双语物理

教学目标：1. 理解光的波动性及其相关概念；2. 掌握光的干涉、衍射和偏振现象；3. 理解光的衍射极限、单缝衍射和双缝干涉等基本原理；4. 能够运用光的波动理论解决实际问题。

教学重点：1. 光的干涉和衍射现象；2. 单缝衍射和双缝干涉的计算。

教学难点：1. 光的衍射极限计算；2. 双缝干涉条纹间距的计算。

教学时间：2课时教学过程：第一课时一、导入1. 回顾光的波动性及其相关概念；2. 引入波动光学的基本原理。

二、讲解1. 光的干涉现象：解释光的干涉原理，展示干涉现象的实验结果，讲解干涉条纹的形成；2. 光的衍射现象：解释光的衍射原理，展示衍射现象的实验结果，讲解衍射条纹的形成；3. 光的偏振现象：解释光的偏振原理，展示偏振现象的实验结果，讲解偏振光的形成。

三、练习1. 讲解单缝衍射和双缝干涉的计算方法；2. 给出几个简单的计算题，让学生当堂解答。

第二课时一、复习1. 回顾光的波动性及其相关概念；2. 回顾光的干涉、衍射和偏振现象。

二、讲解1. 光的衍射极限计算：讲解衍射极限的原理，并给出计算公式；2. 单缝衍射和双缝干涉条纹间距的计算：讲解条纹间距的计算方法，并给出计算公式。

三、练习1. 给出几个关于光的衍射极限和条纹间距的计算题，让学生当堂解答；2. 讲解计算题的解题思路，帮助学生掌握计算方法。

四、总结1. 总结本节课所学内容，强调光的波动性、干涉、衍射和偏振现象；2. 强调光的衍射极限和条纹间距的计算方法。

教学评价：1. 学生能够正确理解光的波动性及其相关概念；2. 学生能够运用光的波动理论解决实际问题；3. 学生能够熟练计算光的衍射极限和条纹间距。

教学反思：1. 教师在讲解过程中，要注意结合实验现象和实际应用，让学生更好地理解波动光学的基本原理；2. 教师要注重培养学生的计算能力，提高学生的实际应用能力；3. 教师要根据学生的学习情况，调整教学方法和进度，确保教学效果。

中学物理双语教学研究摘要：本文分析了高中物理双语教学的必要性，讨论了双语教学教学目标的制定和两种语言使用时应该注意的问题，并从概念、例题、新课引入等方面对双语教学的具体施行进行了探索与研究。

关键词：双语教学物理教学教学目标我们正处在一个经济社会全面发展的新时代，为满足当前社会的发展和科技进步的需要，培养高素质的双语人才，满足新世纪我国经济发展对人才的更高要求，实现中小学与大学在教学语言上的衔接，培养具有科学素质的语言人才，中学学科教育中的双语教学显得非常必要。

1.双语教学的必要性1.1经济社会发展的需要。

随着经济社会的快速发展，英语作为经济活动中的重要交流工具，其作用越来越重要。

但是目前不少学生的英语运用能力还处于低级阶段。

针对英语教学中的实际问题，现代教育理念要求把英语作为一门交际工具看待，高中毕业生的英语应当在交际方面基本过关，能在英语交际需求日益增强的环境中，顺利地为以后的学习，进入社会，及参加社会活动打好基础。

所以双语教学有一定的必要性。

1.2信息和网络技术环境下交流的需要。

新世纪信息和网络技术正改变着我们的学习方式，信息和网络技术营造了一个信息化的环境，它正逐渐改变、发展我们的社会，推动学习和生活效率的提高。

我国要真正融入世界发展的大潮，就必须运用网上的通用语言进行交流与合作。

因此，在信息环境的背景下，英语特别是学科英语的熟练应用就显得尤为重要。

1.3物理学科发展的需要。

传统的教学研究主要是对备课、讲课、命题、考试等教学环节进行综合研究，这显然不适应面向世纪的教学要求，加强国际合作，资源共享，经验互通也成为必然趋势。

物理教材中很多公式、定理的证明都要借助于相关的国外原版教材，这就要求在中学引入双语教学，通过对这门课程的学习，掌握物理定理、定义和一些专有名词的正确英语表达，了解物理的推理过程，同时还可以提高英文文献的阅读能力。

2.上好物理双语课需要注意的方面2.1课堂用语的规范性。

课堂常规用语的标准化有助于学生多听、多说日常交际用语，这样可以培养学生良好的听说能力，增强英语的实际运用能力。

标。
［ 关键词 ］ 高中物理 双语教 学
教 学 内容 三、高 中物理实行双语教学对教学 内容的要求 目前在我国物理双语教学还处在 实验阶段 ，很多学校都是将其作 为选修课开设的 ， 所以教学 内容方 面没 有唯一的界定 ， 有些 学校 引进了 国外的原版物理教材 ， 有的是通过现代 化网络寻找所需 的材料 ， 这些都 为物理双语教学提供了很大 的空 间 ， 但笔 者认为物理双语教学者应在 教学内容的选择上遵循 以下几 点原则 ： ｌ 容选 择要适 度。 内 适度就是指在选择高中物理双语 教学的内容时 应 充分考虑到学生原有的英语 水平和物理基础。一方面 ， 选择的教学内 容 不能过 于简单 ， 所选择的教学 内容在语 言方面没有难度 ， 如果 那就达 不 到双语教 学中提 高学生专业英语能力 的教学 目标 ，如果所选择 的教 学 内容在物理学方 面都 是在以前物理课上学 过的 ， 没有新的 内容 ， 那么 也 达不到物理课程 原本要求的教学 目的 ，而且容易让学生对物理 双语 课 失去兴趣 ； 另一方面 ， 选择的教学 内容也不能 太难 ， 远远超 出学生 的 接受 能力 ， 也就是说教学 内容必须 以学生原有 的知识 结构为基础 ， 与原 有 的知识结构发生联系 ， 这样 学生才能对新 的教学 内容进行建构 ， 将其 纳 为 自己的认知结 构里 ，这样才能促进学生 的知识结构和认知结构 的 发展 。所 以适度原则是 物理双语教学者在选择教 学内容时应该遵循 的 个重要原则 ， 必须首先 了解 学生原有 的英语 和物理水 平 ， 在原有 的知 识结构上才能建构新 的结构 ， 这样才能促进学生有效地发展 。 ２内容选择要符合教学 目标 。 ． 教学 目标是进行教学 内容选择 时必须 要考虑 的， 因为教学 内容就是为实现教学 目标服务 的。 中物理双语教 高 学 的教学 目标主要分为三个 ： 一是物理学专业英语语 言 目标 ， 二是物理 学的学科 目标 ， 三是物理学 的跨 文化 目标 ， 这三个 目 标是进 行物理双语 教学 的重要 目的 ，当物理 双语 教师在选择教学 内容 时应 考虑教学 内容 是否有利于三个 目标的实现 ， 只有 这样才能发挥物理 双语 教学的作用 ， 使学生的学习效果达到双语教学 的要求 。 ３内容选择要符合学生的兴趣发展 。 ． 兴趣是学生进行学习和探索未 知的原动力 ， 如果学生对学习失去 了兴趣 ， 那也就是失去 了继续 学习的 动力 ， 那么教学效果就达不到预期的 目标 。目前在我 国有些高 中物理双 语 课是作为有益的选修课开设 的 ，在高中开设物理双语课 主要 就是想 通 过双语 的学习 ， 激发学生学 习物理 知识 的兴趣 ， 强化学生学 习物理的 内在动机 ， 因此在教 学内容的选择上要符 合学生的兴趣发展 ， 尽量 选择 学生感兴趣 的物理知识 ， 例如物 理原理在生 活 、 生产 中的应用实例 ， 与 物理学有关 的 自然现象 ， 物理学史 ， 的物理学进展等等 。这些 都可 最新 以促 进学 生 对物 理 兴趣 的 产 生 。 只要将 以上三 点相综合 ，选择 的教 学内容就会促进学生 的知识发 展 ，也会达 到物 理双语教学 的教学 目标 ，而且激发学生学 习物理 的兴 趣。 显而易见 ，我 国开展 的高 中物理双语教 学是提高学生英语应用能 力和培养学生 国际交流 能力 的重要举措 ， 然当前还处于初始 阶段 ， 虽 与 美 国、加拿大和新加坡等 已成功实行物理 双语 教学的国家还有一定 的 差距 ，但我们可 以在它们 的探索 中发展 自己国家具有特色 的高 中物理 双语教学 ， 虽然学生 的接受 能力和教师 队伍 的建设 都还有待提高 ， 但通

双语物理教学中的难点及对策摘要：新疆高校目前推行的双语教学是指非语言类课程用汉语进行教学，通过对学科知识的学习，提高汉语应用能力，最终目标是使教师能够以汉语作为教学语言传授专业知识，学习者通过汉语学习专业知识，达到熟练运用两种语言进行交际和学习的目的。

本文结合本人的教学体会研究提高双语教学效果的对策，探素符合少数民族学生实际情况的双语教学之路。

关键词：物理双语教学；问题；难点；对策【中图分类号】g640新疆少数民族学生入学时，汉语基础较差，难以适应中学物理的教学要求。

物理学是基础科学，对少数民族学生来说学习物理的作用十分重要，如何提高学生的学业成绩成为教师首要考虑的问题。

一、少数民族双语教学的意义和任务我国是一个多种民族语言、多方言的国家，汉语不仅是汉民族共同语，也是国家通用语。

随着经济大变革，商品大流通，民族地区越来越多的人认识到：少数民族要走向世界，必先走向全国，必须通过语言关。

物理学是研究最基本的物质结构和相互作用规律及其实际应用的科学，它有自己的完整规律，与使用的语言无关。

然而学习、研究和应用物理的人却是要用语言来表达对物理学的认识，为正确描述物理现象和教育年轻一代，语言也就渗透到了物理学中。

我们必须承认由于历史的原因，内地大城市的科技发展水平，特别是在物理双语教学改革方面远远超过新疆少数民族地区。

要学习和掌握高科技知识，发展新疆地区经济，培养各种人才迫在眉睫，物理作为理科中一门重要学科，要想让中学生尤其是少数民族中学生学好绝非易事。

理科“双语”教学一直是双语教育的薄弱环节，从教以来我遇到了很多问题与困难，但当我看到学生那一双双求知的眼睛，我感到自己责任重大。

通过一段时间教学工作之后，我知道了教学中存在的的问题，在实践与摸索中也想出了一些应对的措施，下面和大家一起来探讨：二、存在问题1、没有语言环境，专业术语学习更加困难学生学习汉语的语言环境少，除了在学校课堂学习时说汉语，生活中民族学生平时很少接触到说汉语的人，从小习惯了用母语交流，如果刻意让学生说汉语也不现实，所以很多学生在校时能说几句汉语，回家了基本就不说了。

