chapter 21: gauss’s law gauss’s law : introduction the total electric flux through a closed...
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Chapter 21: Gauss’s Law
Gauss’s law : introduction
The total electric flux through a closed surface is equal to the total (net) electric charge inside the surface divided by
Gauss’s law is equivalent to Coulomb’s law
Gauss’s law : introduction (cont’d)
Consider a distribution of charge• Surround it with an imaginary surface that encloses the charge• Look at the electric field at various points on this imaginary surface
q
E
E
E
imaginarysurface
0q
FE
0q
F test charge
• To find out charge distribution inside the imaginary surface, we need to measure electric fields especially on the surface• To do that place a test charge of a known charge amount and measure the electric force+
+
Charge and Electric Flux
Electric fields by different charges
+
q q2
qq2
+ +
-
-
-
outwardflux
outwardflux
inwardflux
inwardflux
qE
qq EE
22
qq EE
qq EE
22
Charge and Electric Flux
Electric flux
+
q q2+ +outward
fluxoutwardflux
+
qWhen the distance to the surface doubled,the area of the surfacequadrupled andthe electric field becomes ¼.
qE
qq EE
22
)(4
1)2( rErE qq
Charge and Electric Flux
Definition of electric flux For any point on a small area of a surface, take the product ofthe average component of E perpendicular to the surface andthe area. Then the sum of this quantity over the surface is thenet electric flux
• Whether there is a net outward or inward electric flux through a closed surface depends on the sign of the enclosed charge.• Charges outside the surface do not give a net electric flux through the surface.
• The net electric flux is directly proportional to the net amount of charge enclosed within the surface but is otherwise independent of the size of the closed surface.
A qualitative statement of Gauss’s law
Calculating Electric Flux
Analogy between electric flux and field of velocity vector A good analogy between the electric flux and the field of velocityvector in a flowing fluid can be found.
A (area)
A
velocity vector (flow speed)
A vector area that defines the plane of the area,perpendicular to the plane
A
volume flow rate: Adt
dV
volume flow rate:
A
AAAdt
dV
cos
nAAAA
;cos;cos
A
Calculating Electric Flux
Analogy between electric flux and field of velocity vector
A (area)
A E
Electric field vector
A vector area that defines the plane of the area,perpendicular to the plane
A
electric flux: EAE
electric flux:
AAAdt
dV cos
cos;cos AAEE
Adt
dV
AEAEEA
cos
volume flow rate:
E
E field isperpendicularto this plane
Calculating Electric Flux
A small area element and flux
AdEd E
Total flux for an area
dAnAdAdEdAEdAEd EE
;cos
Example: Electric flux through a disk
E
Ar = 0.10 m
30
C/mN54
30cos)m4N/C)(0.031100.2(cos
m0314.0m)10.0(
2
23
22
EA
A
E
Calculating Electric Flux
Example : Electric flux through a cube
1n̂2n̂3n̂
4n̂
5n̂
6n̂
L
E
221 180cosˆ
1ELELAnEE
222 0cosˆ
2ELELAnEE
090cos2
6543 ELEEEE
06
1
i
i EE i
Calculating Electric Flux
Example : Electric flux through a sphere
+
+q
r=0.20 m
Ad
q=3.0 CA=4r2
C/mN104.3
)(0.20m)N/C)(41075.6(
N/C1075.6
(0.20m)
C100.3)C/mN100.9(
4
//ˆ//,
25
25E
5
2
6229
20
EAEdA
r
qE
AdnEEE
Gauss’s Law
Preview: The total electric flux through any closed surface (a surface enclosing a definite volume) is proportional to the total (net) electric charge insidethe surface.
Case 1: Field of a single positive charge q
+
q
r=R
E
E
204
1
R
qE
A sphere with r=R
at r=R
surfaceE
0
2
0
)4(4
1
q
RR
qEAE
The flux is independent of the radius R ofthe surface.
Gauss’s Law Case 1: Field of a single positive charge q (cont’d)
+
q
r=R
r=2R
RE
RR EE
4
12
dAdAR
dAdA R 42
Every field line that passes throughthe smaller sphere also passes throughthe larger sphere
The total flux through each sphere isthe same
The same is true for any portion ofits surface such as dA
RR ERRRRRRE ddAEdAEdAEd2224
4
1
0q
AdEE
This is true for a surface of any shape or sizeprovided it is a closed surface enclosing the charge
Gauss’s Law Case 2: Field of a single positive charge (general surface)
+
q
E
dA
+
EcosE
EnE
dA
cosdAsurface perpendicularto E
dAEdAEd E cos
0q
AdEE
Gauss’s Law Case 3: An closed surface without any charge inside
0AdEE
+
Electric field lines that go in come out.Electric field lines can begin or end insidea region of space only when there is chargein that region.
