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SPH4U7-Fields 1. Which of the following graphs shows how the electric field strength E varies with distance d from the centre of a charged, conducting, hollow sphere of radius R ? A B E E 0 R 0 d C 2 a. b. c. d R d D E 0 R E R 0 d Which of the following diagrams most accurately depicts the field between two oppositely charged plates? d. e. 3. An electron and proton with the same kinetic energy enter a region containing a uniform magnetic field. Which is deflected more from its original path? Explain. 4. Why are magnetic fields and electric fields used in particle accelerators? 5. This question is about gravitation and orbital motion. (a) Define gravitational field strength at a point in a gravitational field. The diagram below shows three points above a planet. The arrow represents the gravitational field strength at point A. C B planet (b) 6. A Draw arrows to represent the gravitational field strength at point B and point C. An electron accelerates from rest (starting at a position that is 75% across from the negative plate) through an electric field and into a magnetic field as shown in the diagram below. The plates have a potential difference of 85 V and the magnetic field has a magnitude of 0.75 T. (Remember: me = 9.1x10–31 kg and e = 1.6x10–19 C.) (a) Calculate the initial speed of the electron upon entering the magnetic field. (b) Calculate the magnitude and direction of the magnetic force on the electron. (c) Calculate the radius of the electron’s circular path. (d) How would the path change if it was a proton that was fired into the field at that speed? Explain. 7. You are on a strange planet. You note that the stars do not rise and set but circle around in planes parallel to the horizon. You then set out on a journey which takes you in a straight line (really a great circle) for 16,000 km and find that the starts now rise and set perpendicular to the horizon. You now drop a stone on the planet from a height of 3 m and measure the time the stone takes to fall to the ground in which is 0.53 seconds. What is the mass of the planet? 8. In a Millikan-type experiment, two horizontal parallel plates are 1.0 cm apart. A sphere of mass 2.2 10–13 kg remains stationary when the potential difference between the plates is 350 V with the upper plate negative. (a) Is the sphere positively or negatively charged? Explain. (b) Calculate the magnitude of charge on the sphere. (c) How much excess or deficit of electrons does the sphere have? 9. A proton enters a region of uniform electric field between two conducting horizontal metal plates as shown below. +95 V Path of proton P 2.2 cm 2.7 × 107 m s–1 0V 12 cm As the proton enters the region of the electric field, it is travelling parallel to the plates. a) Calculate the force on the proton b) Suggest why gravitational effects are not considered when calculating the net force on the proton. c) In a mass spectrometer, electric and magnetic fields are used to select charged particles of one particular speed. A uniform magnetic field is applied in the region between the plates, such that the proton passes between the plates without being deviated. i) Draw an FBD of the proton ii) State the direction of the Magnetic field iii) Calculate the magnetic field strength Answers 1.A (Charge distributes itself uniformly on the surface so Fnet inside the sphere is 0) 2.B 3. Since the kinetic energies are the same, the proton is moving slower. Because its speed is less the force on it is less (F=qvB). A=F/m, Force is less and mass is more than the electron so its acceleration is less. Therefore its deflection is a lot less than the electron. 4. Magnetic fields accelerate and change direction. Electric fields accelerate the charge 5. a) Gravitational field strength is the gravitational force experienced by the small test mass at that location divided by the mass of the test mass. C B planet A b) 6. a) Ee=Ek (0.25x85)(q)=0.5mv2 v=2.73x106 m/s b) FM is directed upwards (RHR #3) FM=qvBsinθ =3.28x10-13N c)Fc=FM r=(mv2/(qvB) = 2.06x10-5 m 7. –Based on the info about the path of the stars you have moved from the pole to the equator. - The distance (16000 km) is therefore a quarter of the circumference of the earth - Therefore 2πr=64000 km= 64000000m -Therefore r=10185916m -u=0, t=0.53m, s=3m….. 3=0.5a(.53)2 - a=21.35m/s2 - a=g=GM/r2 - M=3.2x1025 kg 8. a) The electric force must be up to balance the downward force of gravity. The upper plate is negative and the bottom plate is positive. The sphere must be positively charged to be repelled by the bottom plate and attracted to the upper plate. (b) and The magnitude of charge on the sphere is 6.2 10–17 C. (c) The sphere has a deficit of 385 electrons since its charge is positive. 9. a) Fe=Vq/d= 6.92x10-16N b) Fg is significantly smaller than Fe c) i) Fm Fe ii) Into the page (RHR #3) iii)Fe=FM 6.92x10-16N=qvBsinθ B=1.6x10-4 T