Chapter 1 Particle KinematicsI) Choose one correct answer among following choices1. An object is moving along the x-axis with position as a function of time given by x=x(t). Point O is at x=0. The object is definitely moving toward O when A. 0<dt dx B. 0>dtdx C. 0)(2<dt x d D. 0)(2>dt x d 2. An object starts from rest at x=0 when t=0. The object moves in the x direction with positive velocity after t=0. The instantaneous velocity and average velocity are related by A. v v < B. v v = C. v v > D. dt dx can be larger than, smaller than, or equal to tx 3. An object is moving in the x direction with velocity )(t v x , anddt dv x is nonzero constant. With 0=x v when t=0, then for t>0 the quantity dtdv v x x is A. Negative. B. Zero. C. Positive.D. Not determined from the information given.4. An object is moving on the xy-plane with position as a function of time given by r = a t 2 i + b t 2 j (a and b are constant). Which is correct?A. The object is moving along a straight line with constant speed.B. The object is moving along a straight line with variable speed.C. The object is moving along a curved path with constant speed.D. The object is moving along a curved path with variable speed. 5. An object is thrown into the air with an initial velocity )8.99.4(0j i v +=m /s . Ignore the air resistance (空气阻力). At the highest point the magnitude of the velocity is ( )(A) 0 (B) 4.9m/s (C) 9.8m/s (D)22)8.9()9.4(+ m/s6. Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an additional horizontal acceleration during its descent, itA. strikes the plane at the same time as the other body.B. strikes the plane earlier than the other body.C. has the vertical component of its velocity altered.D. has the vertical component of its acceleration altered.7. A toy racing car moves with constant speed around the circle shown below. When it is at point A its coordinates are x=0, y=3m and its velocity is 6m/s i . When it is at point B its velocity and acceleration are:A. -6m/s j and 12m/s2i , respectively. B. 6m/s j and -12m/s 2 i , respectively. C. 6m/s j and 12m/s 2i , respectively. D. 6m/s j and 2m/s 2 i, respectively.8. A stone is tied to a 0.50-m string and whirled at a constant speed of 4.0m/s in a vertical circle. Its acceleration at the bottom of the circle is:A. 9.8m/s 2, upB. 9.8m/s 2, downC. 8.0m/s 2, upD. 32m/s 2, up9. A boat is able to move through still water at 20m/s. It makes a round trip to a town 3.0 km upstream. If the river flows at 5m/s, the time required for this round trip is:A. 120 sB. 150 sC. 200 sD. 320 s II) Fill in the empty space with correct answer1. A particle goes from x =-2m, y =3m, z =1m to x =3m, y =-1m, z =4m. Its displacement is : .2. The x-component of the position vector of a particle is shown in the graph in Figureas a function of time. (a) The velocity component v x at the instant 3.0 s is .(b) When is the velocity component zero ? The time is .(c) Is the particle always moving in the same direction along thex-axis? .3. The angle turned through by a wheel is given by θ＝at +bt 2, where a and b areconstants. Its angular velocity ω＝ , and its angular acceleration β＝ .4. When a radio wave impinges on the antenna of your car, electrons in the antenna move back and forth along the antenna with a velocitycomponent v x as shown schematically in Figure . Roughlysketch the same graph and indicate the time instants when(a) The velocity component v x is zero;(b) The acceleration component a x is zero;(c) The acceleration has its maximum magnitude.5. A car is traveling around a banked, circular curve of radius 150 m on a test track. At the instant when t=0s, the car is moving north, and its speed is 30.0 m/s but decreasing uniformly, so that after 5.0 s its angular speed will be 3/4 that it was when t=0s. The angular speed of the car when t=0s is , the angular speed 5.0 s later is , the magnitude of the centripetal acceleration of the car when t=0s is , the magnitude of the centripetal acceleration of the car when t=5.00s is , the magnitude of the angular acceleration is , the magnitude ofthe tangential acceleration is .6. A projectile is launched at speed v 0 at an angle θ (withthe horizontal) from the bottom of a hill of constant slope βas shown in Figure. The range of the projectile up the slopeis .III) Calculate Following Problems:1. An object with mass m initially at rest is acted by a force j t k i k F 21+=, where k 1and k 2 are constants. Calculate the velocity of the object as a function of time.2. You are operating a radio-controlled model car on a vacant tennis court. Your position is the origin of coordinates, and the surface of the court lies in the xy-plane. The car, which we represent as a point, has x- and y-cooridnates that vary with time according to x=2.0m-(0.25m/s 2)t 2 , y=(1.0m/s)t+(0.025m/s 3)t 3 .a. Find the car ’s instantaneous velocity at t=2.0s.b. Find the instantaneous acceleration at t=4.0s.3. An object moves in the xy-plane. Its acceleration has components a x =2.50t 2 and a y =9.00-1.40t. At t=0 it is at the origin and has velocity j i v 00.700.10+=.Calculate the velocity and position vectors as functions of time.4. An automobile whose speed is increasing at a rate of 0.600 m/s 2 travels along a circular road of radius 20.0 m. When the instantaneous speed of the automobile is 4.00 m/s, find (a) the tangential acceleration component, (b) the radial acceleration component, and (c) the magnitude and direction of the total acceleration.5. Heather in her Corvette accelerates at the rate of (3.00i -2.00j) m/s 2, while Jill in her Jaguar accelerates at (1.00i +3.00j ) m/s 2. They both start from rest at the origin of an xy coordinate system. After 5.00 s, (a) what is Heather’s speed with respect to Jill, (b) how far apart are they, and (c) what is Heather’s acceleration relative to Jill? Chapter 2 Newton ’s laws of motionI) Choose one correct answer among following choices1. In the SI , the base units (基本单位) for length, mass, time are ( )(A) meters, grams, seconds. (B) kilometers, kilograms, seconds.(C) centimeters, kilograms, seconds. (D) meters, kilograms, seconds.2. Which one of the following has the same dimension (量纲) as time ( )(A)a x (B) a x 2 (C) xv (D) vx 3.Which of the following quantities are independent (无关) of the choice of inertial frame (惯性系)?(A)v (B)P (C)F (D) W4. Suppose the net force F on an object is a nonzero constant. Which of the following could also be constant?A. Position.B. Speed.C. Velocity.D. Acceleration. 5. An object moves with a constant acceleration a . Which of the following expression is also constant? ( ) (A)dt v d (B) dt v d (C) dtv d )(2 (D) dt v v d )( 6. An object moving at constant velocity in an inertial frame must:A. have a net force on it.B. eventually stop due to gravity.C. not have any force of gravity on it.D. have zero net force on it.7. A heavy ball is suspended as shown. A quick jerk on the lower string will break that string but a slow pull on the lower string will break the upper string. The first result occurs because:A. the force is too small to move the ballB. action and reaction is operatingC. the ball has inertiaD. air friction holds the ball back8. A constant force of 8.0 N is exerted for 4.0 s on a16-kg object initially at rest. The change in speed of this object will be:A. 0.5m/sB. 2m/sC. 4m/sD. 8m/s9. A wedge rests on a frictionless horizontal table top. Anobject with mass m is tied to the frictionless incline of thewedge as shown in figure. The string is parallel to theincline. If the wedge accelerates to the left, when theobject leaves the incline, the magnitude of its acceleration isA. gsin θB. gcos θC. gtan θD. gcot θ10. A crane operator lowers a 16,000-N steel ball with a downward acceleration of 3m/s 2. The tension force of the cable is:A. 4900NB. 11, 000NC. 16, 000ND. 21, 000N11. A 1-N pendulum bob is held at an angle θ from the verticalby a 2-N horizontal force F as shown. The tension in the stringsupporting the pendulum bob (in newtons) is:A. cos θB. 2/ cos θC. 5D. 112. A car moves horizontally with a constant acceleration of 3m/s 2. A ball is suspended by a string from the ceiling of the car. The ball does not swing, being at rest with respect to the car. What angle does the string make with the vertical?A. 17◦B. 35◦C. 52◦D. 73◦13. A 32-N force, parallel to the incline, is required to push a certain crate at constant velocity up a frictionless incline that is 30◦ above the horizontal. The mass of the crateis:A. 3.3kgB. 3.8kgC. 5.7kgD. 6.5kg II) Fill in the empty space with correct answer 1. A 2.5 kg system has an acceleration 2)4(s m i a =. There are two forces acting onthe system, and One of the forces is N j i F )63(1 -=. The other force is .2. Two masses, m 1 and m 2, hang over an ideal pulley and the system is free to move. The magnitude of the acceleration aof the system of two masses is . The magnitude of the tension in the cord is .3. You are swinging a mass m at speed v around on astring in circle of radius r whose plane is 1.00 m abovethe ground as shown in Figure. The string makes anangle θ with the vertical direction.(a) Apply Newton’s second law to the horizontal andvertical direction to calculate the angle θ is .(b) If the angle θ = 47.4° and the radius of the circle is1.50 m, the speed of the mass is .(c) If the mass is 1.50 kg, the magnitude of the tensionin the string is .(d) The string breaks unexpectedly when the mass is movingexactly eastward. The location the mass will hit the groundis . III) Calculate Following Problems:1. A wedge with mass M rests on a frictionless horizontal tabletop. A block with mass m is placed on the wedge, and ahorizontal force F is applied to the wedge. What must be the magnitude of F if the block is to remain at a constant height above the table top?2. The mass of blocks A and B in Figure are 20.0kg and 10.0kg, respectively. The blocks are initially at rest on the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F is applied to the pulley. Find the accelerations a 1 ofblock A and a 2 of block B when F is(a) 124N ; (b) 294N ; (c) 424N.3. An object is drop from rest. Find the function of speed withrespect to time and the terminal speed. Assuming that the drag forceis given by D = bv 2.4. A small bead can slide without friction on a circular hoop that is ina vertical plane and has a radius of 0.100m. The hoop rotates at aconstant rate of 4.00rev/s about a vertical diameter.(a) Find the angle β at which the bead is in vertical equilibrium.(b) Is it possible for the bead to “ride ” at the same elevation as thecenter of the hoop?(c) What will happen if the hoop rotates at 1.00rev/s.Chapter 3 Linear momentum, Conservation of momentum I) Choose one correct answer among following choices1. An object is moving in a circle at constant speed v . The magnitude of the rate of change of momentum of the objectA. is zero.B. is proportional to v .C. is proportional to v 2.D. is proportional to v 3.2. If the net force acting on a body is constant, what can we conclude about its momentum? A. The magnitude and/or the direction of P may change. B. The magnitude of P r remains fixed, but its direction may change. C. The direction of P remains fixed, but its magnitude may change. D. P remains fixed in both magnitude and direction.3. If I is the impulse of a particular force, what is dt I d / ?A. The momentumB. The change in momentumC. The forceD. The change in the force4. A variable force acts on an object from 0=i t to f t . The impulse of the force is zero. One can conclude that A. 0=∆r and 0=∆P . B. 0=∆r but possibly 0≠∆P . C. possibly 0≠∆r but 0=∆P . D. possibly 0≠∆r and possibly 0≠∆P .5. A system of N particles is free from any external forces. Which of the following is true for the magnitude of the total momentum of the system?A. It must be zero.B. It could be non-zero, but it must be constant.C. It could be non-zero, and it might not be constant.D. The answer depends on the nature of the internal forces in thesystem.6. The x and y coordinates of the center of mass of thethree-particle system shown below are:A. 0, 0B. 1.3m, 1.7mC. 1.4m, 1.9mD. 1.9m, 2.5m7. Block A, with a mass of 4 kg, is moving with a speed of2.0m/s while block B, with a mass of 8 kg, is moving in theopposite direction with a speed of 3m/s. The center of mass ofthe two block-system is moving with a velocity of:A. 1.3m/s in the same direction as A.B. 1.3m/s in the same direction as B.C. 2.7m/s in the same direction as A.D. 1.0m/s in the same direction as B.8. A large wedge with mass of 10kg rests on a horizontal frictionless surface, as shown. A block with a mass of 5.0kg starts from rest and slides down the inclinedsurface of the wedge, which is rough. At one instantthe vertical component of the block ’s velocity is3.0m/s and the horizontal component is 6.0m/s. Atthat instant the velocity of the wedge is:A. 3.0m/s to the leftB. 3.0m/s to the rightC. 6.0m/s to the rightD. 6.0m/s to the left9. A 1.0-kg ball moving at 2.0m/s perpendicular to a wall rebounds from the wall at1.5m/s. The change in the momentum of the ball is:A. zeroB. 0.5N · s away from wallC. 0.5N · s toward wallD. 3.5N · s away from wallII) Fill in the empty space with correct answer1. Two objects, A and B , collide (碰撞). A has a mass of kg m A 2=, and B has a mass of kg m B 4=. The velocities before the collision are )32(j i v A +=m /s and )24(j i v B +=m /s . After the collision, )23(j i v A +='m /s . The final velocity of B ='B v m /s .2. A stream of water impinges on(撞击) a stationary “dished ”turbine blade, as shown in Fig.8. The speed of the water is v ,both before and after it strikes the curved surface of the blade,and the mass of water striking the blade per unit time is constantat the value dt dm /=μ. The force exerted by the water on theblade is ___________.3. A 320g ball with a speed v of 6.22m/s strikes a wall at angle θ of 30.0o and then rebounds with the same speed and angle. It is in contact with the wall for 10.4 ms.(a) The impulse was experienced by the wall is .(b) The average force exerted by the ball on the wall is .4. The muzzle speed of a bullet can be determined using a devicecalled a ballistic pendulum, shown in Figure. A bullet of mass mmoving at speed v encounters a large mass M hanging vertically asa pendulum at rest. The mass M absorbs the bullet. The hangingmass (now consisting of M + m) then swings to some height habove the initial position of the pendulum as shown. The initialspeed v ′of the pendulum (with the embedded bullet) after impactis . The muzzle speed v of the bullet is .III) Calculate Following Problems:1. A block of mass m 1=1.60kg initially moving to the right with a speed of 4.00 m/s on a frictionless horizontal track collides with a spring attached to a second block of mass m 2=2.10kg initially moving to the left with a speed of 2.50 m/s, as shown in Figure. The spring constant is 600 N/m.(a) At the instant block 1 is moving to the right with a speed of 3.00 m/s, as in Figure, determine the velocity of block 2.(b) Determine the distance the spring is compressed at that instant.2. A3.00-kg steel ball strikes a wall with a speed of 10.0m/s at an angle of 60.0° with the surface. It bounces offwith the same speed and angle. If the ball is in contactwith the wall for 0.200 s, what is the average forceexerted on the ball by the wall?3. A small ball with mass m is released from rest at thetop of a container which inside wall is semicircle-shapedand frictionless. The container with mass M and radius R rests on a frictionless horizontal surface, as shown. When the ball slides to point B at the bottom of the container, find the normal force exerted by the container on the ball.Chapter 4 Work and EnergyI) Choose one correct answer among following choices1. The work done by gravity during the descent of a projectile:A. is positiveB. is negativeC. is zeroD. depends for its sign on the direction of the y axis2. A particle has a constant kinetic energy E k . Which of the following quantities must also be constant? ( )(A)r (B) v (C) v (D) P3. A 0.2kg block slides (滑行) across a frictionless floor with a speed of 10 m /s . The net work done on the block is ( )(A) -20J (B) -10J (C) 0J (D) 20J4. A 0.50kg object moves in a horizontal circular track with a radius of 2.5m. An external force of 3.0N, always tangent to the track, causes the object to speed up as it goes around. The work done by the external force as the mass makes one revolution is:A. 24 JB. 47 JC. 59 JD. 94 J5. A man pushes an 80-N crate a distance of 5.0m upward along a frictionless slope that makes an angle of 30◦ with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5m/s 2, then the work done by the man is:A. −200 JB. 61 JC. 140 JD. 200 J6. When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude F = a x+b x 2, where a and b are constants. The work done in stretching this rubber band from x = 0 to x = L is:A. a L 2 + b Lx 3B. a L + 2b L 2C. a + 2b LD. a L 2/2 +b L 3/37. An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10cm. While the spring is being extended by the force, the work done by the spring is:A. −3.6JB. −3.3JC. 3.6 JD. 3.3J8. Two objects with masses of m 1 and m 2 have the same kinetic energy and are both moving to the right. The same constant force F is applied to the left to both masses. If m 1 = 4m 2, the ratio of the stopping distance of m 1 to that of m 2 is:A. 1:4B. 4:1C. 1:2D. 1:19. At time t = 0 a 2-kg particle has a velocity of (4m/s)i − (3m/s)j. At t = 3s its velocity is (2m/s)i + (3m/s)j . During this time the work done on it was:A. 4 JB. −4JC. −12 JD. −40 J10. A 2-kg block starts from rest on a rough inclined plane that makes an angle of 60o with the horizontal. The coefficient of kinetic friction is 0.25. As the block goes 2.0m down the plane, the mechanical energy of the Earth-block system changes by:A. 0B. −9.8JC. 9.8JD. −4.9 J II) Fill in the empty space with correct answer1. A chain (链条) is held on a frictionless table withone-fourth of its length hanging over the edge, as shown infigure. If the chain has a length L and a mass m , the work required to pull the hanging part back on the table isJ .2. A 0.1kg block is dropped from a height of 2m onto a spring offorce constant k = 2N/m, as shown. The maximum distance thespring will be compressed is _________m . (g=10m/s 2)3. A single constant force j i F 53+=N acts on a4.00-kg particle.(a) If the particle moves from the origin to the point having the vector positionj i r 32-=m, the work down by this force is .(b) If its speed at the origin is 4.00 m/s, the speed of the particle at r is . 4l(c) The change in the potential energy of the system is .III) Calculate Following Problems:1. A 3.00-kg mass starts from rest and slides a distance ddown a frictionless 30.0° incline. While sliding, it comes intocontact with an unstressed spring of negligible mass, asshown. The mass slides an additional 0.200 m as it is broughtmomentarily to rest by compression of the spring (k=400N/m). Find the initial separation d between the mass and thespring.2. Two masses are connected by a light string passing over alight frictionless pulley as shown. The 5.00-kg mass is released fromrest.(a) Determine the speed of the 3.00-kg mass just as the 5.00-kg masshits the ground.(b) Find the maximum height to which the 3.00-kg mass rises.Chapter 5 Angular momentum and Rigid bodyI) Choose one correct answer among following choices1. A particle moves with position given by j i t r 43+=, where r is measured in meters when t is measured in seconds. For each of the following, consider only t > 0. The magnitude of the angular momentum of this particle about the origin isA. increasing in time.B. constant in time.C. decreasing in time.D. undefined2. A solid object is rotating freely without experiencing any external torques. In this caseA. Both the angular momentum and angular velocity have constant direction.B. The direction of angular momentum is constant but the direction of the angular velocity might not be constant.C. The direction of angular velocity is constant but the direction of the angular momentum might not be constant.D. Neither the angular momentum nor the angular velocity necessarily has a constant direction.3. A 2.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s. The magnitude of its angular momentum about the center of the circle is:A. 6.0kg · m 2/sB. 12 kg · m 2/sC. 48 kg/m 2 · sD. 72 kg · m 2/s 24. A 6.0-kg particle moves to the right at 4.0m/s as shown. Themagnitude of its angular momentum about the point O is:A. zeroB. 288 kg·m 2/sC. 144 kg·m 2/sD.24kg·m 2/s5. Two objects are moving in the x, y plane as shown.The magnitude of their total angular momentum (aboutthe origin O) is:A. zeroB. 6kg · m2/sC. 12kg · m2/sD. 30kg · m2/s6. A 2.0-kg block starts from rest on the positive x axis 3.0m from the origin and thereafter has a constant acceleration given by )/(342s m j i a-=. At the end of 2s its angular momentum about the origin is:A. 0B. (−36 kg·m 2/s)kC. (+48 kg·m 2/s)kD. (−96 kg·m 2/s)k7. As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12rad/s. The circle is parallel to the xy plane and is centered on the z axis, a distance of 0.75m from the origin. The z component of the angular momentum around the origin is: A. 6.0kg · m 2/s B. 9.0kg · m 2/s C. 11kg · m 2/s D. 14kg · m 2/sII) Fill in the empty space with correct answer1. A particle located at the position vector )2(j i r-=m is acted by a force )3(j i F+=N . The torque about the origin should be_______m N ⋅.2. The velocity of a m =2kg body moving in the xy plane is given by )2(j i v-=m /s .Its position vector is )2(j i r+=m . Its angular momentum L about the origin should be___________s m kg 2⋅.3. Two particles each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. The total angular momentum of the system about any origin is .4. A particle is located at r= (0.5m)i+ (−0.3m)j+ (0.8m)k. A constant force of magnitude 2N acts on the particle. When the force acts in the positive x direction, the components of the torque about the origin is , and when the force acts in the negative x direction, the components of the torque about the origin is .5. A uniform beam of length l is in a vertical position with its lower end on a rough surface that prevents this end from slipping. The beam topples. At the instant before impact with the floor, the angular speed of the beam about its fixed end is .6. A disk of mass m and radius R is free to turn about a fixed, horizontal axle. The disk has an ideal string wrapped around its periphery from which another mass m (equal to the mass of the disk) is suspended, as indicated in Figure. The magnitude of the acceleration of the falling mass is , the magnitude of the angular acceleration of the disk is .III) Calculate Following Problems:1. The pulley has radius 0.160m and moment of inertia 0.480kg ·m2. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor.2. A block with mass m slides down a surface inclined 30to the horizontal . The coefficient of kinetic friction is μ. A string attached to the block is wrapped around a wheel on a fixed axis. The wheel has mass m and radius R with respect to the axis of rotation. The string pulls without slipping .a) What is the acceleration of the block down the plane?b) What is the tension in the string?3. A wooden block of mass M resting on a frictionless horizontal surface is attached to a rigid rod of length l and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and normal to the rod with speed v hits the block and becomes embedded in it. What is the angular momentum of the bullet –block system?Chapter 9 Mechanic oscillationI) Choose one correct answer among following choices1. A particle on a spring executes simple harmonic motion. If the mass of the particle and the amplitude are both doubled then the period of oscillation will change by a factor ofA. 4.B. 8.C. 2.D. 2 2. A particle is in simple harmonic motion with amplitude A. At time t=0 it is at x=-A/2 and is moving in the negative direction, then the initial phase is:A. 2π/3 radB. 4π/3 radC. π radD. 3π/2 rad3. A particle is in simple harmonic motion with period T. At time t = 0 it is at the equilibrium point. Of the following times, at which time is it furthest from the equilibrium point?A. 0.5TB. 0.7TC. TD. 1.4T4. A weight suspended from an ideal spring oscillates up and down with a period T. If the amplitude of the oscillation is doubled, the period will be:A. TB. 2TC. T/2D. 4T5. The displacement of an object oscillating on a spring is given by x(t) = A cos(ωt + φ). If the initial displacement is zero and the initial velocity is in the negative x direction, then the phase constant φ is:A. 0B. π/2 radC. π radD. 3π/2 rad6. An object is undergoing simple harmonic motion with period T, amplitude A and initial phase πϕ31-=. Its graph of x versus t is:7. An object of mass m, oscillating on the end of a spring with spring constant k , has amplitude A. Its maximum speed is:A. m k A /B.m k A /2C. k m A /D. k Am /II) Fill in the empty space with correct answer1. The total energy of a simple harmonic oscillator (谐振子) with amplitude A and force constant k is_________.2. Find the initial phases (初相) of the simple harmonic motion as shown in figure.1ϕ= 2ϕ=III) Calculate Following Problems:1. An object oscillates with simple harmonic motion along the x axis. Its displacement from the origin varies with time according to the equation: )4cos()00.4(ππ+=t m x .where t is in seconds and the angles in the parentheses are in radians.－－1/21/2A －1/21/2A A －1/21/2A A －1/21/2A A。