Gauss’s law
The total electric flux through a closed surface is equal to the total(net) electric charge inside the surface divided by
i ii iencl
enclE EEqQ
QAdE
,;
0
Applications of Gauss’s Law Introduction
• The charge distribution the field
• The symmetry can simplify the procedure of application
Electric field by a charge distribution on a conductor
• When excess charge is placed on a solid conductor and is at rest, it resides entirely on the surface, not in the interior of the material (excess charge = charge other than the ions and free electrons that make up the material conductor
A Gaussian surface inside conductor
Charges on surface
Conductor
Applications of Gauss’s Law
Electric field by a charge distribution on a conductor (cont’d)
A Gaussian surface inside conductor
Charges on surface
Conductor
• Draw a Gaussian surface inside of the conductor• E=0 everywhere on this surface (inside conductor)• The net charge inside the surface is zero• There can be no excess charge at any point within a solid conductor• Any excess charge must reside on the conductor’s surface• E on the surface is perpendicular to the surface
Gauss’s law
E at every point in the interior of a conducting materialis zero in an electrostatic situation (all charges are at rest).If E were non-zero, then the charges would move
Applications of Gauss’s Law
Example: Field of a charged conducting sphere with q
+
+
+
+
+ +
++
R
R 2R 3R
Gaussian surface
0: ERr
:rR Draw a Gaussian surfaceoutside the sphere
200
2
2
4
1)(4
:law sGauss'
surface sphere thelar toperpendicu
surface sphere on the const.
4,
r
qE
qrE
E
E
rAqQelcl
204
1:
R
qERr
204
1
R
qER
4/RE9/RE r
E
Applications of Gauss’s Law
Example: Field of a line charge
line chargedensity
Gaussian surfacechosen according to symmetry
EEE ,
Ad
enclQ
surfaceGaussian lcylindrica on theEE
0
)2(
rEE
rE
02
1
Applications of Gauss’s Law
Example: Field of an infinite plane sheet of charge
++
+
+
+
+
+++ +
+
density charge:
Gaussian surface
E E
EEE sheet the
AQencl
0
)(2A
EAE
02
E
two end surfacesarea A area A
Applications of Gauss’s Law
Example : Field between oppositely charged parallel conducting plane
+++++++++
---------
plate 1 plate 2
2E
1E
1E
1E
E
2E
2E
ab
c
S1
S2 S3
S4
Solution 1:
No electric fluxon these surfaces
surface)left (0
surface)(right :00
1
E
EA
EAS
surface)right (0
surface)(left :00
4
E
EA
EAS
Solution 2:
inward flux
outward flux
At Point a : 02121
EEEEE
b :
c :
0021 2
2
EEE
02121
EEEEE
Applications of Gauss’s Law
Example : Field of a uniformly charged sphere Gaussian surface
r=R
R
++
++
+ +
++
+ +
+
+
3
34
density charge
R
Q
03 /)
3
4(: rEARr
30
032
4
1
/)3
4()4(
R
QrE
rrE
204
1:
R
QERr
20
0
2
4
1
)4(:
r
QE
QrErR
Applications of Gauss’s Law
Example : Field of a hollow charged (uniformly on its surface) sphere
R=0.250 m
r=0.300 m
E
Hollow charged sphere
N/C1080.1 2E
nC801.0)4(
)4(
20
0
2
rEq
qrEdAE
EE
E
Gaussian surface
Charges on Conductors
Case 1: charge on a solid conductor resides entirely on its outer surface in an electrostatic situation
++
++
++++
+++++ + +
+
Case 2: charge on a conductor with a cavity
++
++
++++
+++++ + +
+ Gauss surface
The electric field at every point within a conductoris zero and any excess charge on a solid conductor is located entirely on its surface.
If there is no charge within the cavity, the netcharge on the surface of the cavity is zero.
