Set No. 2. Gauss's Law and its applications

About gauss law etc.

1. Lecture Set 3Lecture Set 3
Gauss’s LawGauss’s Law
Spring 2007Spring 2007

2. 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

3. 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.

4. 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.

5. 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

6. Definition – Sort of -
Electric Field Lines

7. Field Lines  Electric Field

8. Last time we showed that

9. Ignore the Dashed Line …
Remember last time .. the big plane?

0

0

0

0

0

0
E=0 
0 E=0
We will use this a lot!

10. 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

11. What would you guess is inside the cube?

12. What about now?

13. 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

15. Which box do you think contains more
charge?

16. 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.

17. 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?

18. Mr. Gauss
answered the
question with..

19. Another QUESTION:
Solid Surface
Given the electric field at EVERY point
on a closed surface, can we determine
the charges that caused it??

20. 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??

21. Still another question
Given a small area, how can
you describe both the area
itself and its orientation with a
single stroke!

22. 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
A
n, 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.

23. The “normal component” of the
ELECTRIC FIELD
E
n
E
n
nEnE
nE


)cos(
)cos(


E
E
n
n

24. DEFINITION FLUX
E
n
E
n
)cos(
)(
AE
nEAE

 AFlux
n

25. We will be considering CLOSED
surfaces

nEnE  )cos(
n
E
The normal vector to a closed surface is DEFINED as positive
if it points OUT of the surface. Remember this definition!

26. “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.

27. This flux is LEAVING the closed
surface.


28. Definition of TOTAL FLUX through
a surface




dA
is surface aLEAVING Field
Electric theofFlux Total
out
surface
d
nE

29. Flux is
33% 33%33%
1.A vector
2.A scaler
3.A triangle

30. Visualizing Flux

ndAEflux
n is the OUTWARD
pointing unit normal.

31. Definition: A Gaussian Surface
Any closed surface that
is near some distribution
of charge

32. Remember

ndAEflux
)cos(nEnE
n

E

A

Component of E
perpendicular to
surface.
This is the flux
passing through
the surface and
n is the OUTWARD
pointing unit normal
vector!

33. Example
Cube in a UNIFORM Electric Field
L
E
E is parallel to four of the surfaces of the cube so the flux is zero across these
because E is perpendicular to A and the dot product is zero.
Flux is EL
2
Total Flux leaving the cube is zero
Flux is -EL
2
Note sign
area

34. Simple Example
0
2
2
0
2
0
2
0
2
0
4
4
1
4
1
4
1
4
1





q
r
r
q
A
r
q
dA
r
q
dA
r
q
dA
Sphere





nE
r
q

35. Gauss’ Law
n is the OUTWARD
pointing unit normal.
0
0


enclosed
n
enclosed
q
dAE
q
ndAE




q is the total charge ENCLOSED
by the Gaussian Surface.
Flux is total EXITING the
Surface.

36. Simple Example
UNIFORM FIELD LIKE BEFORE
E
A AE E
0
0


q
EAEA
N
o
Enclosed C
harge

37. Line of Charge
L
Q
L
Q

length
charge


38. Line of Charge
From SYMMETRY E is
Radial and Outward
r
k
rr
E
h
rhE
q
dAE
n









2
4
2
2
2
00
0
0




39. What is a Cylindrical
Surface??
Ponder

40. Looking at A Cylinder from its END
Circular Rectangular
Drunk

41. Infinite Sheet of Charge

cylinder
E
h
0
0
2





E
A
EAEA
We got this same
result from that
ugly integration!

42. 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!

43. 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.

44. Isolated Conductor
Electric Field is ZERO in
the interior of a conductor.
Gauss’ law on surface shown
Also says that the enclosed
Charge must be ZERO.
Again, all charge on a
Conductor must reside on
The SURFACE.

45. Charged Conductors
E=0
E
-
-
-
-
-
Charge Must reside on
the SURFACE

0
0






E
or
A
EA
Very SMALL Gaussian Surface

46. 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

47. Isolated (Charged) Conductor with
a HOLE in it.
0
0

Q
dAE
n

Because E=0 everywhere
inside the conductor.
So Q (total) =0 inside the hole
Including the surface.

48. A Spherical Conducting Shell with
A Charge Inside.

49. 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.

50. Example – A Spatial Distribution of
charge.
Uniform charge density = charge per unit volume
0
0
3
0
2
0
3
1
3
4
4








r
E
r
V
rE
q
dAE
n



(Vectors)
r
E
O
A Solid SPHERE

51. Outside The Charge
r
E
O
R
2
0
0
3
0
2
0
4
1
3
4
4
r
Q
E
or
Q
RrE
q
dAE
n










Old Coulomb Law!

52. Graph
R
E
r

53. Charged Metal Plate
E is the same in magnitude EVERYWHERE. The direction is
different on each side.
E

+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
E
A
A

54. Apply Gauss’ Law

+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
E
A
A
AEAEAEA
Bottom
E
AEAAEA
Top
0
0
0
22
0









Same result!

55. Negatively Charged
ISOLATED Metal Plate
--
-

E is in opposite direction but
Same absolute value as before

56. Bring the two plates together
A







e
e
B
As the plates come together, all charge on B is attracted
To the inside surface while the negative charge pushes the
Electrons in A to the outside surface.
This leaves each inner surface charged and the outer surface
Uncharged. The charge density is DOUBLED.

57. Result is …..
A

 

e
e
B

E
E=0
E=0
0
1
0
0
2








E
A
EA

58. 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.

59. Problem #1
Trick Question
Consider a cube with each edge = 55cm. There is a 1.8 C charge
In the center of the cube. Calculate the total flux exiting the cube.
CNm
q
/1003.2
1085.8
108.1
25
12
6
0







NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!
Easy, yes??

60. Note: the problem is poorly stated in the text.
Consider an isolated conductor with an initial charge of 10 C on the
Exterior. 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

61. Another Problem


m,q both given as is 
0
0
2
2







E
A
EA
Gauss
Gaussian
Surface
Charged Sheet

62. -2


m,q both given as is 
mg
qE
T
Free body diagram

0
2
)sin(
)cos(




q
qET
mgT



63. -3
2
90
0
1003.5
)tan(2
2
)(
m
C
q
mg
and
mg
q
Tan
Divide








(all given)

64. A Last Problem
A uniformly charged cylinder.


R
r
R
E
hRq
rhE
Rr
r
E
hr
rhE
Rr
0
2
0
2
0
0
0
2
2
)(
)2(
2
)(
)2(
















