Class Note for PHYS 220 at IPFW
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Date Created: 02/06/15
Physics 221 Lecture Notes Introduce myself I II III Hand outs A Homework due as indicated B Quizzes NO MAKE UP and always a surprise C Labs D Exams are much like homework E Introduce myself 1 Write Name office number and office phone number 2 Web page httpusersipfwedugrovet 3 How I view this course a There are two basic ways of treating your brain 1 My brain is a warehouse 2 My brain is a factory I firmly believe this course focuses on this point rather than the first b Many if not all students have beliefs in terms of the physical world that are simply not true c Rene Descartes established rules so that he could determine for himself what was scientific fact from common belief In so doing Descartes helped establish the beginnings of modern scientific thought and reasoning d Rene Descartes rules for his own learning 1 Never accept anything that I did not clearly know to be true 2 Divide every difficult question into parts which can be dealt with separately 3 Think in an orderly way starting with what is simplest and easiest to know 4 Write down what I learned so that nothing would be omitted Quick Review of Phys 220 A B C D Vectors and Scalars Understanding the difference between acceleration velocity and displacement Newton s Laws 1 An object moves in straight line at a constant speed unless compelled to change its motion by a net force 2 E m5 3 If object one applies a force on object 2 then object 2 must apply a force of equal magnitude but opposite in direction to object 1 Energy conservation Periodic Waves Properties 1 A periodic wave is a wave with multiple repeating pulses 2 Period of a wave T typical symbol for period a Imagine you are at the beach or a large lake shore Stand in the water and observe the waves hitting your legs 9 89 10 11 12 13 14 15 16 17 18 19 b When a wave crashes into our legs we start a stop watch and time how long it takes for the next wave to come in This measured time is the period Frequency of a wave f typical symbol for the frequency Again imagine you are at the beach and have waded into the water Frequency is the average number of waves that crash into you per second A way to measure the frequency is to start a stop watch when a wave crashes into you Then count the number of waves that crash into you for a ten minute 600 second time period For example let s assume that 100 waves hit you in this time period f 100 0167Hz 6005 Frequency is highly related to period a If we use 100 waves hit us in 600 seconds we can deduce that the average period between each wave is 6 seconds b In general we nd f Wavelength A typical symbol for wavelength this is the Greek letter lambda Once more you are back at the beach and wading in the water This time a friend moves farther ahead of you in the water Your friend holds one end of a tape measure and you hold the other end When a wave crashes into you you yell Now to your friend Your friend then measures the distance between you and the next wave that will hit you see below Wavelength A FRIEND YOU Amplitude A typical symbol for amplitude Back to the beach again Amplitude is half the distance from the wave crest highest part of the wave to the wave trough lowest part of the wave See the below sketch Double the 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 39 quot amplitude There are two speeds associated with waves propagation velocity v a This is how fast the disturbance moves 20 medium velocity VM 21 This is how fast the medium moves up and down or side to side as the wave passes through it 22 For example consider how an ant oating in the water would move as the wave passed it 23 Relationship between propagation velocity wavelength and frequency a v fl Wavefronts a A wavefront marks a feature of a wave b By observing how wavefronts move we can examine wave motion see below sketch wavefronts Rays a rays describe the direction in which the wave is moving b Groups of rays often can tell us how the amplitude of the wave is changing c Rays are always perpendicular to wave fronts How do these wave properties relate to light 1 Wavelength corresponds to light s spectral color 2 spectral colors refer to all the colors we can see in a rainbow a Red orange yellow green blue violet 3 Can we see colors that are not in the rainbow a YES These colors include pink and brown How humans perceive color is more complicated than just the colors in a rainbow b l 350nm 350 billionths ofa meter 0000000350m is violet one end of the rainbow c l 750nm 750 billionths ofa meter 0000000750m is red other side of the rainbow d The other spectral colors are between these end points Frequency also corresponds to light s spectral color 0 fl If the wavelength changes that is a color change so must the frequency Electromagnetic waves are characterized by either the wavelength or frequency depends on which is easier to measure 8 Amplitude is related to brightness 8094 a The human concept of brightness corresponds to the physical term Intensity b Power is how much energy is delivered per time Standard unit is the Watt c Intensity is the power delivered over area of delivery 1 2 3 4 5 6 7 8 9 10 For example assume that a 60W light bulb produces 60W of light power actually reallife light bulbs use up 60W of electrical power If we completely enclose the light bulb in a spherical box the box will have 60W ofpower delivered to it The 60W of power is distributed evenly over the spherical box The inside surface of the sphere has an area of 4727quot2 where r is the radius of the sphere Thus the intensity is 4727quot2 a Note the intensity that reaches the eye would be even smaller because only a fraction of this total amount of light enters the eye A b Intensity reaching the eye 4 121 J where AMP is the area of one s pupil If we had a small area we could imagine that it is on the surface of the sphere Recall that we can only see light that enters our eyes Only the light that can pass through the iris a small opening in the eye is seen Thus humans judge brightness by the amount of light power passing through the iris Example consider a magnifying glass that is D in diameter focusing an electromagnetic wave to D50 What is the increase in intensity The power in the beam of light does not change The initial intensity is S P 2 7 2 2 The final intensity is S f P 2 102 2 100 sf m 2500 4P J 2500s IID IID The intensity is 2500 times greater Can light have any wavelength or frequency 1 If we define light as only the electromagnetic waves which we can see with our eyes then the wavelength of the light must be between 350nm and 750nm IV However an electromagnetic wave can have any wavelength We just can t see them Examples of electromagnetic waves Gamma rays have the lowest wavelength highest frequency Xrays have the second lowest wavelengths Ultraviolet light has the third lowest wavelengths visible light infrared light microwaves g radio waves Wop96 G Plane wave equation 1 angular frequency a 279 This is expressed in units of inverse seconds lsec While one does not really need this extra definition this definition allows one to express the wave displacement without multiplying by a factor involving 11 27239 wave number k The wave number 1s expressed 1n units of inverse meters While one does not really need this extra definition this definition allows one to express the wave displacement without multiplying by a factor involving 11 phase constant This term allows a wave to have any value at x00m and t 00s Superposition of Waves A What happens when two waves run into each other 1 When the two waves are combined in the same medium the disturbances simply add to each other a Consider two counterpropagating waves moving toward each other at 7 ms triangle moving to the right square moving toward left Waves at t 0 10 12 t 055 4 3 3 2 o 1 0 1 1 1 1 1 0 2 4 6 8 10 12 X0 Waves att1s 25 D arb 15 A 0 2 4 6 8 10 12 X m 2 let D1xt A1 sink1x m1t 1 which is a left moving wave 3 let D2 xt A2 sink2x wit 2 which is a right moving wave 4 Because they are in the same medium they have the same wave speed m1 ml a v k1 k2 5 Dirndlx tD1x tD2x t B What happens when two waves with the same frequency and amplitude run into each other 1 Can the two above waves have different wavelengths a NO Since they are in the same medium same speeds with the same frequency and v fl their wavelengths must also be identical 2 a21a22ak1k2kA1AZl 3 Dmm1xt Asinkx wtAsinkx mt 5 C l 2 3 Musica l 2 D 3 making use of the trig relation sinA i3 sinAcosBi cosAsinB Dmm1xt Asinkx mt Asinkx mt sinkxcosat coskxsinat total 95 A s1nkxcosat coskxs1nat m1ltxt 2A sinkxcosat We now observe that at certain positions the total disturbance is always zero D D mz39 a If x 7 where n 1nteger the s1ne term 1s zero wh1ch makes the total disturbance zero These locations are called nodes c Locations that have the largest possible disturbance are called anti nodes This superposition of traveling waves is called a standing wave wave doesn t move How do you create a standing wave The easiest way is to create a standing wave is through a string rmly held at two locations and allowed to oscillate in any direction besides these two points The re ected waves from either end tend to cancel each other out unless one matches the wavelength to the string boundaries Standing waves of light naturally form in most laser systems 1 notes from stringed musical instruments Consider a string tied down at x 0 and x L The traveling wave in the string must have nodes at x 0 and x L mt nn n1 or rewriting l g with n being any k 27239 71 Therefore L positive integer T The speed of any wave on the str1ng 1s determ1ned by v see page 4 u of the notes Therefore using V fl the frequency is f u a Musical instruments create a dominant frequency in terms of power contributions to the total wave called the fundamental frequency l T l The fundamental frequency 1s usually f 2L u b The distinctive sound of a musical instrument is determined by the relative power levels of nonfundamental frequencies E What happens when two waves with the same frequency the same amplitude but different phase constants are copropagating in the same medium 1 Let D1xtAsinkx at 1 2 Let D2xtAsinkx at z 3 Dmxt Asinkx at 1sinkx at 2 4 Useful trig Identities a sina i sin 2sina i cosa cosa cos 2cosa cosa cosa cos 2sina sina 5 Dmxt2Asinkx at 1 kx at 2 cos kx mt 1 kxat Z 6 Special case 1 2 2n7239 where n is any integer 1 2 n cos 1 2 Dmxt2 1quotAsinkx a2t l Hm a a phase shift difference of an integer times 27239 is equivalent to shifting a plane wave by a wavelength The two