普通物理 General Physics 23 - Finding the Electric Field－II

普通物理 General Physics 23 - Finding the Electric Field－II
郭艷光Yen-Kuang Kuo 國立彰化師大物理系暨光電科技研究所電子郵件: 網頁:

普通物理講義-23/國立彰化師範大學物理系/郭艷光教授
Outline 23-1 What Is Physics? 23-2 Flux 23-3 Flux of an Electric Field 23-4 Gauss’ Law 23-5 Gauss’ Law and Coulomb’s Law 23-6 A Charged Isolated Co 23-7 Applying Gauss’ Law: Cylindrical Symmetry 23-8 Applying Gauss’ Law: Planar Symmetry 23-9 Applying Gauss’ Law: Spherical Symmetry 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-1 What Is Physics? In this chapter, we will introduce the following new concepts: We will then introduce the flux (symbol Φ) of the electric field. We will then apply Gauss’ law and determine the electric field. We will also apply Gauss’ law to determine the electric field inside and outside charged conductors. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-2 Flux 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-2 Flux Note1：Φdepends on θ. It is maximum and equal to vA for θ = 0 ( perpendicular to the loop plane ). It is minimum and equal to zero for ( parallel to the loop plane ). Note2 ： The vector is parallel to the loop normal and has magnitute equal to A. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-3 Flux of an Electric Field
Flux SI unit: 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-1 Figure shows a Gaussian surface in the form of a cylinder of radius R immersed in a uniform electric field , with the cylinder axis parallel to the field. What is the flux Φ of the electric field through this closed surface? 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-1 Solutions: 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-2 A nonuniform electric field given by pierces the Gaussian cube. (E is in new-tons per coulomb and x is in meters.) What is the electric flux through the right face, the left face, and the top face? 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-2 Solutions: Right face 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-2 Solutions: Left face 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-2 Solutions: Top face 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-4 Gauss’ Law The flux of E through any closed surface × ε0 = net charge qenc enclosed by the surface. Examples: Surface Surface Surface Surface Note: We refer to S1,S2,S3,S4 as “Gaussian surfaces” 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-3 Figure shows five charged lumps of plastic and an electrically neutral coin. The cross section of a Gaussian surface S is indicated. What is the net electric flux through the surface if q1 = q4 = +3.1 nC, q2 = q5 = -5.9 nC, and q3 = -3.1 nC? 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-4 What is the net charge enclosed by a (real or mathematical ) closed surface is related to the total electric flux through the surface by Gauss’ law as given by Eq (ε0Φ = qenc). 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-4 Gauss’ Law Solutions: 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-5 Gauss’ Law and Coulomb’s Law
dA 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-6 A charged isolated conductor
1. e 2. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-6 A charged isolated conductor
The electrostatic electric field inside a conductor is equal to zero. No electrostatic charges can exist inside a conductor. All charges reside on the conductor surface. There is no charge on the cavity walls. All the excess charge q remains on the outer surface of the conductor. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-5 Figure shows a cross section of a spherical metal shell of inner radius R. A point charge of –5.0 μC is located at a distance R/2 from the center of the shell. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-5 If the shell is electrically neutral, what are the (induced) charges on its inner and outer surfaces? Are those charges uniformly distributed? What is the field pattern inside and outside the shell? 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-5 Solutions: 1.With a point charge of -5.0μC within the shell, a charge of +5.0 μC must lie on the inner wall of the shell. 2. No. 3.As shown in the figure. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-7 Applying Gauss’ Law: Cylindrical Symmetry
2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Featureless cylinder Rotation axis Observer Rotational symmetry 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Featureless sphere Rotation axis Observer Rotational symmetry Observer Magic carpet Infinite featureless plane Translational symmetry 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-6 Upward streamer in a lightning storm. The woman was standing on a lookout platform in the Sequoia National Park when a large storm cloud moved overhead. Some of the conduction electrons in her body were driven into the ground by the cloud’s negatively charged base, leaving her positively charged. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-6 You can tell she was highly charged because her hair strands repelled one another and extended away from her along the electric field lines produced by the charge on her. Lightning did not strike the woman, but she was in extreme 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-6 danger because that electric field was on the verge of causing electrical breakdown in the surrounding air. Such a breakdown would have occurred along a path extending away from her in what is called an upward streamer. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-6 You can see a bright upward streamer near the top of the tree in the opening photograph for this chapter. An upward streamer is dangerous because the resulting ionization of molecules in the air suddenly frees a tremendous number of electrons from those molecules. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-6 Had the woman in developed an upward streamer, the free electrons in the air would have moved to neutralize her, producing a large, perhaps fatal, charge flow through her body. Let’s model her body as a narrow vertical cylinder of height L = 1.8 m and radius R = 0.10 m. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-6 Assume that charge Q was uniformly distributed along the cylinder and that electrical breakdown would have occurred if the electric field magnitude along her body had exceeded the critical value Ec = 2.4 MN/C. What value of Q would have put the air along her body on the verge of breakdown? 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-8 Applying Gauss’ Law: Planar Symmetry

Example 23-7 Figure a shows portions of two large, parallel, nonconducting sheets, each with a fixed uniform charge on one side. The magnitudes of the surface charge densities are σ(+) = 6.8 C/m2 for the positively charged sheet and σ(-) = 4.3 C/m2 for the negatively charged sheet. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Example 23-7 the electric field (a) to the left of the sheets, (b) between the sheets, and (c) to the right of the sheets. 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

23-9 Applying Gauss’ Law: Spherical Symmetry
Inside the shell: r < R Outside the shell: r > R 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

Outside the sphere: r > R Inside the sphere : r < R 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

O 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

End of chapter 23! 2019/1/1 普通物理講義-23/國立彰化師範大學物理系/郭艷光教授

普通物理 General Physics 23 - Finding the Electric Field－II

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普通物理 General Physics 23 - Finding the Electric Field－II

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