Transcript Chapter 21: Electric Charge and Electric Field

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 e0
 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

E

E
+
+
q0

F
q
imaginary
surface

E
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

 F
E
q0
Charge and Electric Flux
 Electric

E q
outward
flux
fields by different charges


E 2 q  2 E q
q
+
 2q
+ +


Eq   Eq
inward
flux
q
-
 2q
-
-
outward
flux


E2 q  2Eq
inward
flux
Charge and Electric Flux
 Electric
flux


E 2 q  2 E q

E q
outward
flux
q
 2q
+
+ +

1 
E q ( 2r )  E q ( r )
4
q
+
outward
flux
When the distance to
the surface doubled,
the area of the surface
quadrupled and
the electric field
becomes ¼.
Charge and Electric Flux

Definition of electric flux
For any point on a small area of a surface, take the product of
the average component of E perpendicular to the surface and
the area. Then the sum of this quantity over the surface is the
net electric flux

A qualitative statement of Gauss’s law
• 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.
Calculating Electric Flux

Analogy between electric flux and field of velocity vector
A good analogy between the electric flux and the field of velocity
vector in a flowing fluid can be found.
A (area)
volume flow rate:

 
dV
 A
dt
velocity vector (flow speed)
A A vector area that defines the plane of the area,
perpendicular to the plane

A
f

A
volume flow rate:
 
dV
 A cosf    A    A
dt
 A




   cosf ; A  A cosf ; A  An
Calculating Electric Flux

Analogy between electric flux and field of velocity vector
A (area)
volume flow rate:
dV
 A
dt
electric flux:
 E  EA

E
Electric field vector

A A vector area that defines the plane of the area,
perpendicular to the plane
E field is
perpendicular
to this plane
f

A

E
electric flux:
 
dV
 A cos f    A    A
dt
 
  EAcosf  E A  E  A
E  E cosf ; A  A cosf
Calculating Electric Flux

A small area element and flux
 
dE  E  dA

Total flux for an area
   
 E   d E   E dA   E cos f dA   E  dA ; dA  n dA
 Example:
Electric flux through a disk
r = 0.10 m 
A   (0.10 m)2  0.0314m 2
A
f  30

E
 E  EA cosf  (2.0 103 N/C)(0.0314 m 2 ) cos30
 54 N  m 2 / C
Calculating Electric Flux

nˆ3
Example : Electric flux through a cube
nˆ5

E
nˆ2
nˆ1
nˆ4
nˆ6

 E1  E  nˆ1 A  EL2 cos180   EL2

 E2  E  nˆ2 A  EL2 cos0   EL2
E3  E4  E5  E6  EL cos90  0
2
L
 E  i 1  Ei  0
i 6
Calculating Electric Flux

Example : Electric flux through a sphere

dA
r=0.20 m
+
+q


E  E , E // nˆ // dA
6
3
.
0

10
C
9
2
2
E
 (9.0 10 N  m / C )
2
4e 0 r
(0.20m)2
q
 6.75105 N/C
 E   EdA  EA  (6.75105 N/C)(4 )(0.20m)2
q=3.0 mC
A=4r2
 3.4 105 N  m 2 / C
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 inside
the surface.

Case 1: Field of a single positive charge q

E
A sphere with r=R
E
r=R
+
q

E

E  surface
1
q
4e 0 R 2
 E  EA 
at r=R
1
q
q
(4R 2 ) 
4e 0 R
e0
The flux is independent of the radius R of
the surface.
Gauss’s Law

Case 1: Field of a single positive charge q (cont’d)

1 
E2 R  E R
4
dA2 R  4dA
r=R
+
q
r=2R

ER
dAR  dA
Every field line that passes through
the smaller sphere also passes through
the larger sphere
The total flux through each sphere is
the same
The same is true for any portion of
its surface such as dA
d ER  ER dAR 
1
ER 4dAR  E2 R dA2 R  d E2 R
4
  q
 E   E  dA 
e0
This is true for a surface of any shape or size
provided it is a closed surface enclosing the charge
Gauss’s Law

Case 2: Field of a single positive charge (general surface)

E
dA
 
E  n  E

E dA
 E cos f
f
dAcosf
+
q
+
surface
perpendicular

to E
d E  E dA  E cosfdA
  q
 E   E  dA 
e0
Gauss’s Law

Case 3: An closed surface without any charge inside
 
 E   E  dA  0
Electric field lines that go in come out.
Electric field lines can begin or end inside
a region of space only when there is charge
in that region.
+

Gauss’s law
  Qencl


 E   E  dA 
; Qencl  i qi , E  i Ei
e0
The total electric flux through a closed surface is equal to the total
(net) electric charge inside the surface divided by e0
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
E at every point in the interior of a conducting material
is zero in an electrostatic situation (all charges are at rest).
If E were non-zero, then the charges would move
• Draw a Gaussian surface inside of the conductor
• E=0 everywhere on this surface (inside conductor)
Gauss’s law
• 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
Applications of Gauss’s Law

Example: Field of a charged conducting sphere with q
Gaussian surface
E

r  R: E  0
+
+
Rr:
+
+
+
R
+
Draw a Gaussian surface
outside the sphere
Qelcl  q, A  4 r 2
+
E  const.on the sphere surface

E  perpendicular tothespheresurface
+
1
q
ER 
4e 0 R 2
Gauss' s law :
q
E (4 r 2 )   E 
e0
ER / 4
ER / 9
r
R
2R
3R
1
1
q
4e 0 r 2
q
r  R: E 
4e 0 R 2
Applications of Gauss’s Law

