《物理双语教学课件》Chapter 7 Rolling Torque, and Angular Momentum 力矩与角动量

第一篇：《物理双语教学课件》Chapter 7 Rolling Torque, and Angular Momentum 力矩与角动量

Chapter 7 Rolling Torque, and Angular Momentum

7.1 Rolling When a bicycle moves along a straight track, the center of each wheel moves forward in pure translation.A point on the rim of the wheel, however, traces out a more complex path, as the figure shows.In what follow, we analyze the motion of a rolling wheel first by viewing it as a combination of a pure translation and a pure rotation, and then by viewing it as rotating along.1.Rolling as rotation and translation combined

(1)Imagine that you are watching the wheel of a bicycle, which passes you at constant speed while rolling smoothly, without slipping, along a street.As shown in the figure, the center of mass O of the wheel moves forward at constant speed vcm.The point P at which the wheel contacts the street also moves forward at speed

vcm, so that it always remains directly below O.(2)During a time interval t, you see both O and P move forward by a distance s.The bicycle rider sees the wheel rotate through an angle  about the center of the wheel, with the point of the wheel that was touching the street at the beginning of t moving through arc length s.We have the relation where R is the radius of the wheel.(3)The linear speed

vcm

sR，of the center of the wheel is ds/dt，d/dt.and the angular speed  of the wheel about its center is So differentiating the equation with respect to time gives us vcmR.(4)The figure shows that the rolling motion of a wheel is a combination of purely translational and purely rotational motions.(5)The motion of any round body rolling smoothly over a surface can be separated into purely translational and purely rotational motion.2.Rolling as pure rotation

(1)The figure suggests another way to look at the rolling motion of a wheel, namely, as pure rotation about an axis that always extends through the point where the wheel contacts the street as the wheel moves.That is, we consider the rolling motion to be pure rotation about an axis passing through point P and perpendicular to the plane of the figure.The vectors in then represent the instantaneous velocities of points on the rolling wheel.(2)The angular speed about this new axis that a stationary observer assign to a rolling bicycle wheel is the same angular speed that the rider assigns to the wheel as he or she observes it in pure rotation about an axis through its center of mass.(3)To verify this answer, let’s use it to calculate the linear speed of the top of the wheel from the point of view of a stationary observer, as shown in the figure.3.The Kinetic energy: Let us now calculate the kinetic energy of the rolling wheel as measured by the stationary observer.(1)If we view the rolling as pure rotation about an axis through P in above figure, we have

KIp2/2, in which 

is the angular speed of the wheel and Ip is the rotational inertia of the wheel about the axis through P.(2)From the parallel-axis theorem, we have

IpIcmMR2, in which M is the mass of the wheel and

Icm

is its rotational inertia about an axis through its center of mass.(3)Substituting the relation about its rotational inertia, we

2obtainK1Icm21MR221Icm21Mvcm.2222(4)We can interpret the first of these terms as the kinetic energy associated with the rotation of the wheel about an axis through its center of mass, and the second term as the kinetic energy associated with the translational motion of the wheel.4.Friction and rolling

(1)If the wheel rolls at constant speed, it has no tendency to slide at the point of contact P, and thus there is no frictional force acting on the wheel there.(2)However, if a force acting on the wheel, changing the speed vcm of the center of the wheel or the angular speed 

