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Chapter 9 Conductors and Dielectrics in Electrostatic Field
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§9-3 Dielectrics 电介质 §9-4 Gauss’ Law in Dielectric 有电介质时的高斯 定律 Electric Displacement 电位移 §9-5 Energy in Electric Field 电场的能量 §9-2 Capacitance 电容器 §9-1 Conductors 导体 Elecrostatic Induction 静电感应 Elecrostatic Induction 静电感应
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Conductor: § 9-1 Conductors and Electrostatic Induction There are many free electrons in it. can move in the conductor randomly e e e e e e e e
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1. The phenomena of the electrostatic induction The charges of an insulated conductor are redistributed because of external E-field.
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No external E-field The process of electrostatic induction of a conductor
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external E-field is supplied-- the electrons start to move. The process of electrostatic induction of a conductor
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor external induced
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The process of electrostatic induction of a conductor 静电平衡状态
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(2). Electrostatic equilibrium conditions : (1). Electrostatic equilibrium state : There is no any charge moving along a definite direction macroscopically inside the conductor or on the surface of the conductor. 2. Electrostatic equilibrium The distribution of charges does not change with time.
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The E-field equals zero everywhere inside the conductor. The conductor is an equipotential body. The conductor surface is an equipotential surface. The E-field at the surface of the conductor is perpendicular to the surface. E=0 The behavior of E-field The behavior of E-potential
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Electrostatic field influences conductor: --electrostatic induction --make the charges in conductor redistribution. Conductor influences electrostatic field : -- make the electrostatic field redistribution. Example
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The field is uniform before the metal sphere is put in. E
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The field is no longer uniform after the metal sphere is put in. + + + + + + + E
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S--any Gaussian surface inside conductor P--any point inside conductor, S--infinite small , No excess charge inside conductor. (1). Entire conductor : 3. The charge distribution on conductor at the electrostatic equilibrium state. Prove : Use No excess charges inside the conductor. They are found only on the conductor surface.
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(2). A conductor with a cavity : Assume charged Q No charge in the cavity: Prove: Draw a Gaussian surface S surrounding the cavity tightly. No charge inside and internal surface of the conductor. All the excess charges distribute on the outside surface of conductor.
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i.e., --No net charge inside S Question ? Are there any equal magnitude and opposite sign charges on the internal surface of the conductor ? Inside the conductor Not at all.
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There are charges q in the cavity : On the inner surface of the conductor : --induction charges -q On the external surface of the conductor : The original charges Q of the conductor + induction charges +q
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3.The relation about the charge distribution on the conductor surface and the its radius of curvature. Example: Two conducting spheres of different radii connected by a long conducting wire. They are equipotential.
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In a qualitative way, for a conductor of arbitrary shape, the charge density distribution on its surface is inverse proportional with its radius of curvature. - - - - - - - - - -
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4. The relation about the E-field just outside the conductor surface and the charge density on the conductor surface.. p is set up by all charges in the space (on and outside the conductor).
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5. Application of electrostatic induction. (1).Tip discharge Lighting rod( 避雷针 ) (2).Electrostatic shielding + + + + ++ + ++- Electrostatic generator candle Electrostatic generator and electric wind
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A conductor shell can shield the external field
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A conductor shell that is connected with the ground can shield the influence of the fields between inside and outside the conductor.
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[Example 1]A neutral conductor sphere with radius R is put on the side of a point charge +q. Assume the distance between the spherical center and the point charge is d. Calculate: The E-field and potential at point 0 set up by the induced charges on the sphere. If the sphere is connected with the ground, how much is the net charge on the sphere? The examples about electrostatic induction
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Solution The total field at 0 = the field set up by q + the field set up by ±q Assume the induction charges are±q =0 !!
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the potential at 0 set up by ± q : the potential at 0 set up by q :
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the sphere is connected with ground. Assume net charge q 1 is left on the sphere. the potential U 0 at 0 = U q + U q1
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[Example 2] Two large parallel plates with the area S carry charge Q a and Q B respectively. QAQA QBQB Find : The charge and field distribution. solution Assume the charge surface density areσ 1,σ 2,σ 3 andσ 4 on the four surfaces. σ1σ1 σ2σ2 σ3σ3 σ4σ4
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Draw a Gaussian surface ⅠⅡ Ⅲ at point P :
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and
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ⅠⅡ Ⅲ The field distribution: E1E1 E1E1 E2E2 E2E2 E3E3 E3E3 E4E4 E4E4 E=0 inside the plates outside the plates: direction: point to left point to right
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ⅠⅡ Ⅲ E Ⅰ = E Ⅲ = 0 discussion Two plates carry equal magnitude and opposite sign charge the charges distribute on the inner surface only.
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ⅠⅡ Ⅲ Charges distribute on the exterior surface only. Two plates carry equal magnitude and same sign charge
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[Example 3] conductor sphere with radius r 1 carries + q and conductor spherical shell with inner and exterior radii r 2 and r 3 carries + Q. Calculate the E distribution, the potential of sphere and shell U 1 and U 2, potential difference △ U Connect sphere and shell with a wire, find E, U 1 and U 2, △ U =?. If the shell is connected with ground, find E, U 1, U 2, △ U =? If the sphere is connected with ground, find the charge distribution. U 2 =?
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solution : the field distribution:
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sphere potential:
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The shell potential: Potential difference:
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Connect sphere and shell with a wire, All charges are on the exterior surface of the shell.
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The shell is connected with ground, U 2 = 0 , no any charge on the exterior surface of the shell.
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The sphere is connected with ground, U 1 =0. Assume sphere charges q ' , then the inner surface of shell charges - q ' , its exterior surface charges ( Q + q ' )
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As r 3 r 1 < r 3 r 2, q ' < 0 The potential of shell:
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