他还提出了刺激反应说，为生理学做出了一定的 贡献。
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17
轶事:蜘蛛织网和平面直角坐标系的创立
• 据说有一天，笛卡尔生病卧床，病情很重，尽管如 此他还反复思考一个问题：几何图形是直观的，而代数 方程是比较抽象的，能不能把几何图形和代数方程结合 起来？他苦苦思索，拼命琢磨，通过什么样的方法，才 能把“点”和“数”联系起来。突然，他看见屋顶角上 的一只蜘蛛，拉着丝垂了下来。一会功夫，蜘蛛又顺这 丝爬上去，在上边左右拉丝。蜘蛛的“表演”使笛卡尔 的思路豁然开朗。他想，可以把蜘蛛看作一个点。他在 屋子里可以上、下、左、右运动，能不能把蜘蛛的每一 个位置用一组数确定下来呢？他又想，屋子里相邻的两 面墙与地面交出了三条线，如果把地面上的墙角作为起 点，把交出来的三条线作为三根数轴，那么空间中任意 一点的位置就可以在这三根数轴上找到有顺序的三个数。 反过来，任意给一组三个有顺序的数也可以在空间中找 到一点P与之对应，这就是坐标系的雏形。
by describing how to calculate it from other quantities that we can measure.
speed = distance / time
v = d / t (m/s)
An equation must always be dimensionally consistent.
位移
average velocity
平均速度
instantaneous velocity
瞬时速度
acceleration
加速度
confusion
混淆
rectilinear motion
直线运动
.
20
valid
正确的