about the center, then there is a tendency for the wheel to slide, the frictional force acts on the wheel at P to oppose that tendency.(3)Until the wheel actually begins to slide, the frictional force is a static frictional force fs.If the wheel begins to slide, then the force is a kinetic frictional force fk.7.2 The Yo-Yo 1.A yo-yo, as shown in the figure, is a physics lab that you can fit in your pocket.If a yo-yo rolls down its string for a distance h, it loses potential energy in amount mgh but gains kinetic energy in both translational and rotational form.When it is climbing back up, it loses kinetic energy and regains potential energy.2.Let us analyze the motion of the yo-yo directly with Newton’s second law.The above figure shows its free-body diagram, in which only the yo-yo axle is shown.(1)Applying Newton’s second law in its linear form yields FTMgMa, Here M is the mass of the yo-yo, and T is the tension in the yo-yo’s string.(2)Applying Newton’s second law in angular form about an axis through the center of mass yields TR0I, Where R0 is the radius of the yo-yo axle and I is the rotational inertial of the yo-yo about its center axis.(3)The linear acceleration and angular acceleration have relation aR0.So After eliminating T in both equations we obtain ag121I/MR0.Thus an ideal yo-yo rolls down its string with constant acceleration.7.3 Torque revisited In chapter 6 we defined torque  for a rigid body that can rotate around a fixed axis, with each particle in the body forced to move in a path that is a circle about that axis.We now expand the definition of torque to apply it to an individual particle that moves along any path relative to a fixed point rather than a fixed axis.The path need no longer be a circle.1.The figure shows such a particle at point P in the xy plane.A single force F in that plane acts on the particle, and the particle’s position relative to the origin O is given by position vector r.The torque  acting on the particle relative to the fixed point O is a vector quantity defined as

rF.2.Discuss the direction and magnitude of (rFsin).7.4 Angular Momentum 1.Like all other linear quantities, linear momentum has its angular counterpart.The figure shows a particle with linear momentum p(=mv)located at point P in the xy plane.The angular momentum l of this particle with respect to the origin O is a vector quantity defined as

lrpm(rv), where r is the position vector of the particle with respect to O.2.The SI unit

of

angular

momentum

is

the kilogram-meter-square per second(kgm2/s), equivalent to the joule-second(Js).3.The direction of the angular momentum vector can be found to use right-hand rule, as shown in the figure.4.The magnitude of the angular momentum is where lrmvsin, is the angle between r and p when these two vectors are tail to tail.7.5 Newton’s second law in angular form

1.We have seen enough of the parallelism between linear and angular quantities to be pretty sure that there is also a close relation between torque and angular momentum.It is dldt.2.The vector sum of all torques acting on a particle is equal to the time rate of change of the angular momentum of that particle.The torque and the angular momentum should be defined with respect to the same origin.3.Proof of the equation: angular momentum can be written as:

lm(rv)

Differentiating each side with respect to time yields dvdrdlm(rv)m(ravv)r(ma)r(F)(rF) dtdtdt

7.6 The Angular Momentum of A System of Particles Now we turn our attention to the motion of a system of particles with respect to an origin.Note that “a system of particles” includes a rigid body as a special case.1.The total angular momentum L of a system particles is the vector sum of the angular momenta l of the particles:

nLl1l2l3lnlii1, in which

i(1,2,3,)labels the

particles.2.With time, the angular momenta of individual particle may change, either because of interactions within the system(between the individual particles)or because of influences that may act on the system from the outside.We can find the change in L as these changes take place by taking the time derivative of above equation.Thus

nndlidLi.dti1dti13.Some torques are internal, associated with forces that the particles within the system exert on one another;other torques are external, associated with forces that act from outside the system.The internal forces, because of Newton’s law of action and reaction, cancel in pair(give a little more explanation).So, to add the torques, we need consider only those associated with external forces.Then above equation becomes

extdL dtThis is Newton’s second law for rotation in angular form, express for a system of particles.The equation has meaning only if the torque and angular momentum vectors are referred to the same origin.In an inertia reference frame, the equation can be applied with respect to any point.In an accelerating frame, it can be applied only with respect to the center of mass of the system.7.7 The Angular Momentum of a Rigid Body Rotating about a Fixed Axis We next evaluate the angular momentum of a system of particles that form a rigid body, which rotates about a fixed axis.Figure(a)shows such a body.The fixed axis of rotation is the z axis, and the body rotates about it with constant angular speed .We wish to find the angular momentum of the body about the axis of rotation.1.We can find the angular momenta by summing the z components of the angular momenta of the mass elements in the body.In figure(a), a typical mass element

mi

of the body moves around the z axis in a circular path.The position of the mass elements is located relative to the origin O by

position vector ri.The radius of the mass element’s circular path is ri, the perpendicular distance between the element and z axis.2.The magnitude of the angular momentum element, with respect to O, is where pi