Example: Field of a line charge

Gaussian surface
E, E  E 
dA
chosen according
to symmetry
line charge
density
Qencl  
E  E on thecylindrical Gaussian surface
 E  E (2 r) 
1 
E
2e 0 r

e0
Applications of Gauss’s Law

Example: Field of an infinite plane sheet of charge
 : chargedensity

E
+ +
+
+

E
+
area A
+ +
+
+ +
+
Gaussian surface
area A

E  thesheet  E  E
Qencl  A
two end surfaces
A
 E  2( EA) 
e0

E
2e 0
Applications of Gauss’s Law

Example : Field between oppositely charged parallel
conducting plane
  plate 1 
E1 E2
+ b E1

+
a
E2
+
S1 +
+
+
+
S2
+
+

E
plate 2  
Solution 1:
A

- E2 E1 S :
EA 
 E  (right surface)
1
e0
e0
c
outward flux
- S
E0
(left surface)
4
 A

S 4 :  EA 
 E  (left surface)
e0
e0
S3
inward flux
E 0
(right surface)
-
No electric flux
on these surfaces
Solution 2:


   
At Point a : E1  E2  E  E1  E2  0



b : E1  E2  E  2   
2e 0 e 0


   
c : E1  E2  E  E1  E2  0
Applications of Gauss’s Law

Example : Field of a uniformly charged sphere
Gaussian surface
+
+
+
+
+ +
+r=R
+
+
+
+
+
R
4 3
charge densit y r  R : EA   (  r ) / e 0
3
4 3
2
Q
E
(
4

r
)


(
 r ) / e0

4 3
3
R
3
1 Qr
E
4e 0 R 3
1 Q
r  R: E 
4e 0 R 2
Q
2
R  r : E (4 r ) 
e0
E
1
Q
4e 0 r 2
Applications of Gauss’s Law

Example : Field of a hollow charged
(uniformly on its surface) sphere
E  1.80102 N/C
r=0.300 m

E
E   E
 E   E dA   E (4 r 2 ) 
R=0.250 m
q
e0
q   E (4e 0 r 2 )  0.801nC
Hollow charged sphere Gaussian surface
Charges on Conductors

Case 1: charge on a solid conductor resides entirely on
its outer surface in an electrostatic situation
+ + +
+
+
+
+
+
+ ++
+
++
+ +

The electric field at every point within a conductor
is zero and any excess charge on a solid conductor
is located entirely on its surface.
Case 2: charge on a conductor with a cavity
+ + +
+
+
+
+
+
+ ++
+
++
+ +
If there is no charge within the cavity, the net
charge on the surface of the cavity is zero.
Gauss surface
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
charged conducting ball
conductor
(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)
charged conducting ball
conductor
(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  /e0 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/e0= A/ e0,
E=  / e0
Exercises
 Exercise
1
This is the same as the field due to a
point charge with charge +2Q
Exercises
 Exercise
1 (cont’d)
Exercises
 Exercise
Q2=-3Q1
2: A sphere and a shell of conductor
Q2
Q1
R1
R2
• 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.
 inner
Q1

4R22
 outer
Q2  Q1  2Q1


2
4R2
4R22
Exercises
 Exercise
Q2=-3Q1
2: A sphere and a shell of conductor (cont’d)
Q2
Q1
R1
R2
r  R1 :
R1  r  R2 :
R2  r :

E0

Q1
E  k 2 rˆ
r

Q1  Q2
2Q1
ˆ
Ek
r  k 2 rˆ
2
r
r
Exercises
 Exercise
3: Cylinder
An infinite line of charge passes directly through the middle of a hallow
charged conducting infinite cylindrical shell of radius R. Let’s focus on
a segment of the cylindrical shell of length h. The line charge has a linear
charge density , and the cylindrical shell has a net surface charge density
of total.
total

R
inner
outer
h
Exercises
 Exercise
3: Cylinder (cont’d)
The electric field inside the cylindrical shell is zero. Therefore if we choose
as a Gaussian surface a cylinder, which lies inside the cylindrical shell, the
net charge enclosed is zero. There is a surface charge density on the inside
wall of the cylinder to balance out the charge along the line.
total

R
inner
outer
h
Exercises
 Exercise
3: Cylinder (cont’d)
• The total charge on the enclosed portion (length h) of the line charge:  h
• The charge on the inner surface of the conducting cylinder shell: Qinner  h
 inner 

 h


2Rh
2R
total
R
inner
outer
h
Exercises
 Exercise
3: Cylinder (cont’d)
• The net charge density on the cylinder:  total
• The outer charge density :  outer
 outer   total   inner   total 

2R
total

R
inner
outer
h
Exercises
 Exercise
3: Cylinder (cont’d)
• Draw a Gaussian surface surrounding the line charge of radius r (< R)
2rhEr 
qencl
e0
, qencl

 h  Er 
for r  R
2e 0 r
total

R
inner
outer
h
Exercises
 Exercise
3: Cylinder (cont’d)
• Draw a Gaussian surface surrounding the line charge of radius r (>R)
• Net charge enclosed on the line:  h Net charge enclosed on the shell: Q  2Rh total
2rhEr 

qencl
e0
, qencl  Q  h  Er 
total
R
inner
outer
 total R


for r  R
e 0 r 2e 0 r
h