waves are identical Thus the total wave is double the amplitude of either initial copropagating wave b This case is called perfectly constructive interference 7 Special case 1 2 2nl7r where n is any integer cos 1 2 J 0 2 D20 x 0 a a phase shift difference of an odd integer times 7239 is equivalent to shifting a plane wave by a half a wavelength The two waves are exactly opposite each other one is positive while the other is the exact negative Thus the two wave s cancel each other out b This case is called perfectly destructive interference F Interference in 2 dimensions 1 Show result of laser beam passing through 2 close slits 2 Now a single spherical wave disturbance has the form Drt Asinkr wt to where r is the distance from the wave source Consider two spherical wave sources separated by a distance d 2 Qquot K h V Let r1 be the distance from the center of the upper slit to a point on a distance screen a distance L away and shifted a distance y upwards Let r2 be the distance from the center of the lower slit to a point on a distance screen a distance L away and shifted a distance y upwards D1r1t Asinkr1 at do Waves passing through upper slit D2 r2 t Asinkr2 at do Waves passing through lower slit D202 r1r2 t D1r1 tD2r1t Asinkr1 at be Asinkr2 at 0 D20 r1r2 t Asinkr1 wt 0 Asinkr2 wt to Using the trigonometric identity sin a sin 2sina cosa D202 r1r2 t Asinkr1 wt 0 Asinkr2 wt to 2Asinkr1 at 0 kr2 wt 0 gtlt D202r r 1 2 2 k k D202 r1r2 t 2As1n3r1 r2 at g ocos3r1 r2 If cos r1 r2 0 the total disturbance D202 is zero this is the cosikr1 mt 0 kr2 at 0 condition for a darldquiet location If cos r1 r2 i1 the total disturbance D202 has the greatest amplitude this is the condition for a brightloud location We can now make use of the trigonometric identities 2N 1 cosa 0 if a rad1ans 1e i900 i270 i450 cosa i1 if a Nzr radians ie 4500 i180 i360 Where N any integer Therefore we expect to nd the minimum locations when k 2N l 3hrz 2 kr1 r2 2Nl7r 27 42 mm Ar A N l 2 Therefore we expect to nd the maximum locations when k 301 r 2 2 N 27239 5 2 N Ar N2 Consider the result of two rocks being thrown simultaneously into a calm pond a solid lines refer to a crest b dotted lines refer to a trough 10 5 6 7 8 gt0 10 ll 12 Location of maX Above equilibrium 0 Location of min below equilibrium 9 Location of equilibrium Locations where two crests meet are marked with a blue dot Locations where two troughs meet are marked with a violet dot Locations where a crest meets a trough are marked with a yellow dot Where does constructive interference occur a When either two crests or two troughs meet Where does destructive interference occur a When a crest meets a trough How will the crests and troughs move over time a The concentric circles will move out at the same rate such that they are always equidistant from each other If we wanted an ant in the water to get as seasick as possible where should we place it a If we connect the locations of constructive interference we form multiple lines b These lines are called orders of constructive interference l The order actually refers to the arbitrary integer defined above c The ant on any of these lines will see the biggest crests and troughs If we wanted an ant in the water and wanted it to experience minimal waves where should we place it 11 a If we connect the locations of destructive interference we form multiple lines These lines are called orders of destructive interference c The ant on any ofthese lines will see no waves at all V Geometric Optics A Geometric optics is entirely based on rays 1 Rays travel in straight lines in the same material a Waves travel in straight lines so rays represent this property of light 2 Converging Rays indicate increasing light intensity D 3 iverging Rays indicate decreasing light intensity gt 4 If rays from an object do not enter our eye we cannot see it 5 We don t see individual rays we see sources of rays B Re ections 1 Law of Re ection The incident angle and the angle of re ection are always equal to each other Re ected Incident beam of beam of 1 ht light 1g surface a 91 is the incident angle b 9 is the re ected angle 2 The law of re ection follows the same rules you would use to determine how bank shots work in pool or in puttputt 3 Do waves like light re ect a The simple answer is yes b Recall that a wave on a rubber hose will re ect several times back and forth c Light can also re ect off of a surface 4 Whenever a wave hits a new medium 2 different events happen a some of the wave can be re ected b some of the wave can transmit beyond the boundary into the second material 5 Schematic of the typical household mirror 12 Glass Thin silver layer a The silver surface touching the glass re ects the light b The glass protects the silver from scratches and oxidation silver in air over time will tarnish 6 Half 7 Silvered Glass or 2way mirrors a These days they are no longer made using a light coating of silver b Today a thin lm coating on the glass provides the re ection Film K Glass V 7 How we see most objects via re ections a First we require a source of ambient light background light One can observe that in a dark room one cannot see any nonluminous object b The background light white light actually contains all the spectral colors c Light from the ambient source bounces off various objects However different amounts of spectral colors bounce off an object In the below sketch only red spectral light is re ecting off the shirt Hence the observer sees the shirt as red SUN 13 8 How does the law of re ection apply to something like a shirt We all see it and we don t have to be located at angle predicted by the law of re ection a Most objects have rough irregular surfaces b Light rays from a single luminous light source can re ect at widely different angles due to the surface c If the surface is rough enough and enough light is used rays go outward in practically all directions 9 When light re ects off a at mirrorlike surface it is called a specular re ection 10 When light re ects off a rough irregular surface it is called a diffuse re ection C In terms of rays what does the eye see 1 A single ray from a point object allows no perception besides light dark 2 Multiple rays from a point object allow one to locate the object a The eye locates objects by sensing that they are sources of rays b This requires that the rays reaching the eye from a Visible point be nonparallel 3 In geometric optics an object is a light source and it is also a source of rays 4 An image is a perceived source of rays a There are two different types of images 1 Real images can be projected on a screen We can place our finger at the image location 2 Virtual image can only be seen if the Viewer looks at or through the optical deVice The image location is inaccessible to our fingers D Plane Mirror that is its surface lies in a plane 14 3 This is a plain old mirror The image formed is a virtual image a A virtual image is an image that looks like an object inside the mirror b Demonstrate large at mirror c Explain how when they look in it looks like there is a hole in the wall and the person on the other side is rightleft reversed mirror image A real image is in general not rightleft reversed mirror image and the object appears to be on our side of the glass This is used in applications where we wish to project the image on a screen Spherical Mirrors Elk WP 0 Consider a basketball and cutting off a portion of it Then coat one of the sides with a highly re ecting substance If you coat what was the inside surface then we have a concave mirror If you coat what was the outside surface then we have a convex mirror The principal axis is a line that extends on both sides normal to the spherical mirror s center If we consider only rays that are near the principal axis paraxial rays and these rays are parallel with each other plane waves they will all cross at a point the focus The focus is related to the radius of curvature the initial radius of the basketball by these relations a Concave mirror f ER 15 b Convex mirror f i R the negative indicates that the rays cross in back of the mirror However an illuminated object will not have all the rays moving out in parallel straight lines Ray Diagrams for concave mirrors a Assume the bottom of the illuminated object is on the principal axis b Example assume an object is half way between the focus and the center of curvature for the mirror c Draw everything to scale examples appear below A object l6 d Ray 1 Draw a line ray from the top of the object a1row to the mirror such that the line is parallel to the principle axis This ray re ects from the mirror through the focal point A r c f object e Ray 2 Draw a line ray from the top of the object a1row to the mirror such that the line passes through the focal point or would pass through the focal point if the line is extended This ray re ects from the mirror such that the outgoing ray will be parallel to the principle axis f object l7 Ray 3 7 Draw a line ray from the top of the object arrow to the mirror such that the line passes through the center point or would pass through the center point if the line is extended This ray re ects back upon itself object 0 I object The location where the beams cross or would appear to cross if the re ected rays are extended is the image Draw a vertical line from the principal aXis to the crossing point Place an arrow at the crossing point If the a1row is in the opposite direction then the object the image is inverted If one measures the distance from the mirror to the image location on the ray diagram it is the same as where it really would occur The fact that the two arrows are not the same size indicates magni cation 18 j Examples r object image For the above example the image is located between the focal point and center point the image is real the image is inverted upside down and the image is shrunken compared to the object size I 1quot 4 For the above example the image is located behind the mirror the image is virtual the image is noninverted right side up and the image is enlarged compared to the object size 10 Ray diagrams for convex mirrors a Everything to correct size as determined by the scale f object 19 Ray 1 7 Draw a line ray from the top of the object arrow to the mirror such that the line is parallel to the principle aXis This ray re ects such that if we extended the line it would pass through the focal point Ray 2 7 Draw a line from the top of the object arrow to the mirror such that if we continued the line beyond the mirror it would pass through the focal point This ray re ects parallel to the principle aXis Ray 3 Draw a