lecture set 3 gausss law. calendar for the weekj today (monday) –one or two problems...
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Lecture Set 3Lecture Set 3Gauss’s LawGauss’s Law
Calendar for the WeekjCalendar for the Weekj
Today (Monday)Today (Monday)– One or two problemsOne or two problems– Introduction to the concept of FLUXIntroduction to the concept of FLUX
Wednesday, FridayWednesday, Friday– Gauss’s Law & some problemsGauss’s Law & some problems
EXAM DATE WILL BE NEXT EXAM DATE WILL BE NEXT WEDNESDAY through GaussWEDNESDAY through Gauss
Protons are projected with an initial speed vi = 9.83 103 m/s into a region where a uniform electric field E = (-720 j) N/C is present, as shown in Figure P23.49. The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons cross the plane and enter the electric field in Figure 23.49.
Summary from last week
222
,2
2
2
)()()(
r
rdsk
r
rdAk
r
rdVk
r
Qk
q
General
r
Qk
q
r
qQk
unitjj
jjj
unit
unit
E
rF
EE
rF
E
rF
(Note: I left off the unit vectors in the lastequation set, but be aware that they should
be there.)
VECTOR VECTOR
Electric Field
We will now introduce a convenient way to represent the overall electric field in a region of space.
It is kinda sorta a map of the field strength. We will then introduce a new concept
FLUX We will use this new concept to introduce Gauss’s
Law
Look at the “Field Lines” of an infinite sheet of charge …
How do you do that?
Ignore the Dashed Line … Remember last time .. the big plane?
00
00
0
0
E=0 0 E=0
NEW RULES (Bill Maher)
Imagine a region of space where the ELECTRIC FIELD LINED HAVE BEEN DRAWN.
The electric field at a point in this region is TANGENT to the Electric Field lines that have been drawn.
If you construct a small rectangle normal to the field lines, the Electric Field is proportional to the number of field lines that cross the small area. The DENSITY of the lines.
Point Charges
They don’t like each other …
Mr. Gauss …
Recall…
We were given Coulomb’s Law We defined the electric field. Calculated the Electric Field given a distribution
of charges using Coulomb’s Law.
unitVi uniti
r
rdqk
r
qrrE
2210
)(
4
1
(Units: N / C)
A Question:
Given the magnitude and direction of the Electric Field at a point, can we determine the charge distribution that created the field?
Is it Unique? Question … given the Electric Field at a
number of points, can we determine the charge distribution that caused it? How many points must we know??
Another QUESTION:
Solid Surface
Given the electric field at EVERY point
on a closed surface, can we determinethe charges that caused it??
Still another question
Given a small area, how can you describe both the area itself and its orientation with a single stroke!
The “Area Vector”
Consider a small area. It’s orientation can be described by a
vector NORMAL to the surface. We usually define the unit normal vector n. If the area is FLAT, the area vector is given by
An, where A is the area. A is usually a differential area of a small part
of a general surface that is small enough to be considered flat.
The normal component of a vector
nEnE )cos(nEThe normal vector to a closed surface is DEFINED as positiveif it points OUT of the surface. Remember this definition!
ANOTHER DEFINITION:Element of Flux through a surface
EENORMAL
NORMAL
A E=|ENORMAL| | |A|
(a scalar)
“Element” of Flux of a vector E leaving a surface
dAd
also
d NORMAL
nEAE
AEAE
n is a unit OUTWARD pointing vector.
This flux was LEAVING the closed surface
Definition of TOTAL FLUX through a surface
dA
is surface aLEAVING Field
Electric theofFlux Total
out
surfaced
nE
Visualizing Flux
ndAEflux
n is the OUTWARD pointing unit normal.
Definition: A Gaussian Surface
Any closed surface thatis near some distribution
of charge
Remember
ndAEflux
)cos(nEnE
n E
A
Component of Eperpendicular tosurface.
This is the fluxpassing throughthe surface andn is the OUTWARDpointing unit normalvector!
ExampleCube in a UNIFORM Electric Field
L
E
E is parallel to four of the surfaces of the cube so the flux is zero across thesebecause E is perpendicular to A and the dot product is zero.
Flux is EL2
Total Flux leaving the cube is zero
Flux is -EL2
Note sign
area
Simple Example
0
22
0
20
20
20
44
1
4
1
4
1
4
1
qr
r
q
Ar
qdA
r
q
dAr
qdA
Sphere
nE
r
q
Gauss’ Law
n is the OUTWARD pointing unit normal.
0
0
enclosedn
enclosed
qdAE
qndAE
q is the total charge ENCLOSEDby the Gaussian Surface.
Flux is total EXITING theSurface.
