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

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 * 6-6 The Charge and Discharge of the Condensers * 6-7 Application of Electrostatics

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

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

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

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

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

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）

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？

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

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 +

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）

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

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.

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

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

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）

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

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

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

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

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

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

Homework p.229 / 6- 8, 9, 13

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

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（10-13 cm） Macroscopic, statistical average of microscopic Study Dielectrics with General Laws

- - - - - - - - - - - - - - + + + + + + + + + + + + + + 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

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

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

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

- - - - - - - - - - - - - - - - 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

4. The relationship between polarized charges and free charges + + + + + + + + + + + - - - - - - - - - - - - - - - - + + + + +

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

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

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

- - - - - - - - - - - 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

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

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)

Example 2（p.209/[Ex.2]） (2)

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

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

Homework p.230 / 6- 19

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

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 = 10 -6 F pF：1 pF = 10 -12 F

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

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

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

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

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

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

Example（p.215/[Ex.4]）

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

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

- - - - - - - - - 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

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

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

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

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

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 ： + _ ++++ --- -

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

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

Homework p.230 / 6- 18, 21, 25, 31, 32

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