li

of this mass

li(ri)(pi)(sin900)(ri)(mivi), and

vi

are the linear momentum and linear speed

900

is the angle between ri and of the mass element, and pi.3.We are interested in the component of li that parallel to the rotation axis, here the z axis.That z component is lizlisin(risin)(mivi)rimivi.4.The z component of the angular momentum for the rotating rigid body as a whole is found by adding up the contributions of all the mass elements that make up the body.Thus, because

nni1vr, we may write

n

ni12Lzlizmivirimi(ri)ri(miri)I

i1i15.The following table extends the list of corresponding linear and angular relations.11 7.8 Conservation of Angular Momentum So far we have discussed two powerful conservation laws, the conservation of energy and the conservation of linear momentum.Now we meet the third law of this type, the conservation of angular momentum.1.If no net external torque acts on the system, from Newton’s second law in angular form, we have

dL/dt0, or Laconstant.This result, called the law of conservation of angular momentum, can also be written as

LiLf.2.This means if the net external torque acting on a system is zero, the angular momentum of the system remains constant, no matter what changes take place within the system.3.Above equations are vector equations.They are equivalent to three scalar equations corresponding to the conservation of angular momentum in three mutually perpendicular directions.Depending on the torques acting on a system, the angular momentum of the system might be conserved in only one or two directions but not all directions.This is if any component of the external torque on a system is zero, then that component of the angular momentum of the system along that axis cannot change, no matter what changes take place within the system.4.Like the other two conservation laws that we have discussed, It holds beyond the limitation of Newton’s mechanics, It holds for particles whose speeds approach that of light(where the theory of relativity reigns), and it remains true in the world of subatomic particles(where quantum mechanics reigns).No exceptions to the law of conservation of angular momentum have ever been found.5.We now discuss four examples involving this law.P283 13

第二篇：《物理双语教学课件》Chapter 9 Oscillations 振动

Chapter 9 Oscillations

We are surrounded by oscillations─motions that repeat themselves.(1).There are swinging chandeliers, boats bobbing at anchor, and the surging pistons in the engines of cars.(2).There are oscillating guitar strings, drums, bells, diaphragms in telephones and speaker systems, and quartz crystals in wristwatches.(3).Less evident are the oscillations of the air molecules that transmit the sensation of sound, the oscillations of the atoms in a solid that convey the sensation of temperature, and the oscillations of the electrons in the antennas of radio and TV transmitters.Oscillations are not confined to material objects such as violin strings and electrons.Light, radio waves, x-rays, and gamma rays are also oscillatory phenomena.You will study such oscillations in later chapters and will be helped greatly there by analogy with the mechanical oscillations that are about to study here.Oscillations in the real world are usually damped;that is, the motion dies out gradually, transferring mechanical energy to thermal energy by the action of frictional force.Although we cannot totally eliminate such loss of mechanical energy, we can replenish the energy from some source.9.1 Simple Harmonic Motion 1.The figure shows a sequence of “snapshots” of a simple oscillating system, a particle moving repeatedly back and forth about the origin of the x axis.2.Frequency:(1).One important property of oscillatory motion is its frequency, or number of oscillations

that

are completed each second.(2).The symbol for frequency is f, and(3)its SI unit is hertz(abbreviated Hz), where 1 hertz = 1 Hz = 1 oscillation per second = 1 s-1.3.Period: Related to the frequency is the period T of the motion, which is the time for one complete oscillation(or cycle).That is T1f.4.Any motion that repeats itself at regular intervals is called period motion or harmonic motion.We are interested here in motion that repeats itself in a particular way.It turns out that for such motion the displacement x of the particle from the origin is given as a function of time by

x(t)xmcos(t), in which xm,,and

are constant.The motion is called simple harmonic motion(SHM), the term that means that the periodic motion is a sinusoidal of time.5.The quantity

xm, a positive constant whose value depends on how the motion was started, is called the amplitude of the motion;the subscript m stands for maximum displacement of the particle in either direction.6.The time-varying quantity

