chapter 6 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 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

++++ + + + + Induced charges 1. Electrostatic Induction + + 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 E in = 0 everywhere in the conductor 2 Electric field at the surface direction surface 1. Condition of Electrostatic Equilibrium

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

2. Charge Distribution on a Conductor F 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 F Point discharge Ex. lightning rod sharp , E large air ionized breakdown

+ + + + + + + + + + (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

+ + + + + + + + + + an equal but opposite charges 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 Gauss surface (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 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. + q q -q 2. Charge Distribution on a Conductor Gauss surface

Create a thin cylindrical Gaussian surface as shown in the right Fig. 3 . The relationship between the surface charge density and neighboring electric field of the surface of a charged conductor + + + + + + + SS 2. Charge Distribution on a Conductor So electric field at the surface direction surface equipotential magnitude by Gausss

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

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. Phenomenon of Sharp Point Discharge Charge Distribution on a Conductor Charge Distribution on a Conductor

+++ ++ + + + + + Charge Distribution on a Conductor Charge Distribution on a Conductor

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

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

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

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

There is a spherical metal shell with an external R 1 =10 cm and internal radius R 2 =7cm in the shell is a concentric metal ball with radius R 3 = 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? Example 1 p.201/[Ex.] The charges of the inner ball distributes on the surface R 3 Sol. Take a sphere with radius r as the Gaussian surface

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

R 1 =10 cm R 2 =7 cm R 3 =5 cm q=10 -8 C Example 1 p.201/[Ex.]

Sol. q distributes on the surface of the sphere Grounded equipotential body U o = 0 U o = U o1 + U o2 q q R l 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? q O

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

1. The influence of dielectric on electric field, the relative permittivity 2. The polarization of dielectric 3. The intensity of polarization 4. The relationship between polarized charges and free charges 6-2 Dielectrics in Static Electric Fields

Introduction General Laws Chapter 5 Vacuum applied to F Conductors Chapter 6-1 F Dielectrics Chapter 6-2 Microscopically vacuum inside materials Coulombs Law is correct 10 - 13 cm Macroscopic, statistical average of microscopic Study Dielectrics with General Laws

+ + + + + + + - - - - - - - Dielectric is filled in between the two Parallel-plates E gets smaller relative permittivity permittivity + + + + + + + - - - - - - - 1. The influence of dielectric on electric field, the relative permittivity

Non-polar Molecules H 2 CH 4 paraffin etc Polar Molecules H 2 O organic glass etc dielectric 2. The polarization of dielectric

Two kinds of molecules different ways F Non-polar Molecules distance - + E = 0, p = 0 E = 0, p = 0 E 0, p 0 E 0, p 0 - + F Polar Molecules directionPolar Molecules | p | |E | all p tend to line up with E 2. The polarization of dielectric

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. The Intensity of electric Polarization P

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

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

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

+ + + + + + + + + + + - - - - - - - - - - - - - - - - + + + + + permittivity 6-3 6-3 Electric Displacement Gausss Law in Dielectrics

Gausss Law at the presence of the dielectric The flux of electric displacement Electric Displacement Vector 6-3 6-3 Electric Displacement Gausss Law in Dielectrics

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 Example 1 p.209 /[Ex.1] r =3 d=10 -3 m U=10 3 V 0 = 8.85 10 -12 Solution:

E 0 = 10 3 kV m -1 P = 5.89 10 -6 C m -2 0 = 8.85 10 -12 Example 1 p.209 /[Ex.1]

As shown in the right Fig. a cylindrical condenser is made up of a thin and long cylinder conductor of radius R 2 and a long solid cylinder of radius R 1 in 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. Example 2 p.209/[Ex.2] Sol. (1)

(2)(2) Example 2 p.209/[Ex.2]

Example 3 supplement A conducting sphere of radius R, with electric charge q 0, is in a uniform infinite dielectric of . Find the electric field and bound charge density on surface. Sol. O R + ++ + + + + + + + + + - - - - - - - -

Discussion (1) > 0 ( r =1 + >1 ) , q 0 opposite sign (2) | q | < | q 0 | and q = q 0 + q = q 0 0 / = q 0 / r < q 0 (3) Vacuum E = E 0 / r < E 0

Homework p.230 / 6- 19

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

Ex. conducting sphere or a constant depend on size shape 1. Capacitance of an isolated conductor Isolated conductor potential proportional to charge 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 F Isolated Conductor one conductor potential F Capacitor two conductors potential difference F Properties of Capacitors Charge distributed uniformly Same magnitude, opposite sign of charges Voltage proportional to charge F Capacitors often seen Spherical capacitor Spherical capacitor Parallel-plate capacitor Parallel-plate capacitor Cylindrical capacitor Cylindrical capacitor

Capacitance of 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 Capacitors

Parallel-Plate Capacitor p.213 /[Ex.1] d -- S BA F Symmetry Charge distributed uniformly F Gausss Law Same magnitude, opposite sign between two plates

Cylindrical Capacitor p.213 /[Ex. 2 ] F Symmetry Charge distribution F Gausss Law R 1 < r < R 2

Spherical Capacitor p.214 /[Ex. 3 ] F Symmetry Charge distribution F Gausss Law R 1 < r < R 2 R1R1 R2R2

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? Example p.215 /[Ex. 4 ] Assume the charge line density of two wires

Example p.215 /[Ex. 4 ]

3. Connections of Capacitors Parallel Connection U same q = q 1 + q 2 U C2C2 C1C1 U C2C2 C1C1 Increase capacitance Increase working voltage Series Connection q same U = U 1 + U 2

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

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

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

The inner and outer radii of a spherical condenser are R 1 and R 2 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 R 1 < r < R 2 Q -Q Example 1 p.219 /[Ex. 1 ]

spherical condenser Discussion 11 22 isolated conductive Sphere isolated conductive Sphere Example 1 p.219 /[Ex. 1 ]

As shown in the right Fig., there is a cylindrical condenser, in between the cylinder is air and the breakdown electric field intensity is E b =3 10 6 Vm -1 the outer radius of the condenser is R 2 = 10 -2 m. Under the circumstances that the air is not breakdown, what is the maximum value of R 1 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]

E b =3 10 6 Vm -1 R 2 = 10 -2 m Example 1 p.219 /[Ex.2]

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

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chapter 6 conductors and dielectrics in static electric fields

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