line from the top of the object a1row to the mirror such that if we continued the line beyond the mirror it would pass through the center point This ray re ects back upon itself 20 e To find the image we must examine only the 3 rays that are re ected from the mirror If we back track these rays beyond the mirror they will intersect at the image location lt T t f object image f For the above example the image is located between the mirror and the focal point the image is virtual the image is noninverted right side up and the image is shrunken 11 General results for spherical mirrors a If the object is closer to a concave mirror than the focal point the image will always be virtual noninverted and enlarged b If the object is farther from a concave mirror than the focal point but closer than the center point the image will always be real inverted and enlarged 21 c If the object is farther from a concave mirror than the center point the image will always be real inverted and shrunken d When using a convex mirror the image will always be viItual noninverted and shrunken F The Mirror Equation 1 l l l ddf 0 x alt7 object distance or the distance the object is away from the mirror d image distance or the distance the image is away from the mirror 1 NOTE all gt 0 if the image is on the same side of the mirror as the object real image and all lt 0 if the image is on the opposite side of the mirror as the object f focal length of the mirror magni cation is defined as the ratio of the image height to the object height d m When m is negative the image is inverted Any distance that goes behind the mirror is considered negative Example A 20cm high object is situated 150cm in front ofa concave mirror that has a radius of curvature of 100cm Using the mirror equation find a the image location and b the image height R 50cm f 2 d015Ocm l l l 61 d f Liiz 1 1 20133 611 f alt7 500m 1500m cm dl75cm d m 050m 0 The magnification indicates the image is half size and inverted Example2 the same object is located at the same distance from 2 spherical mirrors A and B The magnifications are mA 40 and m8 20 Find the ratio of the focal lengths of the mirrors d m d 0 22 mA40 d mB20 7 dlA 4 0610 dug 20d0 l l l l l l fA d0 dIA f8 d0 6118 Li LL 1 I fA 0 0 0 f8 do 2610 2610 4 fA do f3 2d0 4 d 3 2 f3 2610 3 VI Refraction and Lenses A Refraction The bending of light at it travels from one material into another material 1 The bending of light is entirely due to light traveling at different speeds in one medium compared to another a Suppose we located the points at which the light has its maximum disturbance These lines of maximal disturbance are wavefronts b Now imagine a laser beam traveling from air and going into a liquid The lines across the beams represent wavefronts Air Liquid c Note how some sections of the wavefronts enter the liquid before other sections 1 Because these sections of wavefronts enter the liquid first and have to slow down the whole beam seems to bend 23 2 We should also note that the wavelengths the distance from one wavefront to the next inside the liquid are smaller than the wavelengths in the air a v fl still applies and if the speed of the wave changes something must also change on the right side of the equation The wavelength changes the frequency does not 2 Snell s Law n1 sin 91 M2 sin 92 91 Mater1al l I Material 2 62 3 The subscript 1 refers to the material the light starts in first material 4 The subscript 2 refers to the material the light end in second material 5 The angles are measured with respect to the normal This is why a metal spoon looks like it bends when immersed in water The index of refraction for any material is defined as c speed of light in vacuum E I v speed of light in material quot10227101 The speed of light through all other materials is slower mathematically n 2 l The index of refraction of a vacuum 1000000000 exactly one The index of refraction of air 1000293 However the relationship between the wave s speed frequency and wavelength still holds v fl we defined colors by wavelength we calculated frequency by using c fl or f l quot10227101 But if the light passes from vacuum to a material fmml vmmml at c what changes The frequency of light does NOT change fmml fvmm The wavelength of light from one material to another does change In general the wavelength in a material is related to the wavelength in vacuum by the below equation 24 A A vacuum quot10227101 mammal This means the wavelength in a material is always smaller than the wavelength in a vacuum the index of refraction is greater or equal to one Demonstrate this with a piece of acrylic Entering Light Initial 9i Normal I I Line Exmng Final Light Normal Line Observations a There are two places where the light passes from one material to another air to acrylic and then acrylic to air b Normal lines are difined to make a 90 angle with the boundary between the two materials The initial normal line at the air to acrylic boundary and the nal normal line at the acrylic to air boundary c Light directed along a normal line does not bend For the above con guration the outgoing beam will always lie upon the final normal line d The light bends toward the initial normal line it will always bend this direction when passing from a region of lower index of refraction to a region of higher index of refraction e The light bends more and more as 9 is increased A second configuration ofthe acrylic piece Initial 3939393939 n Emerng Exiting Normal m nght Light Lme Final Normal 9f Line a Now the light coming in is always along a normal line and there is no bending at the air to acrylic interface b The light now bends away from the normal line it will always bend away from the normal line when passing from a region of higher index of refraction to a region of lower index 25 c The light bends more and more as Si is increased Eventually 9f hits 90 at which point no light can pass through the increased B Results of Snell s law 1 When looking through the water the actual position and apparent position differ 2 When looking in from above not only is the sh at a different location it is also at a different height 3 The eyebrain assumes that light still travels in a straight line 4 This causes the fish to look like its in one location but it actually is in another C Total internal re ection 1 Is it possible for the diffracted angle to be greater than 90 2 The answer is no Thus there is the possibility that the light does not propagate through the material when the angle of incidence exceeds a critical value 3 Study through Snell s Law n a s1n 91 2s1n 92 quot1 b If9290 then a 91 n c sm 9 2 quot1 d This indicates that the only way total internal re ection could occur is if n1 gt n2 Otherwise we would have the sine of an angle greater than one an impossibility e Example A glass block nl60 is immersed in a liquid A ray of light within the glass hits a glassliquid surface at a 650 angle of incidence Some of the light enters the liquid What is the smallest possible refractive index of the liquid f If the ray of light is in fact entering the liquid the limiting case is for 650 to be the critical angle nzxgum n sin 95 sin650 n ZM 160 71 glass quot1 mm 160sin650 145 D Dispersion of Light 1 2 3 The index of refraction changes for different input frequencies This means that different colors of light see different indices of refraction Because of Snell s law different colors are refracted with different angles a White light can be separated into different colors 26 b This is how rainbows are formed 4 Example Horizontal rays of red light 7 660 nm in vacuum and violet light 7 410 nm in vacuum are incident on the intglass prism shown in the drawing top angle is 25 bottom angle 90 See table 262 for any necessary data What is the angle of refraction for each ray as it emerges from the prism n 1662 nV 1698 nm 1000 When light first passes into the glass the angle of incidence is 0 This results in the angle of refraction also equaling 0 When light hits the glass to air interface the angle of incidence is 25 for red 9 arcsin 1 662 1000 sin250 4462 for violet 6V arcsin 1 698 1000 sin250 4586 Note These colors will greatly separate given enough distance Lenses a In general lenses come in two varieties a converging l Shaped with convex sides That is DO b diverging 2 Shaped with concave sides That is All lenses have what is known as foci on either side c A focus singular of foci is a location where parallel rays passing through a lens will meet a For a converging lens the parallel input rays converge at the focus Focus location 27 b For a diverging lens the parallel input rays diverge and never meet If we trace the diverging rays backwards to a location before the lens we nd the focus Focus location A focus is NEVER on a lens and represents a point is space away from the lens ie there is nothing there The distance from the center of the lens to either of the foci is known as the focal length The foci are equally spaced from the center of the lens Focal lengths for converging lenses are positive and focal lengths for diverging lenses are negative because they occur on the other side of the lens In general there are two types of images a lens can produce Ray Diagram for a converging lens a Draw the focus on each side of the lens b Draw the object and the distance from lens to the object to scale c Ray 1 1 Horizontal line from top of object to lens Then other side of lens through far focal point focal point on other side of lens compared to the object 1 This ray travels from the top of the object to the lens center along the line that connects the top of the object with the near focal point focal point on the same side of the lens compared to the object Then draw line from other side of lens moving horizontally e Ray 3 1 Draw line from top of object through the center of the lens continuing across to the other side of lens and beyond f All the rays should pass through a point forming an image this intersection of rays represents only the top of the object the bottom of the object will appear on the principal axis g Image distance from lens can be measured to yield di the image distance the distance from the lens to the image 1 Image size determined by distance to principal axis h Examples 28 This is a real inverted image real because the image is on the opposite side of the lens and inverted because it is upside down Here the solid colored lines represent the actual path of the rays from the top of the object If we extend the back end of the outgoing rays the rays going through the lens they all intersect at a point This point is called a virtual image