(t)

is called the phase of the motion, and the constant  is called the phase constant(or phase angle).The value of  depends on the displacement and velocity of the particle at t=0.7.It remains to interpret the constant .The displacement

x(t)

must return to its initial value after one period T of the motion.That is，x(t)

must equal

0x(tT)for all t.To simplify our analysis, we put.So we then have xmcostxmcos[(tT)].The cosine function first repeats itself when its argument(the phase)has increased by that we must have 22fT2 rad, so

(tT)t2orT2.It means.The quantity  is called the angular frequency

of the motion;its SI unit is the radian per second.8.The velocity of SHM:(1).Take derivative of the displacement with time, we can find an expression for the velocity of the particle moving with simple harmonic motion.That is, v(t)dx(t)xmsin(t)vmcos(t/2).(2).The dtpositive quantity

vmxm in above equation is called the velocity amplitude.9.The acceleration of SHM: Knowing the velocity for simple harmonic motion, we can find an expression for the acceleration of the oscillation particle by differentiating once more.Thus we have

a(t)dv(t)vmsin(t/2)amcos(t)dtThe positive quantity

amvm2xm is called the acceleration

a(t)2x(t)amplitude.We can also to get , which is the hallmark of simple harmonic motion: the acceleration is proportional to the displacement but opposite in sign, and the two quantities are related by the square of the angular frequency.9.2 The Force Law For SHM 1.Once we know how the acceleration of a particle varies with time, we can use Newton’s second law to learn what force must act on the particle to give it that acceleration.For simple harmonic motion, we have

Fma(m2)xkx.This result-a force proportional to the displacement but opposite in sign-is something like Hook’s law for a spring, the spring constant here being km2.2.We can in fact take above equation as an alternative definition of simple harmonic motion.It says: Simple harmonic motion is the motion executed by a particle of mass m subject to a force that is proportional to the displacement of the particle but opposite in sign.3.The block-spring system forms a linear simple harmonic oscillator

(linear oscillator for short), where linear indicates that F is proportional to x rather than to some other power of x.(1).The angular frequency  of the simple harmonic motion of the block is oscillator is

9.3 Energy in Simple Harmonic Motion 1.The potential energy of a linear oscillator depends on how much the spring is stretched or compressed, that is, on

k/m.(2).The period of the linear T2m/k.x(t).We have U(t)1212kxkxmcos2(t).222.The kinetic energy of the system depends on haw fast the block is moving, that is on

K(t)v(t).We have 1212mvm2xmsin2(t)22

1k212m()xmsin2(t)kxmsin2(t)2m23.The mechanical energy is

EUK121212kxmcos2(t)kxmsin2(t)kxm 222The mechanical energy of a linear oscillator is indeed a constant, independent of time.9.4 An Angular simple Harmonic Oscillator 1.The figure shows an angular version of a simple harmonic oscillator;the element of springiness or elasticity is associated with the twisting of a suspension wire rather than the extension and compression of a spring as we previously had.The device is called a torsion pendulum, with torsion referring to the twisting.2.If we rotate the disk in the figure from its rest position and release it, it will oscillate about that position in angular simple harmonic motion.Rotating the disk through an angle  in either direction introduce a restoring torque given by Here (Greek kappa)is a constant, called the .torsion constant, that depends on the length, diameter, and material of the suspension wire.3.From the parallelism between angular quantities and linear quantities(give a little more explanation), we have

T2I

for the period of the angular simple harmonic oscillator, or torsion pendulum.9.5 Pendulum We turn now to a class of simple harmonic oscillators in which the springiness is associated with the gravitational force rather than with the elastic properties of a twisted wire or a compressed or stretched spring.1.The Simple Pendulum(1).We consider a simple pendulum, which consists of a particle of mass m(called the bob of the pendulum)suspended from an un-stretchable, massless string of length L, as in the figure.The bob is free to swing back and forth in the plane of the page, to the left and right of a vertical line through the point at which the upper end of the string is fixed.(2).The forces acting the particle, shown in figure(b), are its weight and the tension in the string.The restoring force is the tangent component of the weight

