2020-23-Gauss\' Law - PPT PDF

Preview

Summary

PPT...

Description

Questions and Discussions ☞ A shell of uniform charge attracts or repels a charged particle that is

outside the shell as if all the shell’s charge were concentrated at its center. Point charge uniform charge q2

q1

F1,shell 



F1,shell  F1,point2

Same Point charge



Point charge

q2



q1

F1,point2

☞ If a charged particle is located inside a shell of uniform charge, there

is no net electrostatic force on the particle from the shell.

Point charge inside the shell

q1

A shell uniform charge q2

of

1

The Electric Field Due to an Electric Dipole (电偶极子 电偶极子 电偶极子) r( )  z 

d d , r( )  z  2 2

E  E ( )  E( )  

1

The electric dipole

1

1 q q  2 4 0 r( ) 4 0 r(2)

q



4 0  z  12 d 

2

1

q

4 0 z  12 d 

2

 1 1     2 2 4 0 z 2  1  2dz  1  2dz   q



q



2d z

4 0 z 1   d  2z

Usually

2



2 2



q



d

2 0 z 1   d 2 2z 3



2

d 1 qd  1  E  2z 2 0 z 3

Electric dipole moment(电偶极矩 ): p  qd  E 

p 2 0 z 3 1

2

The Electric Field Due to a Line of Charge Charge distribution is continuous. The linear charge density :  

dq ds

Unit of  : C  m -1

dE 

1 dq 1  ds 1 ds   4 0 R 2  z 2 4 0 r 2 4 0 r 2

cos  



z  r



z R2  z2

 dE z  dE  cos 

z 4 0  z  R 2

2



3

ds 2

3

The Electric Field Due to a Line of Charge

E z   dE z   dE  cos    





z

4 0 z 2  R 2

  3

2

2R

0

z



40 z  R

ds 

z  R  Ez 

2



3

ds 2

z  2R



40 z 2  R 2

q zq  Ez  2R 40 z 2  R 2



2



3



3

2

2

q 4 0z 2

z  0  Ez  0 4

The Electric Field Due to a Charged Disk Charge distribution is continuous The surface charge density : 

dq dq  dA 2πrdr

Unit of  : C  m -2 dE z 



zdq

40 z 2  r 2



3

 2

z 2rdr

z R 2 rdr E z   dE z  4 0 0 z 2  r 2





4 0 z 2  r 2



3

2



3

 2

z 2 rdr



4 0 z 2  r 2

z R dr 2  4 0 0 z 2  r 2





3



3

2

2

Take X  z 2  r 2 , m   32 , dX  dr 2



X z  z 2  r 2 m  X dX  m  1  Ez  4   1 0  2 m 1



 12

R

   z 1     0 2 0  z 2  R2

  

5

The Electric Field Due to a Charged Disk

Ez 

  z 1  2 0  z2  R2

   

 z  0 or R  z  Ez  2 0 z  R 

Ez  0

6

Chapter 23 Gauss’s Laws 高斯定律

7

Flux Example of flux: A uniform air stream flowing through a square area（Images from the web).

Flux Φ is defined as volume flow rate (volume of air flow through the loop per unit time):

    v  A  vA cos

 v : velocity vector  A : vector area

   : angle between v and A 8

Flux of an Electric Field The flux of electric field: the electric field passing through a certain area A. Considering a small area ΔA, the flux passing through it is

    E  A

Note: ΔΦ can be positive, negative or zero depends on the direction of the vectors. Examples in the figure: Area 1: The total flux on the left side of the Gaussian surface is negative . Area 2: The total flux through this surface is zero. Area 3: The total flux through this surface is equal to positive.

9 Image from the web

Gaussian Surface Gaussian surface: an imaginary surface The electric flux through a Gaussian surface: proportional to the net number of field lines passing through that surface.