This type of image can never be seen directly This image is also an upright image since the arrow has the same orientation as the object does 4 Diverging Lens a Draw the focus on each side of the lens b Draw the object to scale proper distance from lens as well as proper size c Ray 1 1 Horizontal line from top of object to lens Then hold a straight edge from near focus to point horizontal line meets the lens Draw a dotted line on the left side and solid line on right side of lens d Ray 2 29 1 Hold a straight edge from the top of the object to the far focus Draw a solid line from the object to the lens On the right side of lens draw a horizontal line extending toward the right from the point the first line hit the lens Extend the horizontal line to the left of the lens but dotted line instead of solid e Ray 3 1 Draw line from top of object through the center of the lens continuing across to the other side of lens and beyond All the rays should pass through a point forming an image Image distance from lens can be measured to yield di Image size determined by distance to principal axis Examples rp qoth Just like the second example for the converging lens we have a upright virtual image To nd the image location we have to extend the back end of the outgoing rays marked by dotted lines The image forms at the gray arrow j One will nd that a single diverging lens can only produce virtual images F General Results for a single lens 1 When using a single diverging lens the image will always be virtual non inverted and shrunken 2 When using a single converging lens and the object is within the focal length the image will always be virtual noninverted and enlarged 3 When using a single converging lens and the object is farther from the lens than the focal length but closer than twice the focal length the image will be real inverted and enlarged 4 When using a single converging lens and the object is farther from the lens than twice the focal length the image will be real inverted and shrunken 30 5 If one uses a single converging as one moves the object farther and farther away from the lens the image will form closer and closer to the focal point on the opposite side It 71 i 7 i R R2 G Lens equations R1 is the radius of curvature of the front surface positive if convex and negative ifconcave as seen by observer on one side ofthe lens R2 is the radius of curvature of the back surface positive if convex and negative ifconcave as seen by observer on same side ofthe lens as before It is the index of refraction ofthe lens f is the focal length ofthe lens 1 Lens Makers equation i 2 Lens Equation 1 1 1 do d f d is negative ifthe image is on the same side ofthe lens as the object d is positive ifthe image is on the opposite side ofthe lens as the object 3 magnification is defined as the ratio of the image height to the object height d m id When In is negative the image is inverted II The eye A Schematic ofthe eye page my 39 mm 39 Cun unclwa 75mm musue Scleva thorium 31 B Parts ofthe eye 1 The back of the eye is called the retina and acts like a screen a The retina contains light sensitive receptors known as rods and cones b Rods are more sensitive to light than cones are distributed everywhere in the retina except a region called the fovea Rods dominate for peripheral vision and night vision c Cones are less sensitive to light and exist for the most part only in the fovea d Cones seem to specialize in wavelengths that they are sensitive to One type is in the range of blue another is in the range of green and a third is in the range of reds 2 Lens a This creates an image on the retina b For direct viewing the image is focused on the fovea 3 Iris controls the light entering the eye much like the aperture in a camera C Parts ofthe eye 1 Cornea 7 this is a clear membrane that protects the lens and iris 2 focusing a a camera focus by adjusting the distance from the lens to the film b the eye adjusts its focus by changing the radius of the lens By changing the radius of the lens one changes the focal length of the eye lens This focus change is controlled by the ciliary muscles which bend the lens 3 blind spot a b information must be carried from the rods and cones to the brain This is done by the optic nerve The optic nerve is located in the back of the eye and it has no light sensitive cells ie no rods nor cones Any image or part of an image that forms on the optic nerve can not be seen The brain typically fills in the blind spot with visual information from around the blind spot Draw a circle about l8inch in diameter Draw a 2inches to the right of the circle the should be approximately 18 of an inch tall and wide Close your left eye and look at the circle with your right eye Move your head toward and away from the paper still looking at the circle At some location typically between 8 and 12 inches from the page the will disappear in your periphery but you can see beyond the sign Repeat with your right eye closed Look at the with your left eye and move your head until the circle disappears D Vision Correction 1 Near point is defined as the closest distance from the eye at which a person can clearly see an object 2 Far point is defined as the farthest distance from the eye at which a person can clearly see an object 32 3 Myopia 7 Near Sightedness a b This occurs for people with an unusually close far point For objects at or beyond the far point the image forms before the retina This occurs when the lens is too curved To correct for this problem a diverging lens is placed in front of the eye or eyes One can tell that a person is near sighted if their eyes through their glasses appear small The smaller their eyes the more near sighted the person 4 Hyperopia 7 Far Sightedness one form of it a This occurs for people with an unusually far near point b This form of far sightedness is NOT a result of aging c For objects at or closer than the near point the image forms behind the retina d To correct for this problem a converging lens is placed in front of the eye or eyes e One can tell a person is far sighted if their eyes through their glasses appear large The larger their eyes the more far sighted the person 5 Presbyopia 7 Far Sightedness another form of it a This form of far sightedness is a result of aging b As one gets older eyelenses get atter and atter 1 This actually helps near sighted people 2 However people who once saw normally can now have trouble with close objects c The attening also causes more of a separation of the eye lens from the humors of the eye d The result is that the eye lens starts focusing behind the retina and the lens is harder for the ciliary muscles to bend the lens 6 Astigmatism a This effect can occur for people with or without myopia or hyperopia b Light passing through the center of the lens from a point object does not form an image at the same point as light passing through a different part of the lens c This effect can lead to blurry vision d The classic effect is street lights at night do not appear as points of light but rather elongated shapes of light e Astigmatism occurs because the lens or cornea is not correctly curved f Astigmatism is corrected by using cylindrical lens Instead of having curves that are spherical like on a ball these lens have curves like those on a pop can E Eye diseases that can cause blindness 33 VII Keratoconus 7 Irregular and cone shaped cornea leads to extreme astigmatism a This eye disease does not completely blind sufferers but it can result in vision that is always blurred b There are some genetic links and may also be caused by excessive eye rubbing c For people with less extreme conditions their vision can be corrected with special contact lenses d More extreme forms can only be corrected with corneal transplants Cataracts 7 A milky white sentiment from the aqueous humor uid between the eye lens and cornea settles on the eye lens If untreated the eye lens will become milky and opaque resulting in a complete loss of vision a This is an age related malady but can also occur as a result of too much exposure to ultra violet light and diabetes b In the 2151 century this is corrected with surgery and an arti cial eye lens Glaucoma 7 a group of eye diseases related to the optic nerve which also cause a side effect of increased pressure in the aqueous humor uid between the eye lens and cornea a Initial vision losses may include blurring and a loss of periphery vision b The main treatment is through the use of drugs and possibly surgery to relieve the pressure in the aqueous humor Macular degeneration 7 eye diseases related to the degeneration of the macula lutea latin for yellow spot and refers to the region around the fovea a Usually caused by fatty tissue building up behind the retina but may also be due to abnormal growth of blood vessels behind the retina 1 Why this happens is not known 2 Increased risk factors include being Caucasian age high blood pressure smoking family history obesity and high cholesterol b No treatment as of yet Electric Charge A B All properties of electricity have their root in the properties of atom s building Conservation of Charge Charge is never created nor destroyed it is simply transferred from one object to another The Atom The atom is composed of electrons neutrons and protons and each has the property of charge q This is a fundamental property like length time temperature etc 34 P39er 9 10 ll Neutrons and Protons are huddled together tightly in the nucleus by the Strong Force Each proton has a charge of 160e 19C Each neutron has a charge of 0C Protons and neutrons have approximately the same mass 167e27kg a To have 1 gram of protons one would need 6e23 protons The number of protons determines the element If one changes the number of protons one changes the element ie like turning lead into gold a This can only be done with a high energy accelerator b We can create microscopic amounts of radioactive gold for millions of dollars The number of protons is equal to the number of electrons in atoms Electrons orbi the nucleus Each electron has 7160e19C and mass of 9 l le3 lkg a It takes roughly 1800 electrons to equal the mass of l proton yet an electron has the exact opposite charge of a proton Ions a When an atom gains or loses electrons the atom is now called an ion b It is a positive ion if it loses electrons c It is a negative ion if it gains electrons What is charged matter a An object has a net electrical charge when some of the atoms that make it up are ionized b Typically very few atoms are ionized The rest are neutral Charge Carriers l 2 3 4 Only electrons are exchanged when objects