mgsin, which is always acts opposite the displacement of the particle so as to bring the particle back toward its central location, the equilibrium(0).We write the restoring force as Fmgsin, where the minus sign indicates that F acts opposite the displacement.(3).If we assume that the angle is small, the

sin is very nearly equal to  in radians, and the displacement s of the particle measured along its arc is equal to FmgmgL.Thus, we have

smg()s.Thus if a simple pendulum swings LLthrough a small angle, it is a linear oscillator like the block-spring oscillator.(4).Now the amplitude of the motion is measure as the angular amplitude m, the maximum angle of swing.The period of

a

simple

pendulum

is T2m/k2m/(mg/L)2L/g.This result hods only if the angular amplitude m is small.2.The Physical Pendulum

(1).The figure shows a generalized physical pendulum, as we shall call realistic pendulum, with its weight mg

acting at the center of mass C.(2).When the pendulum is displaced through an angle  in either direction from its equilibrium position, a restoring torque appears.This torque acts about an axis through the suspension point O in the figure and has the magnitude (mgsin)(h).The minus sign indicates that the torque is a restoring torque, which always acts to reduce the angle  to zero.(3).We once more decide to limit our interest to small amplitude, so that (mgh).sin.Then the torque becomes

T2I/mgh,(4).Thus the period of a physical pendulum is when m is small.Here I is the rotational inertia of the pendulum.(5).Corresponding to any physical pendulum that oscillates about a given suspension point O with period T is a simple pendulum of length L0 with the same period T.The point along the physical pendulum at distance L0 from point O is called the center of oscillation of the physical pendulum for the given suspension point.3.Measuring g: We can use a physical pendulum to measure the free-fall acceleration g through measuring the period of the pendulum.9.6 Simple Harmonic Motion and Uniform circular Motion 1.Simple harmonic motion is the projection of uniform circular motion on a diameter of the circle in which the latter motion occurs.2.The figure(a)gives an example.It shows a reference moving circular particle in

P’

uniform with motion angular speed  in a reference circle.The radius xm of the circle is the magnitude of the particle’s position vector.At any time t, the angular position of the particle is t.3.The projection of particle P’ onto the x axis is a point P.The projection of the position vector of particle P’ onto the x axis gives the location x(t)of P.Thus we find

x(t)xmcos(t).Thus if reference particle P’ moves in uniform circular motion, its projection particle P moves in simple harmonic motion.4.The figure(b)shows the velocity of the reference particle.The magnitude of the velocity is xm, and its projection on the x axis is

v(t)xmsin(t).The minus sign appears because the velocity component of P points to the left, in the direction of decreasing x.5.The figure(c)shows the acceleration of the reference particle.The magnitude of the acceleration vector is projection on the x axis is

a(t)2xmcos(t).2xm

and its 6.Thus whether we look at the displacement, the velocity, or the acceleration, the projection of uniform circular motion is indeed simple harmonic motion.9.7 Damped Simple Harmonic Motion A pendulum will swing hardly at all under water, because the water exerts a drag force on the pendulum that quickly eliminates the motion.A pendulum swinging in air does better, but still the motion dies out because the air exerts a drag force on the pendulum, transferring energy from the pendulum’s motion.1.When the motion of an oscillator is reduced by an external force, the oscillator and its motion are said to be damped.An idealized example of a damped oscillator is shown in the figure: a block with mass m oscillates on a spring with spring constant k.From the mass, a rod extends to a vane(both assumed massless)that is submerged in a liquid.As the vane moves up and down, the liquid exerts an inhibiting drag force on it and thus on the entire oscillating system.With time, the mechanical energy of the block-spring system decreases, as energy is transferred to thermal energy of the liquid and vane.2.Let us assume that the liquid exerts a damped forceFd that is

proportional in magnitude to the velocity v of the vane and

block.Then

Fdbv，where b is a damped constant that depends on the characteristics of both the vane and the liquid and has the SI unit of kilogram per second.The minus sign indicates that

