1. Conductors and Dielectrics in Static Electric Fields
Chapter 6
Conductors and Dielectrics
in Static Electric Fields

2. Conductors and Dielectrics in Static Electric Fields
6-1 Conductors in Static Electric Fields
6-2 Dielectrics in Static Electric Fields
6-3 Electric Displacement, Gauss Theorem at the
Presence of the Dielectric
6-4 Capacity, Condenser
6-5 Energy of Static Electric Fields, Energy Density
* The Charge and Discharge of the Condensers
* Application of Electrostatics

3. 6-1 Conductors in Static Electric Fields
1. Condition of Electrostatic Equilibrium
2. The Charge Distribution of a Conductor at
Electrostatic Equilibrium
3. Electrostatic Screening

4. + 1. Condition of Electrostatic Equilibrium 1. Electrostatic Induction
Induced charges

5. 1. Condition of Electrostatic Equilibrium
2. Electrostatic Equilibrium ：there is no motion of charges in a definite direction internally of a conductor +

6. 1. Condition of Electrostatic Equilibrium
Conditions：
（1）Ein= 0 ，everywhere in the conductor
（2）Electric field at the surface
direction  surface

7. 1. Condition of Electrostatic Equilibrium
A conductor is an equipotential body +
The surface is a equipotential surface

8. 2. Charge Distribution on a Conductor
Charge Distribution Determined by：
Charges Outside
Shape of the Conductor
Charge Distribution on the Surface of an Isolated Conductor （p.47/Fig.2-5 ）
convex（positive curvature）
sharp： larger flat：  smaller
concave（ negative curvature ）  smallest
Point discharge （ Ex. lightning rod ）
sharp   , E large  air ionized（breakdown）

9. 2. Charge Distribution on a Conductor
(1). A charged solid conductor：the charges can only be distributed on the surface of the conductor and there is no net charge internally of the conductor.
Gauss surface +
(2). A conductor with a cavity：
internal surface ？External surface？

10. 2. Charge Distribution on a Conductor
(2). A conductor with a cavity：
If there is no charged body in cavity，the charges must distribute on the external surface, and no charge distributed in the internal surface inside of the cavity +
Gauss surface
If there are charges on the internal surface, they must be charges with an equal but opposite charges － ＋
It contradicts with that a conductor is a equipotential body

11. 2. Charge Distribution on a Conductor
If there is a charged body with +q in cavity, the internal surface will be induced and charged with –q, and the charges distributed on the external surface should be calculated by Principle of Conservation of Charge.
Gauss surface q -q q +

12. 2. Charge Distribution on a Conductor
（3）. The relationship between the surface charge density and neighboring electric field of the surface of a charged conductor
Create a thin cylindrical Gaussian surface as shown in the right Fig. +  S
So electric field at the surface
direction  surface（equipotential）
magnitude   （by Gauss’s）

13. 2. Charge Distribution on a Conductor
(4). The Charge distribution law on the surface of a conductor +

14. Charge Distribution on a Conductor
Phenomenon of Sharp Point Discharge
E near the sharp point is so strong that it can ionize the nearby air and make it a conductor and electric discharge will appear in the form of vehement fireball explosion.

15. Charge Distribution on a Conductor
< electric wind experiment > + + +

16. Charge Distribution on a Conductor+ + + +
The principle of lightning rod +
Charged cloud － － － － － － + +
Electrostatic Induction
corona discharge
Grounded reliably

17. 3. Electrostatic screening
1. Screening the external electric field
Screening the external electric field
by a cavity conductor
Electric field inside the cavity is not affected by charges outside the conductor（no matter grounded or not）

18. 3. Electrostatic screening
2. Screening the internal electric field +

19. 3. Electrostatic screening
3. the grounded cavity conductor screen the internal field
Electric field outside the grounded conductor is not affected by charges inside the cavity

20. Sol.：Take a sphere with radius r as the Gaussian surface
Example 1（p.201/[Ex.]）
There is a spherical metal shell with an external R1=10 cm and internal radius R2=7cm ，in the shell is a concentric metal ball with radius R3 = 5cm. If the shell and ball both have a positive charge of q =10-8 C, how are the charges distributed between the two sphere? What is the potential at the center of the ball?
Sol.：Take a sphere with radius r as the Gaussian surface
The charges of the inner ball distributes on the surface R3

21. Example 1（p.201/[Ex.]）
the internal surface of the shell is charged with –q, and the charges of external surface is 2q

