Transcript Electric Field (E)

Chapter 22 Gauss’s Law
• Gauss’s Law is a relationship between the field at all the
points on the surface and the total charge enclosed within
the surface.
• Gauss’s Law is part of the key to using symmetry
considerations to simplify electric-field calculations.
• In Chapter 21 we asked the question “Given a charge
distribution, what is the electric field produced by that
distribution at point P?”
• The answer from Chapter 21 is: The total field at point P is
the vector sum of the fields due to all the point charges.
Chapter 22
1
– The movement of electrons
can be shockingIntroduction
(pardon the
pun).
– If you look at the girl’s hair
(figure to the right), you’ll see
the electrons coating each
individual hair fiber and then
repelling each other.
– Gauss imagined a flow
through a surface placed
around a charge and then
considered outcomes that we
will study in Chapter 22.
The charge within the box determines the magnitude and
direction of the Electric field outside the box.
Twice the charge, twice as many
lines penetrating closed surface
Chapter 22
3
Electric Flux is the “Flow” of the Electric Field in or
out of an enclosed surface
The net electric flux through the surface of the box
is directly proportional to the magnitude of the net
charge enclosed in the box.
Chapter 22
4
Electric Flux = The net sum of the product of the
perpendicular component of E and the surface area
Gauss’s Law – Qualitative statement
The net electric flux (number of lines) penetrating a closed surface is directly
proportional to the total electric charge enclosed by the closed surface and is
independent of the size of the enclosed surface.
Chapter 22
5
The Electric Flux is the scalar product of E and A
En
E
Et
 E  EA
En  E cos 
 
 E  En A  EA cos   E  A
 E  EA cos   0
If the surface (A) is not flat and/or the electric field (E) is not uniform, A is divided into
small elements dA and integrated over the entire surface (A).
Chapter 22
6
Electric Flux Through a disk - Example 22-1 (p.756)
E = 2 x 103 N/C; the disk of radius 0.1m is at an angle.
(a) Disk angle = 30o; electric flux E through the disk? Answer: E = 54 N-m2/C
(b) Disk angle = 90o ; (the disk parallel to E ) electric flux E through the disk?
Answer: E = 0 N-m2/C
(c) Disk angle = 0o; (the disk is normal to E) electric flux E through the disk?
Answer: E = 63 N-m2/C
Chapter 22
7
Electric Flux through a sphere - Example 22-3 (p.757)
q = 3 µC, r = 0.2 m
Determine electric flux through the sphere
Setup


E   E  dA   E cos dA   EdA  E  dA
2
dA

A

4

r

E
q
40 r
2
E  EA
Execute
E
q
40 r
2
 (9 x109 )(3 x10 6 ) /( 0.2) 2  6.75 x105 N / C
E  EA  (6.75x109 )( 4 )(0.20) 2  3.4 x105 N .m2 / C
Evaluate
The is independent of the radius. Larger r produces a
smaller E and a larger A
Chapter 22
8
Gauss’s Law for a point charge in a sphere
1 q
E
2
40 r
1
q
q
2
 E  EA 
( 4r ) 
2
(22-6)
40 r
0
q
E 
(where E is the total electric flux penetrating surface A)
0
Chapter 22
9
Mathematical Representation of Gauss’s Law - General Case
Figure 22-13
E  E cos 
 
d E  E dA  E cos dA  E  dA
Chapter 22
10
Applications of Gauss’s Law
Excess charge on a solid conductor resides entirely on
the conductor’s surface
Note: A Gaussian surface does
not have to be a physical surface
Total electric flux penetrating
imaginary gaussian surface
Qencl = 0
Therefore:
E and E penetrating gaussian surface = 0
This means that the electric field (E) inside the
charged solid conductor is zero.
Chapter 22
11
Problem 22.8
Chapter 22
12
Electric Field (E) outside a Charged Solid Conductor
Sphere (Ex 22-5, p.762)
(imaginary)
q
  Qencl
 E   E  dA 
0
 
 E  dA   E cos dA  E  dA
E  dA  E (4r ) 
2
E ( 4r ) 
2
1 q
E
40 r 2
Figure 22-18
Chapter 22
Qencl
0

q
0
q
0
(r  R)
13
Electric field of a line charge (Example 22.6, p.763)
Problem: Electric charge is distributed uniformly
along an infinite long, thin wire. The charge per
unit length is λ (assume positive). Find the
electric field.
Identify: The system has cylindrical symmetry. The
field must point away from the + charge
Solution: The field lines are radial and lie in a plane perpendicular to the wire.
The field magnitude depends only on the radial distance from the wire.
Execute: No flux out the ends. All flux out the walls of the cylinder, EA
Total flux:  E  ( E )(2rl )
Total enclosed charge: Qencl  l
l
 ( E )( 2rl )
0
l
( E )( 2rl ) 
0
Qencl
From