Chapter 6 RotationIn this chapter, we deal with the rotation of a rigid body about a fixed axis. The first of these restrictions means that we shall not examine the rotation of such objects as the Sun, because the Sun-a ball of gas-is not a rigid body. Our second restriction rules out objects like a bowling ball rolling down a bowling lane. Such a ball is in rolling motion, rotating about a moving axis.6.1 The Rotational Variables1.Translation and Rotation: The motion is the one of pure translation, if the line connecting any two points in the object is always parallel with each other during its motion. Otherwise, the motion is that of rotation. Rotation is the motion of wheels, gears, motors, the hand of clocks, the rotors of jet engines, and the blades of helicopters.2.The nature of pure rotation: Theright figure shows a rigid body ofarbitrary shape in pure rotationaround a fixed axis, called theaxis of rotation or the rotationaxis.(1). Every point of the body moves in a circle whosecenter lies on the axis of the rotation.(2). Every point moves through the same angle during a particular time interval.3. Angular position : The above figure shows a reference line, fixed in the body, perpendicular to the axis, and rotating with the body. We can describe themotion of the rotating body byspecifying the angular position ofthis line, that is, the angle of theline relative to a fixed direction. Inthe right figure, the angular position θ is measured relative to the positive direction of the x axis, and θ is given byHere s is the length of the arc (or the arc distance ) along a circle and between the x axis and the reference line, and r is a radius of that circle.An angle defined in this way is measured in radians (rad) rather than in revolutions (rev) or degree. They have relations4. If the body rotates about the rotation axis as in the right figure, changing the angular position of the reference line from 1θ to 2θ, the body undergoes an angulardisplacement θ∆ given byThe definition of angular displacement holds not only for the rigid body as a whole but also for every particle within the body. The angular displacement θ∆ of a rotating body can be either positive or negative, depending on whether the body is rotating in the direction of increasing θ (counterclockwise ) or decreasing θ (clockwise ).5. Angular velocity(1). Suppose that our rotating body is at angular position 1θ at time 1t and at angular position 2θ at time 2t . Wedefine the average angular velocity of the body in the time interval t ∆ from 1t to 2t to beIn which θ∆ is the angular displacement that occurs during t ∆.(2). The (instantaneous) angular velocity ω, with which we shall be most concerned, is the limit of the average angular velocity as t ∆ is made to approach zero. Thus If we know )(t θ, we can find the angular velocity ω by differentiation.(3). The unit of angular velocity is commonly the radian per second (rad/s) or the revolution per second (rev/s).(4). The magnitude of an angular velocity is called theangular speed , which is also represented with ω.(5). We establish adirection for thevector of the angularvelocity ωby using aright-hand rule , asshown in the figure.Curl your right hand about the rotating record, your fingers pointing in the direction of rotation. Your extended thumb will then point in the direction of the angular velocity vector . 6. Angular acceleration(1). If the angular velocity of a rotating body is not constant, then the body has an angular acceleration. Let 2ω and 1ω be the angular velocity at times 2t and 1t , respectively. The average angular acceleration of the rotating body in the interval from 1t to 2t is defined asIn which ω∆ is the change in the angular velocity that occurs during the time interval t ∆.(2). The (instantaneous) angular acceleration α, with which we shall be most concerned, is the limit of this quantity as t ∆ is made to approach zero. Thusabove equations hold not only for the rotating rigid body as a whole but also for every particle of that body .(3). The unit of angular acceleration is commonly the radian per second-squared (rad/s 2) or the revolution per second-squared (rev/s 2).(4). The angular acceleration also is a vector. Its direction depends on the change of the angular velocity.7. Rotation with constant angular acceleration:Here we suppose that at time ,0=t 00=θ. We also can get a parallel set of equations to those for motion with a constant linear acceleration.8. Relating the linear and angular variables: They have relations as follow:Angular displacement: r d s d d ⨯=θθAngular velocity ： r v ⨯=ωωAngular acceleration ：va r a n t ⨯=⨯=ωαα 6.2 Kinetic Energy of Rotation1. To discuss kinetic energy of a rigid body, we cannot use the familiar formula 2/2mv K = directly because it applies only to particles. Instead, we shall treat the object as a collection of particles-all with different speeds. We can then add up the kinetic energies of these particles to findkinetic energy of the body as a whole. In this way we obtain, for the kinetic energy of a rotating body,In which i m is the mass of the i th particle and i v is itsspeed. The sum is taken over all the particles in the body. 2. The problem with above equation is that i v is not thesame for all particles. We solve this problem by substituting for v in the equation with r ω, so that we haveIn which ω is the same for all particles.3. The quantity in parentheses on the right side of above equation tells us how the mass of the rotating body is distributed about its axis of rotation.(1). We call that quantity the rotational inertia (or moment of inertia) I of the body with respect to the axis of rotation. It’s a constant for a particular rigid body and for a particular rotation axis. We may now write ∑=2i i r m I(2). The SI unit for I is the kilogram-square meter (2m kg ⋅).(3). The rotational inertia of a rotating body depends not only on its mass but also on how that mass is distributed with respect to the rotation axis .4. We can rewrite the kinetic energy for the rotating object asWhich gives the kinetic energy of a rigid body in pure rotation. It’s the angular equivalent of the formula2/2cm Mv K =, which gives the kinetic energy of a rigid body inpure translation.6.3 Calculating the Rotational Inertia1. If a rigid body is made up of discrete particles , we can calculate its rotational inertia from ∑=2i i r m I .2. If the body is continuous , we can replace the sum in the equation with an integral, and the definition of rotational inertia becomes dm r I ⎰=2. In general, the rotational inertia of any rigid body with respect to a rotation axis depends on (1). The shape of the body, (2). Theperpendicular distance from the axis to t he body’s center of mass, and (3). The orientation of the body with respect to the axis .The table gives the rotational inertias of several common bodies, about various axes. Note how the distribution of mass relative to the rotational axis affects the value of the rotational inertia I . We would like to give the example of rotational inertia fora thin circular plate22242003221)(21241mR R R R d dr r dr rd r I R=====⎰⎰⎰πσπσθσθσπ a thin rod3. The parallel-axis theorem : If you know the rotational inertia of a body about any axis that passes through its center of mass, you can find its rotational inertia about any other axis parallel to that axis with the parallel-axis theorem:Here M is the mass of the body and h is the perpendicular distance between the two parallel axes. 4. Proof of the parallel-axis theorem : Let O be the center of mass of the arbitrarilyshaped body shown in crosssection in the figure. Placethe origin of coordinates at O. Consider an axis through O perpendicular to the plane of the figure, and another axis of P parallel to the first axis, Let the coordinates of P be a and b.Let dm be a mass element with coordinates x and y. The rotational inertia of the body about the axis through P is thenWhich we can rearrange as6.4 Newton’s Second Law for Rot ation1.Torque: The following figure shows a cross section of abody that is free to rotate about an axis passing throughO and perpendicular to the cross section. A force F is applied at point P, whose position relative to O is defined by a position vector r. Vector F and r make an angle ϕwith each other. (For simplicity, we consider only forces that have no component parallel to the rotation axis: thus,F is in the plane of the page). We define the torqueτasa vector cross product of the position vector and the force Discuss t he direction and the magnitude of the torque . 2. Newton’s second law for rotation(1). The figure shows a simple case ofrotation about a fixed axis. Therotating rigid body consists of asingle particle of mass m fastened tothe end of a massless rod of length r .A force F acts as shown, causing the particle to move in a circle about the axis. The particle has a tangential component of acceleration t a governed by Newton’ssecond law: t t ma F =. The torque acting on the particle isαατ)()(2mr r r m r ma r F F r t t ====⨯= . The quantity in parentheses on the right side of above equation is the rotation inertia of the particle about the rotation axis. So the equation can be reduced to ατI =.(2) For the situation in which more than one force is applied to the particle, we can extend the equation asατI =∑. Where ∑τ is the net torque (the sum of all external torques) acting on the particle. The above equation is the angular form of Newton’s second law .(3) Although we derive the angular form of Newton’ssecond law for the special case of a single particle rotating about a fixed axis, it holds for any rigid body rotating about a fixed axis, because any such body can be analyzed as an assembly of single particles.6.5 Work and Rotational Kinetic Energy1.Work-kinetic energy theorem: Let’s again consider thesituation of the figure, in whichforce F rotates a rigid bodyconsisting of a single particle ofmass m fastened to the end of amassless rod. During the rotation,Force F does work on the body. Let us assume that the only energy of the body that changed by F is the kinetic energy. Then we can apply the work-kinetic energy theorem to getAbove equation is the angular equivalent of the work-kinetic energy theorem for translational motion. We derive it for a rigid body with one particle, but it holds for any rigid body rotated about a fixed axis.2.We next relate the work W done on the body in the figureto the torque on the body due to force F. If the particle in Fig. 11-17 were move a differential distance ds alongits circular path, the body would rotate through differential angle θd , with θrd ds =. We would getθτθd rd F ds F s d F dW t t ===⋅= . Thus the work done during a finite angular displacement from i θ to f θ is then⎰=f id W θθθτ. Above equation holds for any rigid body rotating about a fixed axis .3. We can find the power P for rotational motion。

