Electrostatics Problems: Electric Fields, Torque, and Dipoles, Exams of Physics

Download Electrostatics Problems: Electric Fields, Torque, and Dipoles and more Exams Physics in PDF only on Docsity! 1 Name: ________SOLUTIONS_____________________ Student ID: ___________________________ Section #: _________ Physics 208 Exam 1 Feb. 20, 2008 Print your name and section clearly above. If you do not know your section number, write your TA’s name. Your final answer must be placed in the box provided. You must show all your work to receive full credit. If you only provide your final answer (in the box), and do not show your work, you will not receive very many points. Problems will be graded on reasoning and intermediate steps as well as on the final answer. Be sure to include units, and also the direction of vectors. You are allowed one 8½ x 11” sheet of notes and no other references. The exam lasts exactly 90 minutes. Problem 1: _______ / 20 Speed of light in vacuum: ! c = 3"10 8 m /s Problem 2: _______ / 20 Permittivity of free space ! " o = 8.85 #10 $12 C 2 /N %m 2 Problem 3: _______ / 20 Coulomb constant ! k =1/4"# o = 9 $10 9 N %m 2 /C 2 Problem 4: _______ / 20 Magnitude of electron charge ! e =1.6 "10 #19 C Problem 5: _______ / 20 Problem 6: _______ / 20 TOTAL: _______ / 120 2 1) [20 points, 4 points each]. Circle the correct answer, AND write some sentences or show a calculation explaining how you obtained the answer. i) You throw a rock into the water. Water waves spread out from the rock at a speed of 0.5 m/s. You measure the distance between the crests to be 20 cm. As the waves pass the bobber on your fishing line, at what frequency does the bobber move up and down? a) 10 Hz b) 2.5 Hz c) 0.5 Hz d) 0.1 Hz e) 0.025 Hz ii) A glass (index of refraction = 1.6) lens has a focal length in air (index of refraction=1.0) of 10 cm. It’s focal length in water (index of refraction = 1.33) is a) Longer b) Shorter c) The same d) infinite (does not focus) iii) Below is a half-ring of uniformly-distributed positive charge in the x-y plane. In what direction is the electric field at the dot at (0,0,0) (dot at the center of the ring)? a) A, along +z-axis b) B, in x-z plane, angled up c) C, along +x-axis d) D, in x-z plane, angled down e) none of the above Explanation/work: From ! "f = v , ! f = v /" = 0.5m /s( ) /0.2m = 2.5Hz Explanation/work: The amount by which light rays are bent at the lens can be determined by Snell’s law. The rays are bent more when the difference in index of refraction is greater When the lens is put in water, the index difference is less, and so the rays are bent less. So the focal length is longer in water than in air. x y z A B C D Explanation/work: Take a pair of differential charges that are equal angles away from the –x axis. The electric field from each point directly away from the charge. So the y- components of the electric fields from each of these differential charges cancel. The x- components add. The net result from all the differential charges is a field in the x- direction. 5 3) [20 points total] A person is trying to spear a fish through a small hole in the ice. a) [10 points] She can see the fish by sighting along her spear held at an angle of 40˚ with respect to the vertical. At what angle with respect to the vertical should she throw her spear through the center of the hole in order to hit the fish? b) [10 points] Suppose the fish is 5 m beneath the ice, directly below the hole. The fish starts swimming parallel to the surface, keeping a constant depth of 5 m. Using principles of refraction, determine how far the fish must swim before the spear fisher can no longer see the fish at any angle through the hole. nwater=1.33 40˚ nair=1.0 The light ray that enters her eye was bent from its propagation direction in the water. So she should aim her spear so that it will go along the direction of the light ray in the water. This is determined from Snell’s law. ! n air sin" air = n water sin" water ! 1.0( )sin40˚= 1.33( )sin" #" = 28.9˚ In order for the spear fisher to see the fish, the light from the fish must refract through the hole at the water-air interface. This will happen as long as the angle of incidence is not so large that the angle of refraction (remember, measured from the vertical) would be 90˚. At this and greater angles there will be total internal reflection, meaning that the light from the fish does not escape through the hole. This critical angle for total internal reflection is when the refracted ray would be 90˚ from the vertical. ! n water sin" crit = n air sin90˚#" crit = 48.75˚ . So ! ! tan" crit = x /5m# x = 5m( ) tan48.75 = 5.7m θ = = Value Units 28.9 deg Distance = Value Units 5.7 m nwater=1.33 nair=1.0 48.75˚ 5m x ! tan 48.75˚( ) = x 5m " x = 5.7m 28.9 6 4) [20 points total] You use a magnifying lens to enlarge the letters on this page. The focal length of the magnifying lens is 5 cm, and you hold it 4 cm from the page. You position your eye lens 2 cm behind the magnifying lens a) [5 points] Does the magnifying lens alone produce a real or virtual image? How much bigger is the image than the actual letter on the page? (Don’t worry yet about viewing it with your eye) b) [5 points] You hold your eye 2 cm behind the magnifying lens, and look through the magnifying lens at the letters on the page. Your eye lens is able to form a sharp image on your retina of a letter on the page. The retina is 1.7 cm behind your eye lens. Is the image on your retina upright or inverted with respect to the letter on the page? Explain using properties of lenses. The object distance for the lens is 4cm, and the lens focal length is 5 cm. Since the object distance is less than the focal length, a virtual imaged will be formed on the same side of the lens as the object. Can find the location or the virtual image from ! 1 s + 1 s' = 1 f to obtain ! 1 " s = 1 5cm # 1 4cm $ " s = #20cm Image 20 cm to the left of the magnifying lens Magnification is ! " image distance object distance = " "20cm 4cm = 5 Image is: Upright Inverted Since a sharp image is formed on the retina, it is a real image, and hence inverted relative to the object. The object is the image formed by the magnifying glass, which is upright. Hence the image on the retina is inverted with respect to the words on the page. 4 cm 2cm 1.7cm Eye Magnifying lens Page Eye Lens Retina Mag = Image type = Virtual 5.0 7 c) [10 points] Suppose that without the magnifying lens, you view the letter by positioning it at your near point, 20 cm from your eye lens. How much bigger would the letters appear with the magnifying lens? That is, what is the ratio of the image sizes on your retina with and without the magnifying lens? The distance between your eye lens and retina is 1.7 cm. Without the magnifying lens the image is ! image distance object distance = 1.7cm 20cm = 0.085 times bigger than the letter on the page when holding the letter at the near point (the reason for the 20cm distance is that this is the closest you can hold the letter and have it be in focus. That is the definition of the near point). Now determine the image size on the retina when using the magnifying lens as shown in the figure for parts a) and b). The magnifying lens alone produces a virtual image that is 5 times bigger than the letter on the page. The virtual image is upright, and located 20cm to the left of the magnifying lens. The eye lens, located 2 cm to the right of the magnifying lens, uses this virtual image 22 cm away as an object, and produces an image on the retina that is ! image distance object distance = 1.7cm 20cm + 2cm = 0.0772 times bigger than the object, and that object is 5 times bigger than the letter on the page. This means that the image on the retina is ! 5 " 0.0772 = 0.386 times bigger than the letter on the page. The ratio of these image sizes is ! 0.386 /0.085 = 4.54 . So the letters look 4.54 times bigger with the magnifying lens than without. Image size ratio = 4.54