22. Example 1（p.201/[Ex.]）
R1=10 cm，R2=7 cm
R3=5 cm，q=10-8 C

23. Example 2（[supplement ]）
A point charge with q is located outside of a grounded conductive sphere , and the distance from the center of the sphere is l > R，
What is the induced charges q’ on the conductive sphere?
Sol. ： q’ distributes on the surface of the sphere
Grounded，equipotential body，Uo = 0
Uo = Uo1 + Uo2
q ：
q’ ： R q O l

24. Homework
p.229 / , 9, 13

25. 6-2 Dielectrics in Static Electric Fields
The influence of dielectric on electric field, the relative permittivity
The polarization of dielectric
The intensity of polarization
The relationship between polarized charges and free charges

26. Introduction General Laws （Chapter 5，Vacuum） applied to
Conductors （Chapter 6-1）
Dielectrics （Chapter 6-2）
Microscopically ，vacuum inside materials
Coulomb’s Law is correct（ cm）
Macroscopic, statistical average of microscopic
Study Dielectrics with General Laws

27. - - - - - - - - - - - - - - + + + + + + + + + + + + + +
The influence of dielectric on electric field, the relative permittivity
Dielectric is filled in between the two Parallel-plates E gets smaller
relative permittivity
permittivity

28.  2. The polarization of dielectric
Non-polar Molecules：（H2、CH4、paraffin etc）
dielectric
Polar Molecules ：（H2O、organic glass etc）

29. 2. The polarization of dielectric
Two kinds of molecules —— different ways
Non-polar Molecules —— distance - +
E = 0 , p = 0
E  0 , p  0 - +
Polar Molecules —— direction
（ |  p |  |E |， all p tend to line up with E ）
E = 0 ,  p = 0
E  0 ,  p  0

30. 3. The Intensity of electric Polarization P
There are m dipoles within the volume element 
polarized
Polarization vector P ：
P at a point ：V  0
Uniform ： Same P at every point
Unit： C/m 2

31. - - - - - - - - - - - - - - - - 3. Polarization P
Polarization P and the area density of the polarized charges
：The electric dipole of molecule
：The intensity of electric Polarization
： The area density of the polarized charges

32. 4. The relationship between polarized charges and free charges

33. 4. The relationship between polarized charges and free charges
The relationship of polarization P and electric field intensity E
polarizability

34. 6-3 Electric Displacement Gauss’s Law in Dielectrics
permittivity

35. 6-3 Electric Displacement Gauss’s Law in Dielectrics
Electric Displacement Vector
The flux of electric displacement
Gauss’s Law at the presence of the dielectric

36. - - - - - - - - - - - Example 1（p.209/[Ex.1]） d + + + + + + + + + + +
Put a piece of dielectric with r =3 in between the two parallel charged plates with the plate distance d =1 mm. Before putting the dielectric into the two plates is 1000 V. If the surface charge density remains uncharged after the dielectric inserted, what are the E, P, the surface charge densities of the plates and the dielectric, and the D in the dielectric? d U
Solution:
r =3
d=10-3 m
U=103 V
0 = 8.85  10-12

37. Example 1（p.209/[Ex.1]） E0 = 103 kV· m-1 P = 5.8910-6 C · m-2
0 = 8.85  10-12

38. Example 2（p.209/[Ex.2]） Sol. ：(1)
As shown in the right Fig. a cylindrical condenser is made up of a thin and long cylinder conductor of radius R2 and a long solid cylinder of radius R1in the center, between them the dielectric with the relative permittivity r is filled. Assume that the charge per unit length of the outer and inner cylinders are + and - ，What are：(1)E，D, P in the dielectric.；
(2)The surface densities of the polarized charges in the inner and outer surfaces of the dielectric.
Sol. ：(1)

39. Example 2（p.209/[Ex.2]）
(2)

40. Example 3（supplement）
A conducting sphere of radius R , with electric charge q0 , is in a uniform infinite dielectric of  . Find the electric field and bound charge density on surface.
Sol.： - ’ + O R

41. Discussion ∴ | q’ | < | q0 |
(1) ∵  >0 ( r =1 +  >1 ) ∴ ’ , q0 opposite sign
(2)
∴ | q’ | < | q0 |
and q = q0 + q’ = q0 0/  = q0/ r < q0
(3) Vacuum
∴ E = E0/ r < E0

