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The electric field between the two plates shown below has a magnitude of 500N/C. The plates are separated by a distance d of 2 cm. What is the potential difference VBA between the two plates? What is the speed of a proton released from plate B just before it hits plate A if the system is in a vacuum? (assume that the surface area of the plates is much larger than the distance between the plates.) + • • • • • Question 1 Question 2 Question 3 Question 4 Question 5 • • • • • • Question 6 Question 7 Question 8 Question 9 Question 10 Paired Question 1. Which of the following physics principles is the best to use to solve this problem? A: Charge conservation principle B: The principle of conservation of mechanical energy C: Gauss’ Law Choice: A Incorrect There is no exchange of charge in this problem. Choice: B Correct Mechanical energy will be conserved in this system. Choice: C Incorrect This law is used to find the electric field, which is of no interest to us here. 2. Which plate is at the higher potential? A: A B: B C: They are both at the same potential. Choice: A Incorrect Potential decreases in the direction of the electric field E, which is directed from the positively charged plate to the negatively charged plate. Choice: B Correct The positive plate has a higher potential. Choice: C Incorrect The charge separation will cause a potential difference between the two plates. 3. What is meant by an electric potential difference VBA between two points? A: It is the electric potential energy with respect to infinity of a unit positive charge. B: It is the work that must be done to move a unit charge slowly from b to a against an electric force. C: It is the negative of the work done by the electric forces in taking an object from a given reference point to a point p. Choice: A Incorrect Electric potential energy is not the same as electric potential. Electric potential is electric potential energy per unit charge associated with a test charge q at that point. Choice A is the electric potential energy where one has placed the reference point at infinity. Choice: B Correct This is expressed mathematically as: W UB UA (VB VA ) VBA q q Choice: C Incorrect This is the electric potential energy at point p. 4. When released, what type of path will the proton follow?(assume that the electric field between the plates is uniform) A: Parabolic B: Linear C: Oscillatory Choice: A Incorrect The proton is not shot perpendicular to the electric field so the path will not be parabolic. Choice: B Correct This is similar to an object in a free fall in a gravitational field. Choice: C Incorrect The charge will move in the direction of the electric field, which has only one component of direction. The path will be linear. 5. As the proton moves through the potential difference, what kind of energy does it lose? A: Gravitational energy (GPE) B: Kinetic energy (K) C: Electric potential energy (EPE) Choice: A Incorrect The gravitational potential energy of this system is much less than the electric potential energy and can be ignored. Choice: B Incorrect Initially, there is no kinetic energy. The proton gains kinetic energy as it speeds up from rest. Choice: C Correct It loses electrical potential energy, which is the analog of gravitational potential energy when a ball is dropped from the roof of a building. 6. What is the change in electric potential energy UBA equal to, as the proton moves between the plates? A: Ed B: qE C: qVBA UB UA UBA Choice: A Incorrect This is the magnitude of the potential difference. Choice: B Incorrect This is the electrical force. Choice: C Correct If the same reference point is used, potential energy U and potential difference V are related by the following equation: U qV …and since potential energy can be defined as the capacity for doing work, We see that: W UA UB UAB UBA UBA q(VB VA ) qVBA 7. As the proton moves between the charged plates, it gains which type of energy? A: gravitational potential energy B: kinetic energy C: both kinetic and gravitational potential energy Choice: A Incorrect Changes in gravitational potential energy can be ignored, because the gravitational interactions involved are much weaker than the electrical interaction. Choice: B Correct This proton will hit plate A with some speed v. Since it starts at rest and gains speed it also gains kinetic energy. Choice: C Incorrect Changes in gravitational potential energy can be ignored. Notice that the proton will gain kinetic energy as it speeds up from rest. 8. Which of the following equations describes the conservation of mechanical energy for this system? 1 A: qE mv2 2 1 2 B: VBA mv 2 1 2 C: qVBA mv 2 Choice: A Incorrect qE has the units of force, not energy. Using the law of conservation of energy in this situation, we should equate the sum of the kinetic energy and electric potential energy of the proton before it is released, to the sum of the kinetic energy and electric potential energy just before it hits the negatively charged plate. Choice: B Incorrect VBA is potential difference not potential energy. We must have units of energy on each side of the equality. Choice: C Correct There is no initial velocity. 0 1 1 2 2 mvi qVB mv f qVA 2 2 1 2 qVB qVA q(VB VA ) qVBA mv f 2 1 1 2 2 qVBA mv f mv 2 2 9. From the information given in the original problem statement, what is magnitude of the potential difference VBA between the plates? A: 1000 V B: 10V C: 250V Choice: A Incorrect Make sure to convert the distance between the plates to meters. VBA Ed (500N/C)(0.02m) 10J/C 10V Choice: B Correct VBA Ed (500N/C)(0.02m) 10J/C 10V Choice: C Incorrect Try Again using the following equation: VBA Ed 10. Solve the equation found in question 8 for the speed of the proton v. answer Final Answer: 1 2 qVBA mv 2 2qVBA v2 m 2qVBA v m 2(6.02x1019 C)(10.0V) v (1.67x1027 kg) v 8.5x104 m/s