For discrete surface:

For continuous surface:

     E  A      E  dA

10

Example: Figure shows a Gaussian surface in the form of a cylinder of radius R immersed in a uniform electric field E, with the cylinder axis parallel to the field. Find the flux of the electric field through the closed surface of the cylinder. Solution: Total flux = flux through surface a + flux through surface b + flux through surface c

         E  dA   E  dA   E  dA a

 

b

 E dA cos180

c

0

a

+

 E dA cos 0 + E dA cos 90 0

b

  -E  dA + E a



b

0

c

dA +  0 dA c

  - E (cap's area) + E (cap's area) + 0 = 0

11

Example: Find the flux of spherical Gaussian surface of radius r which encloses a positive point charge +q at its center. Solution:  E At every point on the surface:  = 0, E = constant  +q dA

     E  dA  E  dA  EA

+

q A  4 r 2 2 4 0 r 1 q q 2  4 r   EA   0 40 r 2 E

1

Note: It holds for any Gaussian surface that encloses the charge q and for any location of q inside A

Surface 1

A2 +

A1

E on surface Total electric field lines crossing the whole surface

Surface 2 E1 > E2 same

 A1   A2 12

Gauss’ Law If there are positive charges inside the surface, which means field lines are created inside the Gaussian surface. If there are negative charges inside the surface, which means field lines are terminated inside the Gaussian surface. Gauss’ law:

   0  E  dA  qenc A

charge enclosed by the hypothetical closed surface(Gaussian surface) A. ☞ Gauss’ law relates the electric fields at points on a closed Gaussian surface to the net charge enclosed by that surface 13

Illustration of Gauss’ law

The electric field is outward for all points on this closed surface. The flux and charge inside are both positive.

No Charge inside the surface. The electric field lines enter the surface and also leave it.

The electric field is inward for all points on the surface. The flux and charge inside are both negative.

Net charge inside this closed surface is zero. Number of the electric field lines entering the surface = number of lines leaving it. 14

Gauss’ Law and Coulomb’s Law Coulomb’s law:



dA

1

q E 4 0 r 2





E



The Gauss’s law:  0  E  dA  qenc

q

Both describe the electric field distribution around a system of charges.

Point charge

The Coulomb’s law can be derived from the Gauss’s law: (Really?) Consider a spherical Gaussian surface around a point charge q, apply the Gauss’ law

   0  E  dA  q enc because of symmetry, EdA is the same on every point and the integral becomes:

   0  E  dA   0  EdA  qenc

Gaussian spherical surface of radius r

☞ Note that E and dA has the same direction on every point of the surface.

 0 E 4r 2   q

☞ It gives the Coulomb’s law at the end:

E

1

q 4 0 r 2

15

A Charged Isolated Conductor In electrostatic equilibrium, the electric field inside a conductor is zero Conductor inside E = 0

Electrostatic equilibrium

Excess charge of a conductor In electrostatic equilibrium Gaussian surface inside the conductor

inside E = 0 Conductor





F

E

Conductor

-e

Electrons always present in any conductor

Since the electric field inside the conductor = 0 Flux through any Gaussian surface inside the conductor = 0

Net charge inside the conductor = 0 ☞ In electrostatic equilibrium, all If inside E ≠ 0, free electrons will excess charge on a conductor is entirely move. Not electrostatic equilibrium on the conductor's outer surface. 16

An Isolated Conductor with a Cavity No electric charge q in the cavity + +  Eint  0 + cavity + +

An electric charge q in the cavity +  Eint  0

+

+

+ +

+q + + + + Gaussian surface No net charge on the cavity walls, all the excess charge remains on the outer surface of the conductor.   0  E   dA  qin  0 A

 Eint ernal  0

☞ The cavity doesn't change the distribution of the excess charges.

+ + Gaussian surface

+

+

 Eint ernal  0 



0  E  dA  q  qin  0 A

The electric charge on the inner surface of the conductor is:

qin  q

17

The External Electric Field While there is no field inside a conductor, the charges on the surface generate an electric field outside the surface. The external electric field at point p just outside the conductor’s surface:

Eext 

 0

 E



charge per unit area on the conductor’s surface at point p

A

1.Consider a flat surface with uniform charge density σ. 2.Draw a cylindrical Gaussian surface as shown. 3.The net charge enclosed by the Gaussian surface is σA and the flux through the outside surface is EA 4.Note that the field on the left is zero and hence there is no flux on the left         surface. E  dA  E  dA  E  dA  E  dA



A







int ernal

curved

external

 0  0  Eext A

  A  0

E ext 

 0

18

Electric Field from Non-uniform Charge Distribution

Uniform charge distribution

Conductor sphere

Charge distribution is not uniform

Lower charge density

Higher charge density

Conductor

Electric field is very easy to calculate.

In general, electric field is very difficult to calculate. 19

Example: A negative point charge q = -5 μC is located at a distance R/2 to the center of an electrically neutral spherical metal shell, find the distribution of charges on the shell. Solution: - + + + + + Inside the conductor, E = 0. - + + + -q Since there is no field inside the conductor, there is - + + no flux through the Gaussian surface. + + By Gauss’ law, the net charge enclosed by the Gaussian surface must be zero.