我们 正处 在一 个 经 济社 会全 面 发 展 的 新 时 代 ，为 满 足 当 前 社会 的 发展 和科 技 进 步 的需 要 ， 培 养 高 素 质 的双 语 人 才 ， 满 足 新 世 纪 我 国 经 济 发 展 对 人 才 的 更 高 要 求 ，实 现 中 小 学 与 大 学 在 教 学 语 言 上 的 衔接 ． 培 养 具 有 科 学 素 质 的语 言 人 才 ， 中学 学 科 教 育 中 的 双语 教学 显 得 非 常 必 要 。
１ ． ２ 信 息和 网络 技 术 环 境 下 交 流 的 需要 。 新 世 纪 信 息 和 网络 技 术 正 改 变 着 我 们 的学 习方 式 ，信 息 和 网络 技 术 营 造 了 一 个 信 息 化 的 环 境 ， 它正 逐 渐 改 变 、 发 展 我 们 的社 会 ， 推 动 学 习 和生 活 效 率 的 提 高 。 我 国 要 真 正 融人 世 界 发 展 的大 潮 ． 就 必 须 运 用 网上 的 通 用 语 言 进行 交 流 与 合 作 。 因 此， 在 信息 环 境 的背 景 下 ， 英 语 特 别 是 学 科 英 语 的熟 练 应 用 就 显得尤为重要 。 １ ． ３ 物 理 学科 发 展 的 需 要 。 传 统 的教 学 研 究 主 要 是 对 备 课 、 讲课 、 命题 、 考 试 等 教 学 环节 进 行 综 合 研 究 ， 这 显 然 不 适 应 面 向 世纪 的教 学要 求 ， 加 强 国 际合 作 ， 资源 共 享 ， 经 验 互 通 也 成 为 必 然 趋 势 。物 理 教 材 中 很 多公 式 、 定理 的证 明都 要 借 助 于 相关 的 国外 原 版 教 材 ， 这 就 要求 在 中学 引 入 双 语 教 学 ， 通 过 对 这 门课 程 的学 习 ． 掌 握 物 理 定理 、 定 义 和一 些 专 有名 词 的正 确 英 语 表达 ， 了解 物 理 的推 理 过程， 同 时 还 可 以提 高英 文 文 献 的 阅 读 能 力 。 ２ ． 上 好 物 理 双 语 课 需 要注 意 的 方 面

a Vector can never be equal to a Scalar
.
10
1.6 Vector Algebra
a Vector quantity can be expressed by a
• Every physical theory has a range of validity. • No theory has ever regarded as the final truth.
1.2 Idealized Models A model is a simplified version of a physical
Chapter 1 Physical Quantities
and Vectors
.
1
1.1 The nature of physics
• Physics is an experimental science. • Physicists observe the phenomena of nature and try to find patterns and principles that relate these phenomena. These patterns are called physical theories or physical laws, or principles.
ball move in a vacuum.
We ignore the earth’s rotation.
point masses
We make the weight constant.
.
5
1.3 Standards and Units
Physics is an experimental science. Experiments require measurements.

