lecture set 3 gauss’s law spring 2007. calendar for the week today (wednesday) –one or two...
TRANSCRIPT
Calendar for the WeekCalendar for the Week
Today (Wednesday)Today (Wednesday)– One or two problems on One or two problems on EE– Introduction to the concept of FLUXIntroduction to the concept of FLUX
FridayFriday– 7:30 session if wanted7:30 session if wanted– Quiz – Chapter 23 – Electric FieldQuiz – Chapter 23 – Electric Field– Gauss’s Law & some problemsGauss’s Law & some problems
EXAM APPROACHINGEXAM APPROACHING
NoteNote
Pleas read the material on Electric Pleas read the material on Electric DipolesDipoles
We will NOT cover it in class but it is We will NOT cover it in class but it is part of the course.part of the course.
We will use it later in the semester.We will use it later in the semester.
It It couldcould show up on the exam. show up on the exam.
Approximate ScheduleApproximate Schedule
EXAM NUMBEREXAM NUMBER DATEDATE
11 2/2 (Fri)2/2 (Fri)22 3/2 (Fri)3/2 (Fri)
33 4/4 (Wed)4/4 (Wed)Fri. is Good FridayFri. is Good Friday
FINAL EXAMFINAL EXAM Check UCF Cal.Check UCF Cal.
The figure shows two concentric rings, of radii R and R ' = 2.95R, that lie on the same plane. Point P lies on the central z axis, at distance D = 2.20R from the center of the rings. The smaller ring has uniformly distributed charge +Q. What must be the uniformly distributed charge on the larger ring if the net electric field at point P due to the two rings is to be zero?[-3.53] Q
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
How about this??
20% 20% 20%20%20%
1. Positive point charge
2. Negative point charge
3. Large Sheet of charge
4. No charge
5. You can’t tell from this
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?
0% 0%0%0%
1. Q
2. 2Q
3. You can’t tell
4. 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?
Another QUESTION:
Solid Surface
Given the electric field at EVERY point
on 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.
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
dAdd
also
dd NORMAL
nEAE
AEAE
For a CLOSED surface:n is a unit OUTWARD pointing vector.
Definition of TOTAL FLUX through a surface
dA
is surface aLEAVING Field
Electric theofFlux Total
out
surfaced
nE
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
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.
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 conductor.
So Q (total) =0 inside the holeIncluding the surface.
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
Charged Metal Plate
E is the same in magnitude EVERYWHERE. The direction isdifferent on each side.
E
++++++++
++++++++
E
A
A
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.
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??
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