are charged Protons and neutrons never move when dealing with electricity electrons have a negative charge and are thus NEGATIVE CHARGE CARRIERS What is a positive charge carrier An electron hole The space left by a removed electron is called an electron hole and this is the POSITIVE CHARGE CARRIER Charging objects 1 Rubbing two unlike materials a One material gains electrons and thus a net negative charge b The other object has lost electrons and is positively charged c The total charge on any object is equal to the Number of protons minus the number of electrons 160e19C Charging by Contact a When a charged object is touching an uncharged object electrons or electron holes can go from one material to another Charging by Induction a This is a way of charging an object without any touching b Induce charges by sending same charge through a ground line 35 c when ground line removed the charge is xed 4 Example a A metal sphere has a charge of 80 uC What is the charge after 60el3 electrons have been placed on it 1 80 uC means there are amp Se 13 missing 160e 19C electrons 2 If we add 60el3 electrons then there are l0el3 extra electrons 3 The charge of l0el3 electrons is l0el3 160e 19C 16LC Material differences 1 Some materials acceptlose electrons more easily than others 2 Electrical conductors a The atoms that make up these materials very loosely hold on to their electrons b Furthermore conducting materials will easily accept extra electrons c The result is that electrons can move easily through the material 3 Electrical insulators a The atoms that make up these materials tightly hold onto any electron in its presence very tightly even if there are extra electrons b Excess charge forced on an electrical insulator will ideally not move Coulomb s Law The static cling force 1 F 1 r 47239s0 r 2 IF THE CHARGES HAVE THE SAME SIGN BOTH POSITIVE OR BOTH NEGATIVE THE FORCE IS REPULSIVE 3 IF THE CHARGES HAVE OPPOSITE SIGNS THE FORCE IS ATTRACTIVE 4 Consider two charged objects where one has double the charge of the other object a Which is bigger the force on the object with charge q or the force on the object with charge 2q b RECALL NEWTON S Third Law c Demonstrate that one arrives at the same force for both charges using Coulomb s law 5 Example Consider a 780 uC charge located at x 015m and a 20 uC charge located at x 045 In Find the magnitude and direction of the electrostatic force on the 780 uC particle 36 2 76 76 szqlgzz 899X109Nn 80gtlt10 C2x10 C r 030m F160N Since the force must be attractive opposite signed charges the force is directed in the X direction Find the force on the 20 uC particle F 160N directed in the 7X direction Newton s Third Law Find the magnitude and direction of the force on the 60 uC due to the other charges The magnitude ofthe force between 30uC and 60uC is X100cm 300m2 100m2 J 2 J l62N m 2 E 16 Z 899 gtlt109 NW2 Jig0 X 10160 X10761 The magnitude ofthe force between 60uC and 20uC is F k 899x109 N39mz 6390X1962390 X10sixlloocmjz 120N 2 r2 I C2 l30cm2J m These forces are directed as shown below always directly toward or away from other charge 37 A 2 V 131 Flcosacsinqv 162N 0 m y 154Nc51Ny r 13 sz 120Nj 2 13302 131 117 154N51Nj 120N3 13302 154Nc 69Ny 154N2 69N2 168N law a tan 156vv 24 1 relative to the positive X direction Do demonstrations of balloon sticking to wall or picking up paper with balloon Does the balloon have a charge Does the wall have a charge Does the paper have a charge 38 VIII Electric Field a Polarization of materials That is atoms stretch in the presence of electric elds b The electron cloud around atoms changes its spatial distribution in the presence of strong electric elds We have ve objects A B C D E which may or may not be charged We observe the following I Object A attracts Object B II Object E is attracted to Objects B C and D III Object C repulses Object D IV Protons are attracted to Object A V Protons are repulsed from Object D VI Electrons are repulsed from Object B Determine the sign of the charge zero or for all of the objects EXPLAIN YOUR REASONING B negative repels electrons which are negative D positive repels protons which are positive C positive repels D E neutral attracts positive and negative charges A neutral attracts positive and negative charges A What is an electric field 1 An electric eld describes the force a point charge would have at a location in space a Electric elds are descriptions of a region of space No object or matter needs to be present at that location An Electric eld at a location helps describe forces that would occur on a charged object if it were placed at that location The idea behind an electric eld is similar to a vector map of a river s velocities a The vector map does not directly show the force on an object other factors like size mass shape buoyancy will be needed to calculate the force b However the vector map of a river does indicate how an object will respond if it were placed at a particular location c The same basic principle applies to electric elds B If an electric eld just describes the Coulomb force why bother dealing with a eld as opposed to the forces 1 2 3 4 Typically a charged object has a huge number of charge carriers electrons or electron holes In is not so easy calculating all the forces from each charge carrier Electric elds describe the net group effect Electric elds describe how an object would be affected if it were placed in a certain location What do electric elds look like 39 l For a pomt charge E F 2252 qtzst a E the electric eld at the location of a small test charge b qt charge of a point charge c Elm force on the point charge d For a single point charge the magnitude of the electric eld is E kn r 2 1 k 899x109 N m 2 q charge ofthe particle 3 r distance from the particle to the location of interest 4 The electric eld has units of NC e Electric elds point from regions closer to positive charges to regions closer to negative charges 1 Arrows point away from a positive charge 2 Arrows point toward a negative charge f Electric eld lines coming together indicate an increasing magnitude of an electric field Electric eld lines spreading out indicate a decreasing magnitude of an electric eld Graphically sketching an electric eld 1 3 4 The length of the arrow shows the magnitude of the electric eld The direction of the arrow shows direction of the electric eld Locations where the electric eld lines come closer together indicate stronger electric eld strength Locations where the electric eld lines more apart indicate weaker electric eld strength Examples of electric elds of point charges a 0 For each of the above sketches one must remember that there are an in nite number of electric eld lines extending radially from the charge RECALL THAT ELECTRIC FIELDS BY THEMSELVES DO NOT REVEAL HOW ANY PARTICLE MOVES THEY JUST DESCRIBE THE FORCES AND THUS THE MOTION OF AN ADDITIONAL PARTICLE PLACED IN THE FIELD a If we place an additional positive charge in a region of space with a nonzero electric eld the positive charge will feel a force in the 40 direction of the electric eld line The magnitude of the force is equal to the charge of the additional particle times the magnitude ofthe electric eld b If we place an additional negative charge in a region of space with a nonzero electric eld the negative charge will feel a force in the opposite direction of the electric eld line The magnitude of the force is equal to the charge of the additional particle times the magnitude of the electric eld Electric Fields from multiple point charges 1 To calculate an electric eld from multiple point objects one must nd the electric eld at the location of interest magnitude and direction for each point charge and then vectorally add them together 2 Example two negative point charges 2 3 Example a Find the electric eld at the origin due to the 50uC charge b Find the electric eld at the origin due to both of the charges c If we placed a 10uC charge at the origin what is the force on this charge 30uc 30cm 50uc 50cm 2 6 a Ekiz 899x109N S39OXIO 18gtlt107 r C 0050m C 41 Electric elds point away from positive charges Therefore E 18 X 107 directed in the negative Xdir b First nd the electric eld at the origin due to the other charge 2 6 Ekiz 899x109N 3390X10 2C30x107 r C 0030m C The electric eld from the 30uC is E 30 gtlt107 directed in the negative ydir If we add the two electric elds we nd E 202 N l8gtlt107 and E 30gtlt107 C C lazy The magnitude of the total electric eld is E20 2 Eton 2 E202 35 X107 202 E directed at 9 tan391 59 below neg XaXis c Using EqEF 10gtlt10396C 35x107 35N C directed at 59 below neg XaXis Example Multiple Choice At which point or points is the electric eld zero for the two point charges shown on the XaXis Only consider points on the XaXis 4q 2q O O gt x A The electric eld is zero somewhere on the XaXis to the left of the 4q charge B The electric eld is zero somewhere on the XaXis to the right of the 2q charge There is no other location C The electric eld is zero at two points along the XaXis one such point is between the two charges but closer to the 2q charge and the other is to the right of the 2q charge D The electric eld is zero somewhere one the XaXis between the two charges but this point is nearer to the 2q charge There is no other location E The electric eld is never zero in the vicinity of these charges The electric eld could not be zero to the left of both charges except at an in nite distance away because each individual eld would point left and thus they cannot cancel each other out 42 The electric eld could not be zero to the right ofboth charges except at an in nite distance away because each individual eld would point to the right and thus they cannot cancel each other out The answer to the multiple choice question is D Can we locate where the electric eld is zero If we de ne L to be the distance between the two charges and x as the distance from the 4q point charge to the zero eld location we can In between the two charges on the XaXis we nd Ex 0k4ki x L x2 2kg 2 2 0 2L xLxlt x x 2kg Om2LZ 4xLx2 x 4Li39xll6L2 8L2 2 x2iEL As x lt L otherwise we are not between both charges and we cannot trust we have the correct si s for each individual electric eld the only location is x 2 J5 Point Charges in electric elds 1 A positive charge ALWAYS feels a force equal in magnitude to its charge times the magnitude of the electric eld This force is ALWAYS directed in the same direction as the electric eld A negative charge ALWAYS feels a force equal in magnitude to the absolute value of its charge times the magnitude of the electric eld This force is ALWAYS directed in the