Fd

opposes the motion.3.The total force acting on the block is Fkxbvkxbdx.dtSo we have equation

d2xdxm2bkx0，dtdtwhose solution is x(t)xmebt/2mcos('t), where ', the angular frequency of the 12 damped oscillator, is given by 'kb2m4m2.4.We can regard the displacement of the damped oscillator as a cosine function whose amplitude, which is decreases with time.5.The mechanical energy of a damped oscillator is not constant but decreases with time.If the damping is small, we can find E(t)

xmebt/2m, gradually by replacing

xm

with

xmebt/2m，the amplitude of the

E(t)12bt/mkxme, which 2damped oscillation.Doing so, we find tells us that the mechanical energy decreases exponentially with time.9.8 Forced Oscillations and Resonance 1.A person swing passively in a swing is an example of free oscillation.If a kind friend pulls or pushes the swing periodically, as in the figure, we have forced, or driven, oscillations.There are now two angular frequencies with which to deal with:(1)the natural angular frequency  of the system, which is the angular frequency at which it would oscillate if it were suddenly disturbed and then left to oscillate freely, and(2)the angular frequency d of the external driving force.2.We can use the right figure to represent an idealized forced simple harmonic oscillator if we allow the structure marked “rigid support” to move up and down at a variable angular frequency d.A forced oscillator oscillates at the angular frequency d of driving force, and its displacement is given by

x(t)xmcos(dt), where xm

is the amplitude of the oscillations.How large the displacement amplitude and .3.The velocity amplitude

vm xm

is depends on a complicated function of d

of the oscillations is easier to describe: it is greatest when d, a condition called resonance.Above equation is also approximately the condition at which the displacement amplitude

xm

of oscillations is greatest.The figure shows how the displacement amplitude of an oscillator depends on the angular frequency

d of the driving force, for three values of the damped coefficient b.4.All mechanical structures have one or more natural frequencies, and if a structure is subjected to a strong external driving force that matches one of these frequencies, the resulting oscillations of structure may rupture it.Thus, for example, aircraft designers must make sure that none of the natural frequencies at which a wing can vibrate matches the angular frequency of the engines at cruising speed.15

第三篇：神经解剖学双语教学课件

神经解剖学双语教学课件

1． 感觉器官Sensory organs.ppt

2． 神经系统总论、脊髓Spinal cord.ppt

3． 脑干Brain stem.ppt

4． 小脑、间脑Cerebellum.ppt

5． 端脑

6． 神经传导通路

7． 脑和脊髓的被膜和血管

8． 脊神经

9． 脑神经 10．

11． 内脏神经、内分泌系统 帕金森病

第四篇：《物理双语教学课件》Chapter 15 Electric Fields 电场

Chapter 15 Electric Fields

Suppose we fix a positively charged particle q1 in place and then put a second positively charged particle q2 near it.From Coulomb's law we know that q1 exerts a repulsive electrostatic force on q2.Then you may ask how q1 “know” of the presence of q2? That is, since the charges do not touch, how can q1 exert a force on q2? This question about action at a distance can be answered by saying that q1 set up an electric field in the space surrounding it.At any given point P in that space, the field has both magnitude and direction.Thus when we place q2 at P, q1 interacts with q2 through the electric field at P.The magnitude and direction of that electric field determine the magnitude and direction of the force acting on q2.Another action-at-a-distance problem arises if we move q1, say, toward q2.Coulomb’s law tells us that when q1 is closer to q2, the repulsive electrostatic force acting on q2 must be greater, and it is.Does the electric field at q2, and thus the force acting on q2, change immediately? The answer is no.Instead, the information about the move by q1 travels outward from q1 as electromagnetic wave at the speed of light c.The change in the electric field at q2, and thus the change in the force acting on q2, occurs when the wave finally reach q2.15.1 The Electric Field 1.The temperature at every point in a room has a definite value.We call the resulting distribution of temperature as temperature field.In much the same way, you can imagine a pressure field in the atmosphere: it consists of the distribution of air pressure values, one for each point in the atmosphere.Theses two examples are of scalar field, because temperature and air-pressure are scalar quantities.2.The electric field is a vector field