0  = qenc = 0 So the induced charges on the inner surface = +5μC. The distribution of charge on the inner wall is skewed because the negative charge is off-center. outer wall:

q out  qin  q

The distribution of negative charge on the outer wall is uniform. Why?

20

Application of Gauss’ Law Using Gauss’ law to calculate the electric field produced by some symmetric arrangements of charge. The Problem is how to choose a proper Gaussian surface.

     E  dA   E cos  dA

1.The Gaussian surface should match the symmetry of the charge.

+ Image from the web Image from the web

2. θ = constant and E = constant at every point, we can put E and  out of the integral.

  E cos   dA

3.The summation of A must be very simple.

21

Applying Gauss’ Law: Cylindrical Symmetry Problem: An infinite long cylindrical plastic rod with a uniform positive linear charge density  . Find the electric field at a distance r from the axis of the rod. Solution: +  1.Choose a cylindrical Gaussian surface which can match the + cylindrical symmetry of the problem.   + r  0 E  dA  qenc  h + A + h 2.Find from the cylindrical symmetry that E must have the + r same magnitude and ( >0) must be directed radically + outward at any point on the cylindrical surface. + Flux through the end caps: =0 +   +  0 E  dA   0 E dA   0 E2 rh  cylinder cylinder +  + E (infinite line of charge) 2 0r + + Direction: radially outward when >0.







Applying Gauss’ Law : Planar Symmetry of Nonconducting Sheet Problem: A thin, infinite nonconducting sheet with a uniform surface charge density + on one side. Find the electric field E at a distance r in front of the sheet. Solution: ++  1.From symmetry, E must be perpendicular to the +  sheet and have the same magnitude at the same E +++  A distance r. +o  ++ E + 2.Choose a cylindrical Gaussian surface and it pierces to the sheet perpendicularly. + + +r   + ++ 0   0 ( EA  0  EA)   A  E  dA  qenc A



Magnitude: E  2 0 Direction: perpendicular to the sheet. It is a uniform electric field. ☞ It also holds for a finite sheet, at points close to the sheet and not too 23 near its edges.

Two Charged Plates: Conducting vs Nonconducting Put two thin infinite plates with uniform surface charge density ±σ on both faces to be close to each other. ++ - Problem: if the plates are nonconducting, find E . ++ - 1.The distribution of charge won’t change since the excess ++ - charge on an insulator can’t move. ++ - ++ - The net electric field is the vector sum of the fields due to ++ - four infinite plates.





(take rightward as positive direction) Left to the plates:

EL  

Between the plates:

EB 

Right to the plates:

ER 

     (  0 ) 2 0 2 0 2 0 2 0

    2     2 0 2 0 2 0 2 0  0

      ( )  ( )0 2 0 2 0 2 0 2 0 24

Problem: if the plates are conducting, find E The excess charge of the two conductors can move and will move onto the inner faces because the charge attract each other.

The net electric field is the vector sum of the fields due to two inner faces with surface charge density ±2σ. Left to the plates: Between the plates: Right to the plates:

EL  

2 2  0 2 0 2 0

2 2 2   0 2 0 2 0 2 2  ( ER  )0 2 0 2 0 EB 

be directed rightward

25

Applying Gauss' Law: spherical Symmetry (Electric field outside a uniformly charged spherical shell) A thin uniformly charged spherical shell with total charge q and radius R 1.Outside the shell r > R

 0   qenc

     E  dA

Spherical Gaussian surface of radius r 

     E  dA   EdA  E  dA  E  4 r 2 q q  0E  4 r 2  q  E   k 4 0 r 2 r2

dA

r

R



E

r

Similar to the electric field from a point charge placed at the center of the sell. A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell's charge are concentrated at its center.

r R R

 0  qenc E=

1 4  0

r

dA

  E(4 r 2 )



E

q q = k r2 r2

2.Inside the sphere r < R

0  =qenc

1

qenc E= 4  0 r 2

Charge enclosed by r Total charge = Volume enclosed by r Total volume

r3 qenc = 3 q R

E=(

Spherical Gaussian surface of radius r

 qenc q = E 4  r3 4  R3 3 3

q q ) r = k r 3 3 R 4 0 R

R

 dA

r

1

27

Homework (2) Elementary Problem Chapter 23, Problem: 11, 16, 29, 32, 39, 49, 54

28...