opposite direction of the electric eld Example how would a stationary electron react to the below electric eld 43 Based on the previous discussion of electric elds the electric eld is constant and directed upwards This means the electron will a constant force directed downwards in the plane of the paper The electron will move with constant acceleration downwards G Conductors in electric elds 1 If we place a conductor in an electric eld the electrons move as a result of the electric eld 2 Since there are electrons on one side of the conductor and electron holes on the other side an extra electric eld develops inside the conductor This leads to the following a IF WE HAVE STATIONARY CHARGES THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO b IF WE HAVE STATIONARY CHARGES AND A HOLLOW CONDUCTOR THE ELECTRIC FIELD INSIDE THE HOLLOW IS ALSO ZERO 1 This is called the shielding effect of conductors H Insulators in electric elds 1 Due to the stretching of atomsmolecules due to the electric eld we nd that the electric eld inside an insulator is reduced when compared to outside the insulator IX Gauss Law A Gauss Law is a description of eld lines and as applied to Physics is derived from the Coulomb electrostatic force B Guass Law introduces a new term called electric ux 1 Electric ux can be considered the net electric eld that passes perpendicularly through a surface Nonzero electric ux Zero electric ux 1 This surface ie area through which the electric elds pass through is user de ned 2 The surface does not have to be in on or even related to any physical object 44 Gauss Law the net ux passing out of a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space a b P Q closed surface any surface that completely encloses a volume 2 permittivity of free space so 885 gtlt103912 C m N 2 recall that k 899 gtlt109 N n 47239s0 Q20 Gauss Law as an equation QD closed surface so 2 m d3 Flux has units of closed mfm QM total enclosed charge If a field line is exiting a closed surface it will contribute to a positive ux If a field line is entering a closed surface it will contribute to a negative ux One can consider the net ux for a closed surface as a count of the number of field lines exiting positive minus the number of field lines entering the surface General steps to calculate the electric ux 1 2 3 Determine the electric field s magnitude for every location on the surface Find the component of the electric field that passes perpendicularly through the surface Integrate over the closed surface using only the component of the electric field that is perpendicular Example ux calculation 1 Consider a point charge at the center of an imaginary not physically there sphere By direct calculation find the ux through the sphere E kl2 and the field is directed radially outward from the origin r The field is always perpendicular to the surface q E A7 The electric field is constant sphere s have a constant radius and passes perpendicularly through the sphere d EA sphm AW Surface area of a sphere 45 q E47rrz 4nr2 2 r 47qu 47r 1 q 47239s0 1 8 0 Example a hollow conducting sphere with outer radius R0 and inner radius R and a net positive charge q Find the electric as a function of r from the center of the hollow sphere rltRi Enclosed net charge Q 0 d3 E 0 E0 RiltrltR0 E0 inside a conductor Q0 This means that there is no net charge on the inner surface R0ltr The electric eld is always normal to the surface this is now exactly the same a charged solid conducting sphere Thus I 0 The electric field is also symmetric about the sphere thuquE ZEcosgpAl EZAA EA The total enclosed charge Qq A 411r2 surface area of a sphere k EA i or E q th1s 1s the same as a pomt charge 8 47230quot2 r 0 Example a hollow conducting sphere with outer radius R0 and inner radius R and a net positive charge q Unlike the preVious problem there is now a 7q point charge at the center of the sphere Find the electric as a function of r from the center of the hollow sphere rltRi Enclosed net charge Q q q l rinsed surface so We would expect spherical symmetry for the field E kl2 directed toward the center of the point charge r RiltrltR0 E0 inside a conductor 510524 0 g surface so 46 Q qqw 0 surface qmnzr q mfm R0ltr qmnzr qoum q surface mfm qoum 0 surface 5105 0 net enclosed charge is zero mfm Any electric eld should be directed radially from the center E0 X Electric Potential Energy Electric Potential and Voltage Recall that an electric eld describes the force on a small charged object providing that we know the charge of this object Can we do something similar B with Energy 1 The answer is yes and this new concept is the electric potential V 2 Electric potential Energy is the potential energy that eXists between two charged particles a EPE k qlqz r EPE potential energy between to charges ql and qz Z k 1 899gtlt109 NZ 47239s0 C ql and qz are the magnitude of the charges r is the distance between the charges b Note that the distance is not squared and we have now dropped the absolute value sign around the charges 1 potential energies can be both positive and negative 2 objects with no initial kinetic energy move from higher potential energy to lower potential energy c Like all energies electric potential energy has units of Z Joules JNm kg 2m 3 3 Example suppose a stationary electron is 100cm from an unmovable 60nC point charge What is the speed of the electron when it is 500cm from the point charge Conservation of Energy 47 5 V 21661qu 0K f mnrf Nm2 2 2899 gtlt109 V f 911x10 3lng100m050m 60 gtlt10 9 C 16 X10 19CX050m 10m vf44gtlt106 S Electric potential is a description of the potential energy one would have if a small charged object were placed at a location a V E k1 q r V electric potential of a location due to the vicinity of nearby charge q N m2 2 k 899 gtlt109 47239s0 q is the magnitude of the charge r is the distance from the charge J kg m2 E C s2 The electric potential is measured in volts V C How is the electric potential related to the electric eld 1 AV As E avz where AS is a small step in length in the direction of the electric eld The Electric eld points from high electric potential to low electric potential If one knows the electric potential at all locations one can nd the electric eld at all locations a It is this feature that makes the electric potential so useful 1 To exactly de ne a vector at a particular location one requires three scalar values a scalar value for each dimension of space X y 2 48 2 By using electric potential we can replace the three needed values for electric eld by a single value for the electric potential 3 Knowing the electric potential at all locations allows us to calculate the electric eld at all locations Equipotential surfaces 1 An equipotential surface is a surface where the electric potential does not change 2 The electric field is ALWAYS perpendicular to equipotential surfaces 3 If we have two equipotential surfaces the average E field magnitude between them is EaVg As In the macroscopic world what does a difference in electric potential mean 1 The most common item which ideally has a constant potential difference between to locations is a battery 2 Differences in electric potential occur because of charge separation b positive charges and negative charges are separated from each other c Electric potential does not only address the two locations with excess charge Electric potential describes the potential energies of additional charges placed in the vicinity of the separated charges How do point charges move in a region where the electric potential is known 1 Positive point charges move from high electric potential to lower electric potential 2 Negative point charges move from low electric potential to higher electric potential The electric potential and conductors 1 Recall that Em K As 2 Since the electric eld is zero inside a conductor there can be no change in the electric potential on an ideal conductor Example problems 1 Two charges are fixed in place with a separation d One charge is positive and has twice the magnitude of the other charge which is negative The positive charge lies to the left of the negative charge Relative to the negative charge locate the two spots on the line through the charge where the electric potential is zero First we will see if there is a solution between the two charges V k k o 20201 Ci V r 49 XI kq rdr3r d0 r 3 Now we will see if there is a solution to the right of both charges 2 meI k k 0 k q d2r dr 0 Similarly rr r d Before leaving this problem there are other locations where V0 These points form a sphere centered at d3 to the right of the negative charge on the line between the two charges Capacitance A Imagine two parallel conducting square plates with opposite charges separated by distance d 4 d gt39 Since electric potential is caused by charge separation there must be an electric potential difference from one plate to another RECALL THAT THE ELECTRIC POTENTIAL IS THE SAME EVERYWHERE ON AN IDEAL CONDUCTOR Because the electric field is zero inside a conductor the electric potential must be the same inside a conductor We would expect that the electric potential difference between the two plates would be proportional to charge on the positive plate Imagine a small cylindrical Gaussian surface with one side embedded in the middle of the positively charged plate 4 d gt39 50 We would expect the electric eld to go from one plate directly into the other plate 1 Recall that the electric potential is the same everywhere on the plates an Equipotential surface 2 The electric eld must pass perpendicularly through the surface 3 Thus the electric eld looks like i The electric ux could only result from the circular area of the cylinder in the middle of the two plates 1 The electric eld is always parallel to the sides of the cylinder Thus no ux through this area 2 There is no electric eld inside a conductor so the circular area inside the conductor has no net ux contribution By Guass Law 80 chxrclz ant A plum where q the net charge on the entire plate Apia the area of the entire plate Amlz the area of the circular part of the cylinder The length of this cylinder is irrelevant in terms of electric ux 1 This indicates that the electric eld is constant everywhere between the two plates A q ZEAEWZ q we 80Apla22 E 1 80Aplatz Now use EaVg As 51 XII Electric Circuits A B 814 d d 0 plate AVV V 1 0 plum q V CAV 814 C 0 plaza d j Big capacitance means more charge is stored for an equivalent electric potential difference Dielectric Material 1 Usually the gap between the conducting plate is lled with an insulating material a dielectric a Otherwise the two plates