课程名称：物理学基础（Physics Basics）课时： 2课时教学目标：1. 知识与技能目标：- 学生能够理解并掌握牛顿第一定律的概念。

- 学生能够运用牛顿第一定律解释日常生活中的现象。

- 学生能够使用英语描述牛顿第一定律及其相关物理量。

2. 过程与方法目标：- 通过小组讨论和实验活动，培养学生的合作能力和问题解决能力。

- 通过英语交流，提高学生的英语听说能力。

3. 情感态度价值观目标：- 激发学生对物理学的兴趣，培养学生对科学探索的热情。

- 培养学生的科学精神和批判性思维能力。

教学重难点：- 重点：牛顿第一定律的概念和物理意义。

- 难点：牛顿第一定律在生活中的应用和英语表达。

教学过程：第一课时一、导入新课1. 教师用英语简要介绍牛顿第一定律的背景和重要性。

2. 提问：What do you know about the law of inertia?（你们知道什么关于惯性定律？）二、新课讲解1. 牛顿第一定律的定义：- 教师用英语讲解牛顿第一定律的定义：“An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.”（静止的物体将保持静止，运动的物体将保持匀速直线运动，除非受到非平衡力的作用。

）- 用中文解释并举例说明。

2. 牛顿第一定律的物理意义：- 教师用英语解释牛顿第一定律的物理意义，并引导学生思考其重要性。

- 用中文总结并讨论。

三、小组讨论1. 将学生分成小组，讨论以下问题：- What are some examples of objects at rest and in motion in ourdaily life?- How does the law of inertia affect our daily life?2. 各小组用英语汇报讨论结果，教师点评并纠正发音和语法错误。

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r C
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Coordinate system can be chosen freely.
Components of vectors (50)
A component of a vector is the projection of the vector on an axis. This component may be in 2 or 3 dimensional (3D) coordinate system.
gh = 1 v 2 2
L T 2 L
L 2 T
[L T ] = [L T ]
2 2 2 2
一个量的量纲是该量所描述的物理特性
The dimensions on both sides of an equation must be equal. ----check the validity(正确性)of a calculation
Watt and Angstrom ( A ) are derived units.
Derived unites are defined in terms of the basic units. Some have given names, some do not.
speed force m/s kgm/s2 Newton
Scientific notation Mass of the Earth:

学物 理 课 程 Ｐ ｅ－ ｔ ｃｏ ｅ Ｉ ｒ ｔ ｎ教 学模 式 的 特 点 和 教 学效 果 。 ｒ ｎ ｕｉ ｓ
关键词 ： 双语 教 学 ； ｅｒＩｓ ｕｔ ａ教 学模 式 ； 学物 理 Ｐｅ－ｎｔ ｃｉ ｒ ｏ 大
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航 海教 育研 究 ２ １ ． ０２ １
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要 ： 概 述 Ｐ ｅ—ｎｔ ｃｏ 在 ｅｒ ｓｒ ｔ ｎ教 学模 式 的 基 础 上 ， Ｉ ｕｉ 分析 开展 双 语 物 理 课 程 Ｐ ｅ—ｎｔｕｔ ｎ 学模 式 实践 基 础 ， 讨 大 ｅｒＩ ｒｃｉ 教 ｓ ｏ 探
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加州 大学 伯克利 分校 、 圣地 亚哥 分 校 以及 我 国香港 路 。 。 大学 、 北京 师范大 学 、 浙江 大学 等多 所 高校 ， 班 的 大

大学物理双语教学的探讨摘要：本文介绍了作者几年来进行大学物理双语课程教学积累的经验与体会。

从大学物理双语课程教学的意义、面临的困难、教材选择、课程体系设计、英语运用，以及与科研结合等几方面，探讨了在高校开展大学物理双语课程教学可行的措施与方案。

关键词：打学物理双语课程；双语教学；课程设计北京邮电大学从2005年面向国际学院04级开始开设大学物理双语课程，迄今3年，对2004级——2006级学生的教学中，取得了一些成功的经验。

在2007年3月，我们申请获得北京邮电大学教学研究与改革项目“大学物理双语课程教学的探索与实践”，面向2006届学生开设大学物理双语课程试点A班，研究在各个专业中基础课程取语教学的模式。

本文介绍我们在理工科大学生的主要基础课——大学物理课程双语教学的实践中的研究思路、具体做法和改革实践，就大学物理双语教学提出一些抛砖引玉的看法和建议。

一、大学物理双语教学的意义和面临的困难物理学是一门自然科学，研究自然界物质的运动形式和运动规律。

有一套完整的研究认识规律，和使用的语言无关，语言只是研究物理和描述自然的工具，不能替代科学的思维。

但是由于社会和历史的原因，近几百年来科学技术的成果大多是用英语发表的。

不论是科学技术专业杂志，还是各种国际学术会议，大都将英语作为交流语言。

从这个意义上可以说“现代科学技术的研究主要是用英语思考”。

从一个狭义的层面看，我国理工科硕士生、博士生的培养中，英语是必不可少的获取课题背景的重要工具。

然而国内大学的英语教学与专业课程脱钩的现状，造成我们多数硕士生和少量博士生还不能熟练运用英语开展科学研究。

比如论文开题阶段，学生宁愿去看国内杂志上有关课题“再加工”的综述文章，也不愿去查阅国外杂志上有关本课题的最原始的和最新近的相关文章，他们实际掌握的是第二手而不是第一手研究背景资料。

另外学生们运用英语表述困难，使他们的研究成果很难在国际知名杂志或国际会议上发表，以至使研究水平的展现大打折扣。

《物理双语教学课件》Chapter14ElectricCharge电荷Chapter 14 Electric ChargeWe now begin a study of the branch of physics concerned with electric and magnetic phenomena. The laws of electricity and magnetism play central roles in operation of many devices such as radios, televisions, electric motors, computers, and high-energy accelerators. More fundamentally, we now know that the inter-atomic and inter-molecular forces that are responsible for the formation of solids and liquids are electric in origins. Furthermore, such forces as the pushes and pulls between objects and the elastic force in a spring arise from electric forces at the atomic level.The early Greek philosophers knew that if you rubbed a piece of amber, it would attract bits of straw. The Greeks also recorded the observation that some natura lly occurring “stone” known today as the mineral magnetite, would attract iron. From these modest origins, the sciences of electricity and magnetism developed separately for centuries-until 1820.The new science of electromagnetism,the combination of electrical and magnetic phenomena, was developed further by workers in many countries. One of the best was Michael Faraday, a truly gifted experimenter with a talent for physical intuition and visualization. In the mid-19th century, James Clerk Maxwellput Fa raday’s ideals into mathematical form, introduced many new ideas of his own, and put electromagnetism on a sound theoretical basis.14.1 Electric Charge1.The definition of electric charge: If you walk across a carpet in dry weather, you can produce a spark by bring your finger close to a metal doorknob. On a grander scale, lighting is familiar to everyone. Each of those phenomena represents a tiny glimpse of the vast amount of electric charge that is stored in the familiar objects that surround us. Electric charge is an intrinsic characteristic of the fundamental particles making up those objects.2.The classification of charge(1)The vast amount of charge in an everyday object is usually hidden because the object contains equal amounts of two kinds of charge: positive charge and negative charge. With such equality-or balance-of charge, the object is said to be electrically neutral; that is, it contains no net charge to interact with other objects.(2)If the two types of charge are not in balance, then there isa net charge that can interact with other objects. And webecome aware of the existence of the net charge. We say that an object is charged to indicate that it has a charge imbalance, or net charge. The imbalance is always very small compared to the total amount of positive charge and negative charge contained in the object.3.To see the demonstration ofthe figure, we come to theconclusion that charges withthe same electrical sign repeleach other, and charges with opposite signs attract each other.4.The SI unit of charge is derived from the SI unit of electric current, the ampere (A). The SI unit of charge is the coulomb (C).14.2 Conductors and Insulators1.In some materials, such as metals, tap water, and the humanbody, some of the negative charge can move rather freely. We call such materials conductors. In other materials, such as glass, chemically pure water, and plastic, none of the charge can move freely. We call these materials nonconductors or insulators.2.The properties of conductors and insulators are due to thestructure and electrical nature of atoms. Atoms consist of positively charged protons, negatively charged electrons, and electrically neutral neutrons. The protons and neutrons are packed tightly together in the central nucleus.3.The charge of a single electron and that of a single protonhave the same magnitude but are opposite in sign. Hence an electrically neutral atom contains equal numbers of electrons and protons. Electrons are held near the nucleus because they have the electrical sign opposite that of the protons in the nucleus and thus are attracted to the nucleus.4.When atoms of a conductor likecopper come together to form the solid,some of their outmost electrons do notremain attracted to the individualatoms but become free to wanderabout within the solid, leaving behind positively charged atoms. We call the mobile electrons conduction electrons. The experiment of the figure demonstrates the mobility of charge in a conductor.5.Semi-conductor, such as silicon and germanium, are materialsthat are intermediate between conductors and insulators. The microelectronic revolution that has changed our lives in so many ways is due to devices constructed of semi-conducting materials.6. Finally, there are superconductors , so called because they present no resistance to the movement of electric charge through them . When charge moves through a material, we say that an electric current exists in the materials. Ordinary materials, even good conductors, tend to resist the flow of charge through them, In a superconductor, however, the resistance is not just small; it is precisely zero .14.3 Coulomb ’s law1. Let two charged particles (also called point charges ) have charge magnitudes q 1 and q 2 and be separated by a distance r.(1) The electrostatic force of attraction or repulsion between them has the magnitude )'(412210221slaw Coulomb r q q r q q k F πε==,in which k is a constant with the value 229/1099.8C m N ??, called the electrostatic constant , and the quantity0ε, called the permittivity constant , with the value 2212/1085.8m N C ??-.(2) Each particle exerts a force of this magnitude on the other particle; the two forces form an action-reaction pair . If the particles repel each other, the forces on each particle point away from the other particle as in figures (a) and (b). If theparticles attract each other, the forceon each particle points toward theother particle as in figure (c).2.Coulomb’s law has the same form as that of Newton’sgravitational law.(1)Both describe inverse square laws that involve a propertyof the interacting particles-the mass in one case and the charge in the other.(2)The laws differ in that gravitational forces are alwaysattractive but electrostatic forces may be either attractive or repulsive, depending on the sign of the two charges. This difference arises from the fact that, although there is only one kind of mass, there are two kinds of charge.3.Coulomb’s law has survived every experimental test; noexceptions to it have ever been found. It holds even within the atom, correctly describing the force between the positively charged nucleus and each of the negatively charged electrons, even though classical Newtonian mechanics fails in that realm and is replaced there by quantum physics. This simple law also correctly accounts for the forces that bind atoms and molecules together to form solids and liquids.4. Still another parallel between the gravitational force and the electrostatic force is the both obey the principle of superposition . If we have n charged particles, they interact independently in pairs, and the force on any one of them, let us say particle 1, is given by the vector sum n F F F F F 11413121 ++++=, in which, for example, 14F is theforce acting on particle 1 owing to the presence of particle 4. An identical formula holds for the gravitational force.5. The two shell theorems that we found so useful in our study of gravitation have analogs in electrostatics: (1) A shell of uniform charge attracts or repels a charged particle that is out side the shell as if all the shell ’s charge were concentrated at its center. (2) A shell of uniform charge exerts no electrostatic force on a charged particle that is located inside the shell .6. Spherical conductors : If excess charge is placed on a spherical shell that is made of conducting material , the excess charge spreads uniformly over the external surface . This arrangement maximizes the distances between all pairs of the excess charge. According to the first shell theorem, the shell then will attract or repel an external charge as if the excess charge on the shell were concentrated at its center.14.4 Charge is quantized and conserved1.The experiment shows that electric charge is made up ofmultiples of a certain elementary charge. That is, any positive or negative charge q that can be detected can be written as = =nq, in which e, the elementary charge, has ±ne±,1±,3,2the value C19. The elementary charge is one of the .1-6010important constants of nature. The electron and proton both have a charge of magnitude e.2.When a physical quantity such as charge can have onlydiscrete values rather than any value, we say that the quantity is quantized. Energy, angular momentum are quantized; Charge adds one more important physical quantity to the list.3.If you rub a glass rod with silk, a positive charge appears onthe rod. Measurement shows that a negative charge of equal magnitude appears on the silk. This suggests that rubbing does not create charge but only transfers it from one body to another, upsetting the electrical neutrality of each body during the process. This hypothesis of conservation of charge has stood up under close examination, both for large-scale chargedbodies and for atoms, nuclei, and elementary particles. No exceptions have ever been found. Thus we add electric charge to our list of quantities-including energy and both linear and angular momentum-that obey a conservationlaw. We have a lot of examples: Radioactive decay of nuclei, annihilation process of electron and positron, and the pair production, the converse of annihilation.。

