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A.P. Physics Electrostatics Review 2 Figure 1: An electron source fires electrons through the holes of a parallel plate capacitor. The apparatus shown in the Figure 1 above consists of two oppositely charged parallel conducting plates, each with area of 0.25 m2 , separated by a distance d =0.011 m . Each plate has a hole at its center through which electrons can pass. High velocity electrons produced by an electron source enter the top plate with speed of 5.40 106 m/s, take 1.49 ns to travel between the plates, and leave the bottom plate with speed of 8.02 106 m/s . 1. Which of the plates, top or bottom, is negatively charged? Support your answer with a reference to the direction of the electric field between the plates. 2. Calculate the magnitude of the electric field between the plates. 3. Calculate the magnitude of the charge on each plate. 4. Determine the potential difference between the plates. 5. Determine the change in potential energy that the electrons experience between the plates. Figure 2: A positively charged sphere is on a light string and is pushed to an angle due to the electric field between two metal charged plates. 1. Determine which plate is positive and which plate is negative in Figure 2. 2. Derive an equation in order to determine the value of the charge on the sphere in terms of the electric field (E) between the plates, the angle (), the mass of the sphere (m), and any universal constants. 3. Explain any changes in Figure 2 that would occur if the metal plates were moved closer together. Justify your answer. 4. Now, assume that the distance between the plates is 5.6 cm. What is the potential difference required to be between the plates to keep a 5.2 gram sphere with a charge of 17.3 µC at an angle of 250? Two students were having a discussion at lunch about the Rutherford Gold Foil Experiment. They were interested in understanding how an alpha particle could be deflected backwards by a gold nucleus. They both understood that an alpha particle can have a reasonably large velocity and wanted to see how close the alpha particle could get to the nucleus before being deflected. Below are the summaries of each student’s discussion. Student A: The alpha particle is positive and will approach the nucleus of the gold atom and then be repelled away after it gets close enough to the nucleus. This is due to the fact that the electric potential of the alpha particle is completely converted to electric potential energy. Below is a graph of the beginning velocity of the alpha particle (at infinity) vs. the closest distance (R ) that the alpha particle gets to the gold nucleus before it is deflected backwards. Observe that the graph is a straight line indicating a direct relationship between the beginning velocity and the closest distance. Student B: The alpha particle is positive and will approach the nucleus of the gold atom and then be repelled away after it gets close enough to the nucleus. This is due to the fact that the kinetic energy of the alpha particle is completely converted to electric potential energy. Below is a graph of the beginning velocity of the alpha particle (at infinity) vs. the reciprocal closest distance (1/R) that the alpha particle gets to the gold nucleus before it is deflected backwards. Observe that the graph is a straight line indicating a direct relationship between the beginning velocity and the reciprocal of the closest distance. 1. Critique the ideas of Student A by stating ideas that you agree with and any ideas that you disagree with. Make sure that you cite specific evidence based on physics principles that you have been discovering in class. 2. Critique the ideas of Student B by stating ideas that you agree with and any ideas that you disagree with. Make sure that you cite specific evidence based on physics principles that you have been discovering in class. 3. If you disagree with both graphs done by the students above, make a sketch of the graph of the relationship between the beginning velocity of the alpha particle and the closest distance below. Identify what the slope of the line is equal to if you decide that it is possible to calculate.