will naturally attract each other and discharge themselves 2 The d1electr1c constant 1s de ned as K E a E0 is the electric eld magnitude just outside the dielectric material b E is the electric eld magnitude inside the dielectric material 4 Since we know that the electric eld inside a capacitor is constant L E K 80A d K K q K8014 V d Why use capacitors l Capacitors are electrical energy storage devices 2 Many devices charge a capacitor stores energy to be released at a later time a for example Medical de brillator 3 Capacitors are also used to filter a nonconstant electric potential Voltage to make it closer to a constant value How much energy is stored in a capacitor 2 EPE q lCV2 2C 2 Electromotive Force 1 E the electromotive force is the maximum potential difference or voltage a usually the E in a circuit is the battery or power supply 2 Note that the electromotive force is NOT a force Current 52 Power 59 1 current is the amount of charges that pass a surface per unit time Aq At Despite it actually being caused by the motion of negative charge carriers ie electrons conventional current goes from positive to negative or high electric potential to low electric potential Law The potential difference across a material is equal to the resistance times the current VIR In today s world Ohm s Law would not be considered a law That is many materials do NOT obey it In particular these materials are doped semiconductors which are used in diodes transistors integrated circuits etc Resistance is the difficulty experienced by electrons travel from high to low potentials That is the harder it is for the electrons to travel through the material the higher the resistance a Resistance depends on the size and shape of the material L b R p2 where p is the res1st1v1ty c Resistivity does NOT change with the size or shape 1 Resistivity changes from one material to another 2 Resistivity also can change with temperature 3 p p0 1 aT To where at is the temperature coefficient of resistivity P EVIPower At Typically the power is the power lost to the system through heat Since vIR PV2R and PIZR Example Compare the power lost in a do diameter wire with 10 amps of current with a 2d0 diameter wire of the same length drawing the same current L L a R P A P d 2 1 2 4L L b R p R p 1 doz 2 doz 4 L I L c PlIle 0 PZIZR2 p2 do do d The power lost in the smaller diameter wire is 4 times greater than the power lost in the larger diameter wire What happens to a wire with too much power lost in it 53 a The wire s temperature increases b If the wire s temperature becomes hot enough the nonconducting insulation may melt or burn off Resistors in circuits 1 2 For there to be any current in a particular element that element must be in a complete loop Series Wiring a The resistors are connected head to tail with nothing else between them b Resistors in series act like a single larger resistor Equivalent resistor l RZSWWM R1 R2 R3 c Resistors in series have the same current 1 If one resistor burns out all the resistors will have no power 2 Example blinking Christmas lights Parallel Wiring a The resistors are connected head to head and tail to tail b Resistors in parallel act like a single smaller resistor Equivalent resistor 1 RLR equivalent 1 2 3 parallzl c Resistors in parallel have the same voltage 1 If one resistor burns out it will not effect the others 2 Example At home watching TV reading and listening to the stereo a If one is turned off the others are not effected Mixed Wiring a often the circuit elements are neither parallel nor series This is called mixed wiring 309 409 b Find the equivalent resistance 909 2009 g 809 609 va VWV 909 2009 E 809 1009 54 Kirchoff s Rules 1 The sum of the magnitudes of current into any point is equal to the sum of the magnitudes of current out of the point a The current into a point current out of a point In any closed circuit loop the sum of the voltage drops equals the sum of the voltage rises a Example nd the current in all sections of this circuit 55 1112 13 160V 12800 2I34OOQ 600V 11200g2 13 4009 600V 11200g21112400g2 600V 116OOQ12400 2 160V 128OOQ11 12 4009 160V 12120 211400g2 11 40021 30012 600V 400A 30012 600 2 12 4009 600V 400A 30012 600 2 12 4009 180V 12140g2 12 2A 12914 7 11 40021 30012 A m 0141 3 1 1 1 12 Amr42A Now suppose we want to find the magnitude of the voltage between the following points of the circuit 600V 11 2009 1 What is the magnitude of the electric potential difference between points A and B 56 2 3 4 Example Answer Zero For the electric potential to be nonzero there must be a voltage drop or rise between the points The simplest path from A to B has no voltage rise or drop What is the magnitude of the electric potential difference between points B and C Answer 571V The simplest path from B to C contains the 4009 resistor Thus there is a voltage drop equal to 134009 gAlt400Qm 571V What is the magnitude of the electric potential difference between points C and D Answer Zero There is no current going from C to D because this is not part ofa closed circuit Ifthere is no current there can be no voltage drop across the resistor What is the magnitude of the electric potential difference between points A and D Answer 571V The simplest path from A to D contains the 4009 resistor and 1009 resistor Thus there is a voltage drop equal to 13400g2 0100 2 gAlt400Qm 571V 2009 8009 WW 4 1609 9009 1809 A The current in the 8009 resistor is 0500 A Find the current in the 200 9 resistor The voltage across the 8009 is also across the 1609 This voltage is 80090500A400V Therefore the current in the 1609 is 0250A The current in the 2009 resistor is the sum of the currents in the 8009 and 1609 Therefore 10750A B Find the current in the 9009 resistor 57 The voltage across the 8009 and 2009 resistor is equal to the voltage across the 9009 resistor This voltage is 190V Thus the current is 21 lA XIII Magnetic Forces and Fields Where does this eld come from A l A magnetic eld comes about due to the movement of charged particles That is charged particles with a nonzero velocity a ONE CAN ONLY HAVE A MAGNETIC FIELD IF THERE ARE MOVING CHARGES 2 The smallest magnet comes about due to electrons circling a nucleus 3 Bar magnets come about when the atom s elds align with each others elds producing a macroscopic effect 4 Bar magnets were originally made by heating iron until it is red hot then hitting lightly it with a hammer The idea is that the heat loosens the atoms and allows them to align their elds with each other and the Earth s eld The magnetic eld B l Historically it was noticed that if one reduces the friction on a magnet one end will point toward the north 2 Opposite poles of a magnet attract Like poles of a magnet repel 3 DO NOT ASSUME MAGNETIC POLES ARE POSITIVELY OR NEGATIVELY CHARGED 4 The part of the compass magnet that points North is called the north pole Initially it was called the north pointing pole a This means that a compass always points toward the south pole of a magnet 5 The magnetic eld arrows start at the North pole and curves toward the South pole 6 DRAW FIELDS in class of a bar magnet and a C magnet 7 UNLIKE ELECTRIC FIELDS MAGNETIC FIELDS ALWAYS FORM COMPLETE LOOPS 8 Observation If one considers the magnetic eld lines continuing in the material MAGNETIC FIELD LINES ALWAYS FORM COMPLETE LOOPS Forces from magnetic eld 1 2 3 As with the electric eld opposites attract and the same repels North poles are attracted to south poles and viceversa A magnetic eld will not apply a force to a neutral point charge a It should be noted that the reason a magnet attracts iron and other metal is that magnetic eld aligns the minimagnets elds inside the iron Fqusin0 is the force on charged particles due to the magnetic eld B a Charge is q N s C m b B is the magnetic eld has units of Tesla T 58 j The Mass Spectrometer l A strong magnet may have a eld of 3X10394T velocity is V sine is the angle between the velocity and magnetic eld The direction is determined by the Right Hand Rule number 1 for positively charged particles When dealing with magnetic fields velocities and forces you have to be using 3 spatial directions Point your ngers on your right hand in the direction of the magnetic eld with your thumb pointing in the direction of the velocity Your palm will now face the direction of the force If the particle is negatively charged your knuckles will point in the direction of the magnetic force A force that is perpendicular to the velocity means that the object travels in a circle for a constant magnetic eld this occurs for 9 is 90 mv2 mv qVB or r qB Magnetic elds applied to point charges produce a force that is perpendicular to the velocity This leads to circular trajectories quotquotquot quot a 39 1 F M N 1 This is how Chemists and Physicists determine the mass of atoms in an unknown substance 2 First the unknown material is put in a gaseous state 3 The atoms are then ionized by stripping off a single electron 4 The ions are now accelerated through an electric eld a The speed of the atom can be determined this way b At rst all the energy is potential energy Energy EPE qV c At the end ofthe energy is all kinetic Energy mv2 d v 4 m 5 Now the ions no longer see an electric eld but see a perpendicular magnetic field B 6 By varying the magnetic eld we can control the mass of the ions that get into our detector 59 E b rquz m 2V The force on a current ca1rying wire Aq 1 F qus1nt9EvAtBs1nt9 2 FILBsin9 a b c d I is the current L is the length of wire in the eld B is the magnetic eld Sine is the sine of the angle between the wire and the magnetic eld Right hand rule 1 still in effect 1 Fingers point in the direction of the current thumb points in the direction of the magnetic eld palm will now point in direction of the force 3 This force is due to the magnetic eld produced by the current moving charges aligning with the external magnetic eld B Torque on a charge carrying loop of wire 1 Consider a square loop each side has length L in a magnetic eld B A A I M V 2 Find force on each segment a upper segment 1 F ILB sin90 ILB directed out of page b right segment 1 F ILBsin0 0 c lower segment 1 F ILB sin90 ILB directed into page d left segment 1 F ILBsin0 0 3 From the above forces and directions there is a net torque on the loop of wire 4 What happens when the current rotates so that we have the below picture B O O O O O O B A O 0 IO 0 O O 0 yo 0 0 Ac 0 O O 9 O O O O O O O O O 60 Original View Top View 5 Find the forces on each segment of wire in above case a Top Segment from Top View picture 1 F ILB sin90 ILB directed up in top View picture b Right Segment from Top View picture 