E
Gauss’s
0
law:
Evaluation: Not valid for short wire. For a
short wire use the method of Example
21.11(p.731).
Chapter 22
Solve for E:
E
1 
20 r
note: independent of length
14
Field between oppositely charged parallel conducting plates (Ex 22-8)
Two charged parallel plates . Charge per unit area is +σ and -σ
Identify: Opposite charges attract. “Fringing” on the edges of plates. Neglect for
large plates.
Setup: Exploit symmetry. S1, S2, S3, S4 are cylinders with end area A
Execute: Consider S1, flux through the end is EA, net charge enclosed is σA
Therefore from Gauss’s Law, the total flux = enclosed charge/Єo
A
EA 
0
Solve for E:
E

0
Evaluate: The same result
was obtained with
superposition of fields .
Example 21.13, p.732
E  E1  E2 
E=0

 


2 0 2 0  0
E=0
Chapter 22
15
There are practical applications
– Figure 22.17 treats excess charge as residing on the surface of a
conductor.
– Consider Example 22.5.
– Figure 22.18 illustrates Example 22.5.
The field of a line or plane of charge
– Consider Example 22.6 and Figure 22.19.
– See also Example 22.7 and Figure 22.20.
The field of a uniformly charged
sphere
Consider Example 22.9.
–
– Figure 22.22 illustrates the example.
– Follow Example 22.10.
Charges on conductors
– The electric field within a charged conductor may be found.
– Consider Figure 22.23.
– Follow Example 22.11 and Figure 22.24.
Excess charge on solid conductors are located
on the surface
The electric field at every point within a conductor is zero
and any excess charge on a solid conductor is located
entirely on the surface.
Chapter 22
20
Experimental tests of Gauss’s Law
– Regard Figure 22.25.
– A metal container on an insulating stand.
The Van de Graaff generator
– The source of all the static on the child’s hair in our introduction.
– Consider Figure 22.27 below.
Prob. 22.17
R =15.0 cm radius sphere
On a Humid day E=2.00 x 104 N/C will produce a one
inch spark
(a)Using Gauss’ Law what is the charge stored on the
surface of the sphere.
6 in spark
(b)Assume all charge is concentrated in the sphere
center, us e Coulomb’s law to calculated the electric
field.
Note : P0 = Є0
Identify: The electric field required to produce a spark 6 in. long is 6 times as strong
as the field needed to produce a spark 1 in. long.
Set Up:
Gauss’s law,
P0 = Є0
By Gauss’s Law solve for the enclosed charge: q  P0 EA
the electric field is the same as for a point-charge,
E
1 q
4 P0 r 2
Execute: (a) The electric field for 6-inch sparks is six times the electric field of a one inch spark
E  6  2.00 104 N/C  1.20 105 N/C
The charge to produce this field is
(b) Using Coulomb’s law gives
q  P0 EA  P0 E (4 r 2 )  (8.85  1012 C2 /N  m 2 )(1.20  105 N/C)(4 )(0.15 m) 2  3.00  107 C
3.00 107 C
E  (9.00 10 N  m /C )
 1.20 105 N/C
2
(0.150 m)
9
2
2
Evaluate: It takes only about 0.3  C to produce this field
Chapter 22
23
A Faraday cage blocks flow
– Refer to Figure 22.28 below.
– Science-fiction movies always place alien transmitters in these
to prevent them from calling for help.
– Follow Examples 22.12 and 22.13.
Chapter-22 Important Equations
Chapter 22
25
Chapter-22 Important Concepts
Chapter 22
26
Chapter-22 Important concepts and results
Chapter 22
27
Q22.1
A spherical Gaussian surface (#1) encloses and
is centered on a point charge +q. A second
spherical Gaussian surface (#2) of the same
size also encloses the charge but is not centered
on it.
+q
Compared to the electric flux through surface
#1, the flux through surface #2 is
Gaussian
surface #1
A. greater.
B. the same.
C. less, but not zero.
D. zero.
E. not enough information given to decide
Gaussian
surface #2
Q22.4
Conductor
A solid spherical conductor has a spherical cavity in
its interior. The cavity is not centered on the center of
the conductor.
If a positive charge is placed on the conductor, the
electric field in the cavity
A. points generally toward the
outer surface of the conductor.
B. points generally away from the outer surface of the conductor.
C. is zero.
D. not enough information given to decide
Cavity