1. A particle moves along the x axis. Its position as a function of time is given by x=6.0t+8.5t 2, where t is in seconds and x is in meters. What is the acceleration as a function of time? Solution:We find the velocity and acceleration by differentiating x = (6.0 m/s)t + (8.5 m/s 2)t 2:v = d x /d t = (6.0 m/s) + (17 m/s 2)t ;a = d v /d t = 17 m/s 2.2. What is the centripetal acceleration of a child3.6m from the center of a merry-go-round?The child ’s speed is 0.85m/s.Solution: The centripetal acceleration i sa R = v 2/r = (0.85 m/s)2/(3.6 m) = 0.20 m/s 2 ,direction: toward the center.3. Huck Finn walks at a speed of 1.0m/s across his raft(that is,he walks perpendicular to the raft ’s motion relative to the shore).The raft is traveling down the Mississipppi River at a speed of 2.5m/s relative to the river bank.What is the velocity(speed and direction)of Hack relative to the river bank? Solution:If v HR is the velocity of Huck with respect to the raft, v HB the velocity of Huck with respect to the bank, and v RB the velocity of the raft with respect to the bank, then v HB = v HR + v RB , as shown in the diagram.From the diagram we get v HB = (v HR 2 + v RB 2)1/2 = [(1.0 m/s)2 + (2.5 m/s)2]1/2 = 2.7 m/s .We find the angle fromtan θ = v HR /v RB = (1.0 m/s)/(2.5 m/s) = 0.40, which gives θ = 22° from the river bank .4. A 75-m-long train accelerates uniformly from rest.If the front of the train passes a railwayworker 140m down the track at a speed of 25m/s,what will be the speed of the last car as it passes the worker?Solution: We use a coordinate system with the origin at theinitial position of the front of the train. We can find the acceleration of the train from the motion up to the point where the front of the train passes the worker: v 12 = v 02 + 2a (D – 0);(25 m/s)2 = 0 + 2a (140 m – 0),which gives a = 2.23 m/s 2.Now we consider the motion of the last car, which starts at – L , to the point where it passes the worker: v 22 = v 02 + 2a [D – (– L )]= 0 + 2(2.23 m/s 2)(140 m + 75 m), which gives v 2 = 31 m/s .5．what is the maximum speed with which a 1200-kg car can round a turn of radius 80.0m on a flat road if the coefficient of friction between tires and road is 0.55?Is this result independent of theRBHRv = 0mass of the car?Solution: If the car does not skid, the friction is static, with F fr ≤ μs F N .This friction force provides the centripetal acceleration. We take a coordinate system with the x -axis in the direction of the centripetal acceleration.We write ∑F = m a from the force diagram for the auto: x -component: F fr = ma R = mv 2/R ;y -component: F N – mg = 0.The speed is maximum when F fr = F fr,max = μs F N .When we combine the equations, the mass cancels, and we get μs g = v max2/R ;(0.55)(9.80 m/s 2) = v max 2/(80.0 m), which gives v max = 21 m/s .The mass canceled, so the result is independent of the mass .6. A thin rod of length l carries a total charge Q distributed uniformly along its length.Determine the electric field along the axis of the rod starting at one end-that is,find E(x) for 0x ≥.Solution: We choose a differential element of the rod d x ' a distance x ' from the origin of the coordinate system, as shown in the diagram. the limits for x ' are0 to l . The charge of the element is d q = (Q /l ) d x '. We find the electric field by integrating along the rod:''2'2'000000001111()4()4()4()44()|lll dq Q dx Q Q QE i i i i i x x l x x l x x l x l x x x l πεπεπεπεπε--====-=+++++⎰⎰7. The resistance of a packing material to a sharp object penetrating it is a force proportional to the fourth power of the penetration depth,x:F=kx 4i.Calculate the work done to force the sharp object a distanced.Solution: The resisting force opposes the penetration. If we assume no acceleration, the applied forcemust be equal to this in the direction of the penetration. For a variable force, we find the work by integration:W =F ·d s =kx 4d x 0d =kx55d =kd 55.8. A ball of mass 0.540kg moving east (+x direction) with a speed of 3.9m/s collides head-on with a0.320-kg ball at rest.If the collision is perfectly elastic,what be the speed and direction of each ball after the collision? Solution:For the elastic collision of the two balls, we use momentum conservation for this one-dimensionalmotion: m 1v 1 + m 2v 2 = m 1v 1' + m 2v 2';(0.540 kg)(3.90 m/s) + (0.320 kg)(0) = (0.540 kg)v 1' + (0.320 kg)v 2'.Because the collision is elastic, the relative speed does not change:F Nm gF fryxa Rv 1 – v 2 = – (v 1' – v 2'), or 3.90 m/s – 0 = v 2' – v 1'. Combining these two equations, we getv 1' = 0.998 m/s , and v 2' = 4.89 m/s .9. Derive the formula for the moment of inertia of a uniform thin rod of length l about an axis throughits center,perpendicular to the rod. Solution:We select a differential element of the rod of length d x a distance x from the center of the rod. The element is equivalent to a point mass with a mass of d m = (M /l) d x . We integrate from x = – l/2 to x = l/2 to find the moment of inertia of the rod:/22223/23/2/22|()33212l l l l MM M l Ml I x dm x dx x ll l --=====⎰⎰ 10. what is the magnitude of the electric force of attraction between an iron nucleus(q=+26e) and itsinnermost electron if the distance between them is 121.510m -⨯?Solution:The magnitude of the Coulomb force isF = kQ 1Q 2/r 2=(9229.010/N m C ⨯•)(26)( 191.610C -⨯)( 191.610C -⨯)/(1.5 ⨯ 10–12m)2= 2.7 ⨯ 10–3 N .11. The total electric flux from a cubical box 28.0 cm on a side is 321.4510/N m C ⨯•.What charge is enclosed by the box?Solution: The total flux is depends only on the enclosed charge:0QεΦ=or 12223280(8.8510/)(1.4510/) 1.281012.8Q C N m N m C C nC ε--=Φ=⨯•⨯•=⨯=12. A boat can travel 2.20m/s in still water.(a)If the boat point its prow directly across a stream whose current is 1.2m/s,what is the velocity(magnitude and direction) of the boat relative to the shore?(b)What will be the position of the boat,relative to its point of origin,after 3.00s? Solution:(a ) If v BS is the velocity of the boat with respect to the shore,v BW the velocity of the boat with respect to the water, and v WS the velocity of the water with respect to the shore, then v BS = v BW + v WS , as shown in the diagram.From the diagram we getv BS = (v BW 2 + v WS 2)1/2 = [(2.20 m/s)2 + (1.20 m/s)2]1/2 = 2.51 m/s. We find the angle fromtan θ = v BW /v WS = (2.20 m/s)/(1.20 m/s) = 1.83, which givesWSv BWθ = 61.4° from the shore.(b ) Because the boat will move with constant velocity, the displacement will bed = v BS t = (2.51 m/s)(3.00 s) = 7.52 m at 61.4° to the shore.13. A charge Q creates an electric potential of +125V at a distance of 15cm.What is Q?Solution:We find the charge from V = Q /4πÅ0r ;125 V = (9.0 ⨯ 109 N · m 2/C 2)Q /(15 ⨯ 10–2 m), which gives Q = 2.1 ⨯ 10–9 C = 2.1 nC.14. (a)What is the force per meter of length on a straight wire carring a 7.40A - current when perpendicular to a 0.90T -uniform magnetic field?(b)What if the angle between the wire and field is 45.0?Solution: a ) The maximum force will be produced when the wire and the magnetic field areperpendicular, so we have F max = ILB , or F max /L = IB = (7.40 A)(0.90 T) = 6.7 N/m . (b ) We find the force per unit length fromF /L = IB sin 45.0° = (F max /L ) sin 45.0° = (6.7 N/m) sin 45.0° = 4.7 N/m .15. An electron experiences the greatest force as it travels 62.910/m s ⨯ in a magnetic field when it is moving northward.The force is upward and of magnitude 137.210N -⨯.What is themagnitude and direction of the magnetic field?Solution:The greatest force will be produced when the velocity and the magnetic field are perpendicular. We point our thumb down (a negative charge!), and our fingers north. We must curl our fingers to the east, which will be the direction of the magnetic field. We find the magnitude from F = qvB ;7.2 ⨯ 10–13 N = (1.60 ⨯ 10–19 C)(2.9 ⨯ 106 m/s)B , which gives B = 1.6 T east .16. Two different dielectrics each fill half the space between the plates of a parallel-plate capacitor.Determine a formula for the capacitance in terms of 1K ,2K ,the area A of the plates,and the separation d. Solution:The potential difference must be the same on each half of the capacitor, so we can treat the system as two capacitors in parallel: C = C 1 + C 2 = [K 1Å0(1/2A )/d ] + [K 2Å0(1/2A )/d ] = (Å01/2A /d )(K 1 + K 2) = 1/2(K 1 + K 2)(Å0A /d )= Å0A (K 1 + K 2)/2d.17. Let two long parallel wires,a distance d apart,carry equal currents I in the same direction.One wireis at x=0,the other at x=d.Determine B between the wires as a function of x.Solution:Because the currents are in the same direction, between the wires the fields will be in opposite directions. For the net field we have B = B 1 – B 2 = [(μ0/4π)2I 1/x ]j – [(μ0/4π)2I 2/(d – x )]j = (μ0/4π)2I {[(d – x ) – x ]/x (d – x )}j= [(μ0/4π)2I (d – 2x )/x (d – x )]j .18. A 32-cm-long solenoid,1.8cm in diameter,is to produce a 0.30T - magnetic field at its center.If the maximum current is 5.7A ,how many turns must the solenoid have? Solution: We find the number of turns in the solenoid from B = μ0nI = μ0NI /L ;0.30 T = (4π ⨯ 10–7 T · m/A)[N /(0.32 m)](5.7 A), which gives N =1.3 ⨯ 104 turns .19. A spherical cavity of radius 4.50 cm is at the center of a metal sphere of radius 18.0 cm.A point charge 5.50Q C μ= rests at the very center of the cavity,whereas the metal conductor carries no net charge.Determine the electric field at a point(a)3.0 cm from the center of the cavity and (b)6.0 cm from the center of the cavity.Solution: From the symmetry of the charge distribution, we know thatthe electric field must be radial, with a magnitude independent of the direction.(a ) For a spherical Gaussian surface within the sphericalcavity, we have204enclosedQ QE dA E r πεε•=•==⎰,so we have692220(5.5010)(9.010/)4(0.030)QC N m C E r m πε-⨯⨯•==75.510/N C =⨯(away from the center)(b ) The point 6.0 cm from the center is inside the conductor,thus the electric field is 0.Note that there must be a negative charge of – 5.50 μC on the surface of the cavity and a positivecharge of + 5.50 μC on the outer surface of the sphere.20. How strong is the electricfield between the plates of a 0.80F μair-gap capacitor if they are2.0mm apart and each has a charge of 72 C μ? Solution: We find the potential difference across the plates fromdI 1B 1B 2I 2xxyQ = CV ;72 μC = (0.80 μF)V , which gives V = 90 V . We find the uniform electric field between the plates fromE = V /d = (90 V)/(2.0 ⨯ 10–3 m) = 4.5 ⨯ 104 V/m .21. Try to find out I of a uniform thin rod of mass m , length L about C （central ）？dx dm λ=, dm x dI 2= dx x λ2= ⎰-=222L L c dx x I λ 3121L λ=2121mL =22. Protons move in a circle of radius 5.10cm in a 0.725-T magnetic field.What value of electric fieldcould make their paths straight?In what direction must the electric field point? Solution: For the circular motion, the magnetic force provides the radial acceleration: qvB = mv 2/r , or v = qBr /m .To make the path straight, the forces from the electric field and the magnetic field balance:qE = qvB = q (qBr /m )B , or E = qB 2r /m = (1.60 ⨯ 10–19 C)(0.725 T)2(0.0510 m)/(1.67 ⨯ 10–27 kg)= 2.57 ⨯ 106 V/m perpendicular to B .23.infinte line wire (E and B)。