1 F ILB sin90 ILB directed toward right in top View c Bottom Segment from Top View picture 1 F ILB sin90 ILB directed downward in top View d Left Segment from Top View picture 1 F ILB sin90 ILB directed toward left in top View 6 NO NET TORQUE AT THIS ANGLE 7 r NAB sin go a A is area of the wire loop b I is the angle between the loops normal perpendicular to area defined by the loop c N is the number ofwire loops 8 DC Electric Motor a I can not draw this very well so see figure on page 817 of your text b There is a break in the current at 0 c Angular momentum of the rotating motor carries it past the zero torque point G The magnetic field produced by current in wires 1 The magnetic field will circle the current in the wire 2 The direction of the magnetic field is determine by right hand rule 2 a Point thumb of right hand in the direction of the current b The magnetic field circles in the same direction as your fingers will curl to make a fist 3 Magnetic field formulas a Long straight wire b BH if 27ml 1 Distance from wire to measurement point is d 2 see sketch on page 796 of your text c Example two long straight current carrying wires are parallel and near to each other with the current heading in opposite directions Will the wires attract or repel each other A 61 The magnetic eld from the top wire is heading into the paper Using the right hand rule the bottom wire will feel a force downward in the plane of the paper Hence the bottom wire is repelled from the top wire b Circular Loop of wire 1 The magnetic eld magnitude at the center of a loop of wire is B i 2 2 See sketch on page 797 of your text c Solenoid technically in nitely long 1 This is the winding design one would get wrapping a nail with wire 2 B Mm a Field is calculated at the center of the solenoid b n is the number of windings per unit length 4 Why do permanent magnets stick to other magnets a First think about current carrying loops inside the permanent magnets 1 Each current loop may represent something as small as electrons orbiting a nucleus l 39 l h I II M l l n I I39 f 39I l 39 39 39 N a a s I LL 39 I I I I I I f I I I i l E E n39 T 39I 39 i i i X I i i I I i V i I I I I I I I z 6 I I 39 39 39 I39 I 39 f 39 39 an s 3939 KM Itquot 39x M I 39 b Now consider the force of the magnetic eld of one permanent magnet on a current loop in the other 62 5 l The magnetic eld from the right permanent magnetic on the rightmost current loop in the left permanent magnet will generally move toward the left a the part of the magnetic eld moving to the right tries to expand the current loop this is not responsible for the attraction b The other components of the magnetic eld are responsible for the attraction Why do magnets stick to ferromagnetic material such as iron a Atoms magnetic moments align themselves with the magnetic eld b This is very similar to the current loops used in magnetic motors c The atoms align themselves until the net torque is zero d This effectively causes the opposite pole of a magnet to form 1 A ferromagnetic material left in a magnetic eld long enough will eventually become permanently magnetized XIV Electromagnetic Induction A A bar moving perpendicularly with the magnetic eld P39er As the bar moves across the magnetic eld the electrons in the bar will experience a force downward Electrons naturally repel each other thus they will spread out away from the end of the bar A ow of electrons WILL lead to a current opposite direction Current through a resistor should lead to a voltage drop VIR The resistor has power dropped across it exactly as if it was powered by a battery or power supply Thus emf 5 is induced in the bar BUT if there is current running through the bar then there must be a force on the bar a F ILB sin 9 ILB b By RHR 1 the force is exactly opposite the velocity s direction c TO KEEP IT MOVING AT A CONSTANT VELOCITY AN EXTERNAL FORCE MUST BE APPLIED TO THE BAR d Consider the work done 1 Wforcedistance 2 WILBvAt e The work done must be related to a change in energy 1 WAEPE 63 2 ILBvAt AqV M 3 LBvAt AqV At 4 LBV V 5 Thus a voltage or emf is induced in the bar 8 5LBV 9 NOTE Mechanical energy is being converted into electrical energy B Magnetic Flux and Faraday s Law 1 d3 BA cos go E Magnetic Flux a B magnitude of the magnetic field b A area of the circuit loop a complete circuit must form a closed path which must run along the perimeter of an area 0 0 angle between the magnetic eld and a line that passes perpendicularly though the area of the loop Magnetic Flux is in units of Webers Wb Tmz e Magnetic Flux is the number of eld lines passing through a surface 2 Faraday s Law AV N a AV Induced emf b N number of windings number of loops c Ad change in magnetic ux d At change in time 3 One must have a changing magnetic ux to induce a current This means to induce a current at least one of the following must happen a The strength of the magnetic field passing through the loop must change b The area of the loop must change c The angle between the magnetic field and the loop must change 4 In the previous example a AV B B LVN LBV At b Why the negative sign c Negative indicates that the voltage produced causes a magnetic force that opposes the motion d Lenz s Law The induced emf resulting from a changing magnetic ux has a polarity that leads to an induced current whose direction is such that the induced magnetic field opposes the original ux change 5 Now consider a falling conducting ring through a magnetic field a When either the ring is entering or leaving the magnetic field region there is an induced emf Otherwise there is none b This is because there is no change in the magnetic field or area anywhere else in the motion C The electric generator 64 D 1 In general it is inconvenient to move anything in a straight line for a long distance through a magnetic eld 2 Much easier to spin the coil in the eld a AVN NBAM At At b AV 2 NBA cosat cosat c AV 2 NBA cosatcosaAt sinatsinaAt cosat At d For the greatest accuracy we will assume a small At e Then cosaAt m l and sinaAt m wAt f AV 2 NBA cosat wAtsinat cosat At g AV NBA mAts1nat h AV NBAmsinlta2tgt i What is n 1 Since sine is in radians when oat211 we have gone one complete revolution 27239 2 Period T a 3 F f 1 a re uenc q y T 27239 j b c d To make our standard AC voltage 03 21160 Hz and NBAoa 120 J5 How do we actually generate electricity a Spinning coils in a magnetic eld Hydroelectric power uses water to turn the coil Windmills also turn the coil Coal Oil and Nuclear power heat steam to turn the coil Example A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 40 ms perpendicular to a 05 T magnetic eld The resistance of the rod and tracks is negligible The rod maintains electrical contact with the tracks at all times and has a length of 13 m and makes an angle of 300 with respect to the horizontal A 075 Q resistor is attached between the tops of the tracks a What is the mass ofthe rod 1 V LBV13mX050T40 26V s 2 VIR or11 347A R 0759 3 21Fx FE mg sint9 ILB mg sin 9 0 no acceleration 65 4 m ILB 347A13m050T gSing 98sin30 b What is the change in gravitational potential energy over 20 seconds AGPE mgAh mg vsin30 At AGPE 046kg9840 jsin30 20s180 S S 046kg c What is the change in electrical potential energy over 20 seconds AEPE VIAI 26V347A20s180 the gravitational potential energy is converted to electrical energy E Mutual Inductance 39 quot 3amp0 1 Because are using AC and have a changing current we would expect to have a changing magnetic eld in the vicinity of the 2quotd coil Thus we would expect that an induced voltage in coil 2 2 We would expect that N2q32 M11 3 AVZ N2 32 M F Self Inductance n n n n Ml 1 There also is an induced voltage produced in a coil that fights the current that produced it 66 2 Nd L then 3 AV N E L At At 4 The constant L is called Inductance and this device is called an inductor or an RF choke 5 The inductor like a capacitor can also store energy 1 6 Energy 3L2 G Transformers 1 One side has an AC power source primary side the other side has a voltmeter secondary side 2 AVS NS AV N E At F F At 3 The above is due to the same magnetic eld and same area in iron core 4 AV V N AVp Vp N p 5 Thus by having different windings on primary and secondary coils we can change the AC voltage Hence the name transformer 6 If we consider an ideal transformer the power in the primary coil would equal the power in the secondary coil a Pprxmary Psecmidary b 1pr 15V 1p VS NS LWM XV Electromagnetic Waves A What is it 1 A transverse oscillation verses a longitudinal oscillation for sound waves a use a spring to demonstrate the difference between longitudinal and transverse oscillations Electric and Magnetic fields are both transverse to the direction in which it travels Draw a picture showing this property Many Examples a radio waves radar waves microwaves light all varieties Xrays gamma rays and others B How do we make them 67 1 These are created by accelerating a charged particle 2 This can be most easily done by humans by using AC currents C Detection 1 Not all the examples listed above are visible to the naked eye 2 Electric Field pickup through the use of antennas a Electric eld pickup Antenna m 5 39 b Tuning accomplished via the LC combination and the use of resonance c Electric eld ideally parallel to the antenna 3 Magnetic eld pickup through the use a ring antenna a Magnetic eld pickup Antenna We 6 sz 1m b Tuning accomplished via LC combination and resonance c To induce an emf it is ideal that normal of the coil is parallel with the magnetic eld D Properties of all EM waves 1 James Clerk Maxwell was the rst to realize that he could fully describe all electromagnetic behavior with four mathematical equations a Maxwell s Equations 1 Gauss Law 68 2 Ampere s Law 3 No magnetic monopoles 4 Faraday s Law b He found that he could create waves out of electric and magnetic elds which always had the same velocity 1 V SOIuO Experimental Measurements of c veri ed Maxwell s prediction Consider how long it takes the sun s light to get from there to here an average distance of 149X1011m 11 a dvtct or t i 497s 8 minutes and 16 seconds 0 300x108ms Also consider what the unit Light year means c This velocity is c 300X108ms in a vacuum a A light year is the distance light travels in a year b d vt ct 300x108 j yr 36525 24E 60 60i s yr day hr mm d 947x1015m which is equivalent to running around the Earth s equator roughly 237 million times Because light takes some time to move from one point to another the light from distant objects in the universe has traveled for millions of years to reach us 69
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