m2) on the outer surface of the
cylinder?
b
a

View end on:
Draw Gaussian tube contained within the conducting cylinder
o
The field within the conducting cylinder is zero
q
 E dS  
q0
0
A charge equal and opposite to the line charge is induced on the
inner conductor surface that cancels the line charge
b
Charge density on a conducting cylinder
•Now draw Gaussian tube which surrounds the
outer edge
•The tube still contains the line charge and the
field is the same as calculated before
b
a

•The charge inside the Gaussian tube also now
Eoutside 

2 0 r
=( charge on the outer surface = 02bL)
+ (charge on inner surface + line charge =0 )
Therefore by Gauss’ Law
o
r
b
•A charge equal to the line charge is induced on the
outer surface of the cylinder
Question 2
Consider the following two topologies:
A)
2
A solid non-conducting sphere
carries a total charge Q = -3 mC
distributed evenly throughout. It
is surrounded by an uncharged
conducting spherical shell.
1
-|Q|
E
B)
Same as (A) but conducting shell removed
•Compare the electric field at point X in cases A and B:
(a) EA < EB
(b) EA = EB
(c) EA > EB
Question 2
1. a
2. b
3. c
0%
c
0%
b
0%
a
0
of
5
10
Question 2
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.
B)
2
1
-|Q|
E
Same as (A) but conducting shell removed
•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.
• It encloses charge -|Q|, whether or not the uncharged shell is present.
(The field at point X is determined only by the objects with NET CHARGE.)
Question 3
Consider the following two topologies:
A)
2
A solid non-conducting sphere
carries a total charge Q = -3 mC
distributed evenly throughout. It
is surrounded by an uncharged
conducting spherical shell.
1
-|Q|
E
B)
Same as (A) but conducting shell removed
•What is the surface charge density 1 on the inner surface of
the conducting shell in case A?
(a) 1 < 0
(b) 1 = 0
(c) 1 > 0
Question 3
1. a
2. b
3. c
0%
c
0%
b
0%
a
0
of
5
10
Question 3
2
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.
1
-|Q|
E
B) Same as (A) but conducting shell removed
•What is the surface charge density 1 on the inner surface of the
conducting shell in case A?
(a) 1 < 0
•
•
•
(b) 1 = 0
(c) 1 > 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, 1 must be positive, to cancel the charge -|Q|
By the way, to calculate the actual value: 1 = -Q / (4  r12)
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
• Chapter 23 of Fishbane
• Try Chapter 23 problems 25, 29, 33, 47, 51, 56