Post on 25-Feb-2016
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Flux andFlux andGauss’s LawGauss’s Law
Spring 2008Spring 2008
Last Time: Definition – Sort of –
Electric Field Lines
DIPOLE FIELD LINKCHARGE
Field Lines Electric Field
Last time we showed that
Ignore the Dashed Line … Remember last time .. the big plane?
00
00
0
0
E=0 0 E=0
We will use this a lot!
NEW RULES (Bill Maher)
Imagine a region of space where the ELECTRIC FIELD LINES 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. We won’t use this much
What would you guess is inside the cube?A. A positive
chargeB. A negative
chargeC. Can’t tellD. An AlienE. A car
What about now?
A. A positive charge
B. A negative charge
C. Can’t tellD. An AlienE. A car
How about this??
A. Positive point chargeB. Negative point chargeC. Large Sheet of chargeD. No chargeE. You can’t tell from this
Which box do you think contains more charge?
A. TopB. BottomC. Can’t tellD. Don’t care
All of the E vectors in the bottom box are twice as large as those coming from the top box. The top box contains a charge Q. How much charge do you think is in the bottom box?
1. Q2. 2Q3. You can’t tell4. Leave me alone.
So far …
The electric field exiting a closed surface seems to be related to the charge inside.
But … what does “exiting a closed surface mean”?
How do we really talk about “the electric field exiting” a surface? How do we define such a concept? CAN we define such a concept?
Mr. Gauss answered the question with..
Another QUESTION:
Solid Surface
Given the electric field at EVERY pointon a closed surface, can we determinethe charges that caused it??
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??
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 the ELECTRIC FIELD
E
nEn
nEnEnE
)cos(
)cos(
E
E
n
n
DEFINITION FLUX
E
nEn
)cos(
)(
AE
nEAE
AFlux n
We will be considering CLOSED surfaces
nEnE )cos(nEThe normal vector to a closed surface is DEFINED as positiveif it points OUT of the surface. Remember this definition!
“Element” of Flux of a vector E leaving a surface
dAddalso
dd NORMAL
nEAE
AEAE
For a CLOSED surface:n is a unit OUTWARD pointing vector.
This flux is LEAVING the closed surface.
Definition of TOTAL FLUX through a surface
dA
is surface aLEAVING FieldElectric theofFlux Total
out
surfaced
nE
Flux is
A. A vectorB. A scalerC. A triangle
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
14
14
1
41
qrrq
ArqdA
rq
dArqdA
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
qEAEA
No
Enclosed Charge
Line of Charge
L
Q
LQ
lengthcharge
Line of Charge
From SYMMETRY E isRadial and Outward
rk
rrE
hrhE
qdAEn
24
22
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=0E
- --
-
-
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
0QdAE
n
Because E=0 everywhereinside the conductor.
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
1344
rE
rVrE
qdAEn
(Vectors)r EO
A Solid SPHERE
Outside The Charge
r
EO
R
20
0
3
0
2
0
41
344
rQE
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 /1003.21085.8108.1 25
12
6
0
NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!
Easy, yes??
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
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(
qqET
mgT
-3
290
0
1003.5)tan(2
2)(
mC
qmg
andmgqTan
Divide
(all given)
A Last ProblemA uniformly charged cylinder.
R
rRE
hRqrhE
Rr
rE
hrrhE
Rr
0
2
0
2
0
0
0
2
2
)()2(
2
)()2(