Simple ExampleUNIFORM FIELD LIKE BEFORE
E
A AE E
00
q
EAEA
No
Enclosed Charge
Line of Charge
L
Q
L
Q
length
charge
Line of Charge
From SYMMETRY E isRadial and Outward
r
k
rrE
hrhE
qdAEn
2
4
2
2
2
00
0
0
What is a Cylindrical Surface??
Ponder
Looking at A Cylinder from its END
Circular RectangularDrunk
Infinite Sheet of Charge
cylinderE
h
0
0
2
E
AEAEA
We got this sameresult from thatugly integration!
Materials
Conductors Electrons are free to move. In equilibrium, all charges are a rest. If they are at rest, they aren’t moving! If they aren’t moving, there is no net force on them. If there is no net force on them, the electric field must be
zero.
THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO!
More on Conductors
Charge cannot reside in the volume of a conductor because it would repel other charges in the volume which would move and constitute a current. This is not allowed.
Charge can’t “fall out” of a conductor.
Isolated Conductor
Electric Field is ZERO inthe interior of a conductor.
Gauss’ law on surface shownAlso says that the enclosedCharge must be ZERO.
Again, all charge on a Conductor must reside onThe SURFACE.
Charged Conductors
E=0
E
---
-
-
Charge Must reside onthe SURFACE
0
0
E
or
AEA
Very SMALL Gaussian Surface
Charged Isolated Conductor
The ELECTRIC FIELD is normal to the surface outside of the conductor.
The field is given by:
Inside of the isolated conductor, the Electric field is ZERO.
If the electric field had a component parallel to the surface, there would be a current flow!
0
E
Isolated (Charged) Conductor with a HOLE in it.
0
0Q
dAEn
Because E=0 everywhereinside the surface.
So Q (total) =0 inside the holeIncluding the surface.
A Spherical Conducting Shell with
A Charge Inside.
Insulators
In an insulator all of the charge is bound. None of the charge can move. We can therefore have charge anywhere in
the volume and it can’t “flow” anywhere so it stays there.
You can therefore have a charge density inside an insulator.
You can also have an ELECTRIC FIELD in an insulator as well.
Example – A Spatial Distribution of charge.
Uniform charge density = charge per unit volume
0
0
3
0
2
0
3
1
3
44
rE
rV
rE
qdAEn
(Vectors)
r EO
A Solid SPHERE
Outside The Charge
r
E
O
R
20
0
3
0
2
0
4
1
3
44
r
QE
or
QRrE
qdAEn
Old Coulomb Law!
Graph
R
E
r
Charged Metal Plate
E is the same in magnitude EVERYWHERE. The direction isdifferent on each side.
E
++++++++
++++++++
E
A
A
Apply Gauss’ Law
++++++++
++++++++
E
A
A
AEAEAEA
Bottom
E
AEAAEA
Top
0
0
0
22
0
Same result!
Negatively ChargedISOLATED Metal Plate
---
E is in opposite direction butSame absolute value as before
Bring the two plates together
A
ee
B
As the plates come together, all charge on B is attractedTo the inside surface while the negative charge pushes theElectrons in A to the outside surface.
This leaves each inner surface charged and the outer surfaceUncharged. The charge density is DOUBLED.
Result is …..
A
ee
B
EE=0
E=0
0
1
0
0
2
E
AEA
VERY POWERFULL IDEA
Superposition The field obtained at a point is equal to the
superposition of the fields caused by each of the charged objects creating the field INDEPENDENTLY.
Problem #1Trick Question
Consider a cube with each edge = 55cm. There is a 1.8 C chargeIn the center of the cube. Calculate the total flux exiting the cube.
CNmq
/1003.21085.8
108.1 2512
6
0
NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!
Easy, yes??
Problem #2(15 from text)Note: the problem is poorly stated in the text.
Consider an isolated conductor with an initial charge of 10 C on theExterior. A charge of +3mC is then added to the center of a cavity.Inside the conductor.
(a) What is the charge on the inside surface of the cavity?(b) What is the final charge on the exterior of the cavity?
+3 C added+10 C initial
Another Problem from the book
m,q both given as is
0
0
2
2
E
AEA
Gauss
GaussianSurface
Charged Sheet
-2
m,q both given as is
mg
qE
T
Free body diagram
02)sin(
)cos(
q
qET
mgT
-3
290
0
1003.5)tan(2
2)(
mC
q
mg
and
mg
qTan
Divide
(all given)
A Last ProblemA uniformly charged cylinder.
R
r
RE
hRqrhE
Rr
rE
hrrhE
Rr
0
2
0
2
0
0
0
2
2
)()2(
2
)()2(
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