42. Homework
p.230 /

43. 6-4. Capacitor and Capacitance
1. Capacitance of a Single Isolated Conductor
2. Capacitors and Capacitance
Spherical Capacitor
Parallel-Plate Capacitor
Cylindrical Capacitor
3. Connections of Capacitors
Parallel Connection
Series Connection

44. 1. Capacitance of an isolated conductor
Isolated conductor，potential proportional to charge
a constant，depend on size、shape
Ex.：conducting sphere or ：
Capacitance ：C
Unit ：Farad（F） 1 F = 1 C/1 V
F：1 F = F
pF：1 pF = F

45. 2. Capacitors and Capacitance
Isolated Conductor：one conductor，potential
Capacitor ：two conductors，potential difference
Properties of Capacitors
Charge distributed uniformly
Same magnitude, opposite sign of charges
Voltage proportional to charge
Capacitors often seen
Spherical capacitor
Parallel-plate capacitor
Cylindrical capacitor

46. Capacitors The steps for calculating the Capacitance of Capacitors
（1）Assume the two plates are charged with  Q
（2）Calculate the E between the two plates
（3） Calculate the electric potential difference between the two plates U
（4）Calculate the C from C = Q /U

47. Parallel-Plate Capacitor （p.213/[Ex.1]）
Symmetry —— Charge distributed uniformly
Gauss’s Law —— Same magnitude, opposite sign
between two plates ：  - A B S d

48. Cylindrical Capacitor （p.213/[Ex.2]）
Symmetry —— Charge distribution
Gauss’s Law ——  
R1 < r < R2 ：

49. Spherical Capacitor （p.214/[Ex.3]）
Symmetry —— Charge distribution
Gauss’s Law ——  
R1 < r < R2 ： R2 R1

50. Assume the charge line density of two wires
Example（p.215/[Ex.4]）
Assume there are two infinitely long parallel straight wires each with radius R , the distance between their centers is d，and d  R, What is the capacity per unit length?
Assume the charge line density of two wires

51. Example（p.215/[Ex.4]）

52. 3. Connections of Capacitors
Parallel Connection（U same） q = q1 + q2 U C2 C1
—— Increase capacitance
Series Connection（q same） U = U1+ U2 U C2 C1
—— Increase working voltage

53. 6-5 Energy of Static Electric Fields, Energy Density
1. Energy in a Capacitor 2. Energy of Static Electric Fields, Energy Density

54. - - - - - - - - - 1. Energy in a Capacitor
A parallel-plate air capacitor is charged—— Charges are moved from cathode to anode +
U fixed when connected to a battery
Q fixed when the battery is disconnected

55. Example
A uniformly charged sphere of radius R , with electric charge Q ,. Calculate the energy of Static electric field W.
Sol.：

56. 2. Energy of Static Electric Fields, Energy Density
Where is the energy stored ? —— charge？field？
（ Theories and experiments support the latter. ）
Ex. ：electromagnetic waves （carry energy，not charge）
Energy density w ：energy per unit volume
For parallel-plate capacitor：
（charges on plates，energy in between them） E
（Valid for any electric fields）  -
The total energy stored in the whole electric field space

57. Example 1（p.219/[Ex.1]）
The inner and outer radii of a spherical condenser are R1 and R2，respectively, and their charges Q， respectively. If a dielectric with permittivity  is filled in between the two shells, what is the electric field energy stored in this condenser.
Solution ： R1< r < R2 　 -Q Q

58. Example 1（p.219/[Ex.1]） （1） （2） （ spherical condenser ）
Discussion
（1）
（ spherical condenser ）
（2）
（ isolated conductive Sphere ）

59. Example 1（p.219/[Ex.2]） + --- - ++++ _
As shown in the right Fig. , there is a cylindrical condenser, in between the cylinder is air and the breakdown electric field intensity is Eb=3106 V·m-1，the outer radius of the condenser is R2= 10-2 m. Under the circumstances that the air is not breakdown, what is the maximum value of R1 so that the energy stored in the condenser reaches maximum?
Solution ： + _
++++
--- -

60. Example 1（p.219/[Ex.2]） + --- - ++++_
++++
--- -
the electric field energy stored per unit length

61. Example 1（p.219/[Ex.2]）
Eb=3106 V·m-1 ，R2= 10-2 m
END

62. Homework
p.230 / , 21, 25, 31, 32

63. The end