(1)It consists of a distribution of vectors, one for each point in the region around a charged object.(2)In principle, we can define the electric field at some point near the charged object by placing a positive charge q0, called a test charge, at the point.(3)We then measure the electrostatic force test charge.The electric field object is defined as

FEq0E

F

that acts on the

at point P due to the charged

.(4)We represent the electric field at point P with a vector whose tail is at P, as shown in the figure.(5)The SI unit for the electric field is the newton per coulomb(N/C).15.2 Electric Field Lines 1.Michael Fraday, who introduced the idea of electric fields in the 19th century, thought of the space around a charged body as filled with lines of force.Although we no longer attach much reality to these lines, now usually called electric field lines, they still provide a nice way to visualize patters in electric fields.2.The relation between the field lines and electric field vectors(1)At any point, the direction of a straight field line or the direction of the tangent to a curve field line gives the direction of E at that point.(2)The field lines are drawn so that the number of lines per unit area, measured in a plane that is perpendicular to the lines, is proportional to the magnitude of

E.This second relation means that where the field lines are close together, E is large;and where they are far apart, E is small.3.Some Electric Field lines(1)The electric field lines of a sphere with uniform charge as shown in the right figure.(2)Right figure gives the electric field lines of an infinitely large, non-conducting sheet(or plane)with a uniform distribution of positive charge on one side.(3)The figure shows the field lines for two equal positive point charges.(4)The figure shows the pattern for two charges that are equal in magnitude but opposite sign.(5)From above figures, we can come to the conclusion: Electric field lines extend away from positive charge and toward negative charge.15.3 The Electric Fields for some cases 1.The electric field due to a point charge q

(1)If we put a positive test charge q0 at any point a distance r from the point charge, the magnitude of the electrostatic force acting on q0, from Coulomb’s law, is magnitude of the electric field vector is direction of E

F1qq0240r.The.The

EF1qq040r2is the same as that of the force on the positive test charge: directly away from the point charge, as shown in right figure, if q is positive, and toward it if q is negative.(2)We can find the net, or resultant, electric field due to more than one point charges with the aid of the principle of superposition.If we place a positive test charge q0 near n point charges q1, q2,…,qn, then the net force charges acting on the test charge is

F0

from the n point

F0F01F02F0n.So the net electric field at the position of the test charge is F0F01F02F0nEE1E2En.Here Ei q0q0q0q0is the electric field that would be set up by point charge i acting alone.5 2.The electric field due to an electric dipole: Figure(a)shows two

charges

of magnitude q but of opposite sign, separated by a distance d.We call this configuration an electric dipole.Let us find the electric field due to the dipole at a point P, a distance z from the midpoint of the dipole and on its central axis, as shown in the figure.(1)The magnitude of the electric field is Eqd1p20z320z31, in which the product qd is the magnitude

p p of a vector quantity known as the electric dipole moment of the dipole.3.The electric field due to a line of charge

(1)So far we have considered the electric field that is produced by one or, at most, a few point charges.We now consider charge distributions that consist of great many closed spaced point charges(perhaps billions)that are spread along a line, over a surface, or within a volume.Such distributions are said to be continuous rather than discrete.When we deal with continuous charge distributions, it is most convenient to express the charge on an object as a charge density rather than as a total charge.For a line of charge, for example, we would report the linear charge density(or charge per length)whose SI unit is the coulomb per meter.(2)The figure shows a thin ring of radius R with a uniform positive linear charge density , around its circumference.What is the electric field at point P, a distance z from the plane of the ring along its central axis?(3)We can get the magnitude of the electric field as Eqz40(z2R2)3/2.4.The electric field due to a charged disk(1)The figure shows a circular plastic disk of radius R that has a positive surface charge uniform density  on its upper surface.What is the electric field at point P, a distance z from the disk along its central axis?