Charges on Conductors
Case 3: charge on a conductor with a cavity and a charge q inside the cavity
++
++
++++
+++++ + +
+
Gauss surface
+--- --
--
• The conductor is uncharged and insulated from charge q.• The total charge inside the Gauss surface should be zero from Gauss’s law and E=0 on this surface. Therefore there must be a charge –q distributed on the surface of the cavity.• The similar argument can be used for the case where the conductor originally had a charge qC. In this case the total charge on the outer surface must be q+qC after charge q is inserted in cavity.
Charges on Conductors
Faraday’s ice pail experiment
conductor
charged conducting ball
(1) Faraday started with a neutral metal ice pail (metal bucket) and an uncharged electroscope. (2) He then suspended a positively charged metal ball into the ice pail, being careful not to touch the sides of the pail. The leaves of the electroscope diverged. Moreover, their degree of divergence was independent of the metal ball's exact location. Only when the metal ball was completely withdrawn did the leaves collapse back to their original position.
Charges on Conductors
Faraday’s ice pail experiment (cont’d)
conductor
charged conducting ball
(3) Faraday noticed that if the metal ball was allowed to contact the inside surface of the ice pail, the leaves of the electroscope remained diverged (4) Afterwards, when he completely removed the ball from the inside of the ice pail, the leaves remained diverged. However, the metal ball was no longer charged. Since the leaves of the electroscope that was attached to the OUTSIDE of the pail did not move when the ball touched the inside of the pail, he concluded that the inner surface had just enough charge to neutralize the ball.
Charges on Conductors
Field at the surface of a conductor
• The electric field just outside a conductor has
magnitude /0 and is directed perpendicular to
the surface.
Draw a small pill box that extends
into the conductor. Since there is
no field inside, all the flux comes
out through the top.
EA=q/0= A/ 0,
E= / 0
Exercises
Exercise 1
This is the same as the field due to apoint charge with charge +2Q
Exercises
Exercise 1 (cont’d)
Exercises
Exercise 2: A sphere and a shell of conductor
R1
R2
Q1
Q2
Q2=-3Q1 • From Gauss’s law there can be no net charge inside the conductor, and the charge must reside on the outside surface of the sphere
• There can be no net charge inside the conductor. Therefore the inner surface of the shell must carry a net charge of –Q1 , and the outer surface must carry the charge +Q1+Q2 so that the net charge on the shell equals Q2 . These charges are uniformly distributed.
22
122
1222
1
4
2
44 R
Q
R
R
Qouterinner
Exercises
Exercise 2: A sphere and a shell of conductor (cont’d)
R1
R2
Q1
Q2
Q2=-3Q1
rr
Qkr
r
QQkErR
rr
QkERrR
ERr
ˆ2
ˆ:
ˆ:
0:
21
221
2
21
21
1
Exercises
Exercise 3: Cylinder
outer
R
inner
h
total
An infinite line of charge passes directly through the middle of a hallowcharged conducting infinite cylindrical shell of radius R. Let’s focus ona segment of the cylindrical shell of length h. The line charge has a linearcharge density , and the cylindrical shell has a net surface charge densityof total.
Exercises
Exercise 3: Cylinder (cont’d)
outer
R
inner
h
total
The electric field inside the cylindrical shell is zero. Therefore if we chooseas a Gaussian surface a cylinder, which lies inside the cylindrical shell, thenet charge enclosed is zero. There is a surface charge density on the insidewall of the cylinder to balance out the charge along the line.
Exercises
Exercise 3: Cylinder (cont’d)
outer
R
inner
h
total
• The total charge on the enclosed portion (length h) of the line charge:
• The charge on the inner surface of the conducting cylinder shell:
hhQinner
RRh
hinner
22
Exercises
Exercise 3: Cylinder (cont’d)
outer
R
inner
h
total
• The net charge density on the cylinder:
• The outer charge density :
total
Rtotalinnertotalouter
2
outer
Exercises
Exercise 3: Cylinder (cont’d)
outer
R
inner
h
total
• Draw a Gaussian surface surrounding the line charge of radius r (< R)
Rrfor2
,200
r
Ehqq
rhE renclencl
r
Exercises
Exercise 3: Cylinder (cont’d)
outer
R
inner
h
total
• Draw a Gaussian surface surrounding the line charge of radius r (>R)
Rrfor2
,2000
rr
REhQq
qrhE total
renclencl
r
• Net charge enclosed on the line: Net charge enclosed on the shell:h totalRhQ 2