1. Newton’s law for the gravitational force and Coulomb’s lawfor the electrostatic force are mathematically identical. Thus the general feature we have discussed for the gravitational force should apply to the electrostatic force.2. In particular, we can infer that the electrostatic force is aconservative force . Thus when that force acts between two or more charged particles within a system of particles, we can assign an electric potential energy U to the system.3. Moreover, If the system changes its configuration from aninitial state i to a different final state f, the electrostatic force does work W on the particles. The resulting change U in thepotential energy of the system is U = U U = W . As withf iother conservative forces, the work done by the electrostatic force is path independent.4. For convenience, we usually take the reference configurationof a system of charged particles to be that in which the particles are all infinitely separated from each other. And we usually set the corresponding reference potential energy to be zero.17.2 Electric Potential1. The potential energy of a charged particle in an electric field depends on the magnitude of the charge. However, the potential energy per unit charge has a unique value at any point in the electric field. Thus the potential energy per unit charge, which can be symbolized as U/q, is independent of the charge q of the particle and is characteristic only as the electric field we are investigating. The potential energy per unit charge at a point in an electric field is called the electric potential V(or simply the potential) at that point . Thus V = U Electricpotential is a scalar, not a vector.2. The electric potential difference V between any two points i and f in an electric field is equal to the difference in potential energy per unit charge between the two point : V = V f V i = q U = q W. The potential difference between two point is thus the negative of the work done by the electrostaticforce per unit charge that move from one point to the other .3. The SI unit for electric potential is the joule per coulomb. This combination occurs so often that a special unit, the volt (abbreviated V) is used to represent it.4. One electron-volt (eV) is the energy equal to the workq .required to move a single elementary charge e through a.potential difference of exactly one volt, so 1eV 1.60 1019 J 17.3 Equi-potential Surfaces1. Adjacent points that have the same electric potential form anequipotential surface, which can be either an imaginary surface or a real, physical surface. No net work W is done on a charged particle by anelectric field when theparticle moves betweentwo points i and f on thesame equipotential surface. See the Figure.2. Figure shows the electric field lines and cross sections ofequipotential surface for several cases. We can find that equipotential surfaces are always perpendicular to electric field lines and thus to E which is always tangent to these lines.17.4 Calculating the Potential from the Field1. We can calculate the potential difference between any twopoints i and f in an electric field if we know the field vectorat all positions along any path connecting those points.2. Consider an arbitraryelectric field, representedby the field lines in theright figure, and a positivetest charge q that moves along the path shown from point ito point f. The differential work done on the particle by theelectrostatic force during a displacement is Thus the total work done on the particle by dW = F . ds = q E . ds .the fields is theintegration of the differential work done on the charge for all the differential displacement along the path. W = q ϕf . d.i Therefore, V V = W = ϕ f . d.f i q i3. If we choose the potential V at point i to be zero, theniV = ϕf . d, in which we have drooped the subscript f. It gives ius the potential V at any point f in the electric field relative to the zero potential at point i.4. Potential due to a point charge: V = 1 q .4几r5. Potential due to a group of point charges: We can find the netpotential at a point due to a group of point charges to sum up the potential resulting from each charge at the given point . V = n V =1 n q i . Here q is the value of the ith charge, and i 4χ r ii =1 0 i =1 ir is the radial distance of the given point from the ith charge. i6. Potential due to an electric dipole : (1) V = 1 p cos 9 . (2)4χ r 2Induced dipole moment: Many molecules such as water(Seeing right figure) have permanentelectric dipole moments. In othermolecules (nonpolar molecules) andin every atom, the centers of thepositive and negative charges coincide and thus no dipole moment is set up. However, if we place an atom or nonpolar molecule in an external electric field, the field distorts the electron orbits and separates the centers of positive and negative charge and sets up a dipole moment that points in the direction of the field. This dipole moment p said to be induced by the field, and the atom or molecule is said to be polarized by the field. When the field is moved, the induced dipole moment and thepolarization disappear. See the figure.7. Potential due to a continuous charge distribution :V = ϕ dV = ϕ 1 dq . Here the integral is to be taken over the4 r 0entire charge distribution. (1) lines of charge; (2)Charged disk.1. The component of E in any direction is the negative of the rate of change of the electric potential with distance in that direction . Ex = ?V ?x ; E y = ?V ?y ; E z = ?V ?z.1. The electric potential energy of a system of fixed point-charges is equal to the work that must be done by external agent to assemble the system, bringing each charge in from an infinite distance .1. An excess charge placed on anisolated conductor will distributeitself on the surface of that conductorso that all points of theconductor-whether on the surface or inside-come to the same potential. This is true regardless of whether the conductor has an internal cavity.2. If an isolated conductor is placed in an external electric field,as in the right figure, all points of the conductor still come toa single potential regardless of whether the conductor has anexcess charge.。

How complicated?
1.2 Idealized Models
We neglect the size and shape of the ball by representing it as a point object, or particle. We neglect air resistance by making the ball move in a vacuum. point We ignore the earth’s rotation. We make the weight constant.
Chapter 1 Physical Quantities and Vectors
1.1 The nature of physics
• Physics is an experimental science. • Physicists observe the phenomena of nature and try to find patterns and principles that relate these phenomena. These patterns are called physical theories or physical laws, or principles. • Every physical theory has a range of validity. • No theory has ever regarded as the final truth.
Unit prefixes: kilo-, centi-, milli-, micro-, nano
1.3 Standards and Units
Operational definition: kilogram meter second Derived unit: In other cases we define a physical quantity by describing how to calculate it from other quantities that we can measure. speed = distance / time