(2)Our plan is to divide the disk into concentric flat rings and then to calculate the electric field at point P by adding up the contributions of all rings.The magnitude of the electric field is Ez(1).2220zR

15.4 A point charge in an electric field 1.We now want to determine what happens to a charged particle that is in an electric field that produced by other stationary or slowly moving charges.This force is given by which q is the charge of the particle and

E

FqE, in

is the electric field that other charges have produced at the location of the particle.2.Measuring the elementary charge: Figure shows the Millikan

oil-drop apparatus for measuring the elementary charge e.3.Ink-jet printing: Figure shows the

essential features of an ink-jet printer.15.5 A Dipole in an Electric Field 1.The figure shows why the molecule of water(H2O)is an electric dipole.If the water molecule is placed in an external electric field, it behaves as would be expected of the more abstract electric dipole.2.The torque can be written as

pE.3.Potential energy of an electric dipole: The expression for the potential energy of an electric dipole in an external electric field is simplest if we choose the potential energy to be zero when the angle  in the figure is 900.It can be written as

UpE.Microwave cooking: the explanation it.

第五篇：《物理双语教学课件》Chapter 23 Interference 干涉理论

Chapter 23 Interference

Sunlight, as the rainbow shows us, is a composite of all the colors of the visible spectrum.Soup bubbles and oil slicks can also show striking colors, produced not by refraction but by constructive and destructive interference of light.23.1 Young’s Interference Experiment 1.In 1801 Thomas Young experimentally proved that light is a wave, contrary what most other scientists then thought.He did so by demonstrating that light undergoes interference.2.Figure gives the basic arrangement of Young’s double-slit interference experiment.On

the viewing screen C, Points of interference maxima form visible bright rows-called bright bands, bright fringes, or maxima.Dark regions-called dark bands, dark fringes, or minima-result from fully destructive interference and are visible between adjacent pairs of bright fringes.The pattern of bright and dark fringes on the screen is called an interference pattern.Figure is a photograph of the interference pattern.3.To find

what

exactly determines the locations of the fringes in Young’s double-slit interference experiment, let us see the

figure.The

L

r2path-length-difference between rays be written as

r1 and can Ldsin, where d is the separation of the two slits.(1)For a bright fringe, number of wavelengths.It is dark fringe, It is L

L

must be zero or an integer

dsinmm0,1,2,.(2)For a

must be an odd multiple of half a wavelength.2m1,2,.(3)Using above two equations, dsin(2m1)we can find the angle  to any fringe and thus locate that fringe;further, we can use the values of m to label the fringes.4.We now wish to derive an expression for the intensity I of the fringes as a function of .(1)Let us assume that the electric field components of the light waves arriving at point P in the figure from the two slits are not in phase and vary with time as

E1E0sint and

E2E0sin(t2dsin)respectively.(2)So we have where 2dsin.(3)Thus

11EE1E22E0cos()sin(kx)，22the intensity is I(2E0cos1)2

214I0cos2.2

5.Combining more than two waves:

23.2 Interference from Thin Films 1.The colors we see when sunlight illuminates a soap bubble or an oil slick are caused by the interference of light waves reflected from the front and back surfaces of a thin transparent film.The thickness of the soap or oil film is typically the order of magnitude of the wavelength of the light involved.2.Figure shows a thin transparent film of uniform thickness L and index of refraction n2, illuminated by bright light of wavelength  from a distant point source.For now, we assume that air lies on both sides of the film and thus that

n1n3

in figure.For simplicity, we also assume that the light rays are almost perpendicular to the film.3.Reflection phase shifts: Refraction at an interface never causes a phase change.But reflection can, depending on the indices of refraction on two sides of the interface.When an incident wave travels in the medium of lesser index of refraction(with greater speed), the wave that is reflected at the interface undergoes a phase shift of  rad, or half a wavelength.4.The optic path-length difference in the case of thin film is 2nL2.2nL2mm1,2,.5.For a bright film, we have 6.For a dark film, we have

2nL2(2m1)2m0,1,2,.23.3 Michelson’s Interferometer 1.An interferometer is a device that can be used to measure lengths or changes in length with great accuracy by means of interference fringes.2.Figure shows

Michelson’s interferometer.