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Unformatted text preview: Phys 11120, page 1 Final Exam solutions - Physics 1120 - Fall, 2007 1. A balloon of mass M has been positively charged ( its charge is + Q ) and is held up to the ceiling. The ceiling polarizes, so the balloon sticks to it. The friction coefficient between the balloon and the ceiling is μ , and we will assume that the electric field, E, created by the ceiling is uniform and directed upwards. What is the (minimum) magnitude E of this electric field? A) E = μ Q/Mg B) E = Mg/ μ Q C) E = μ Mg/Q D) E = Q/Mg E) E = Mg/Q The upward force would be F=QE, the downward force is just Mg. Those are the only forces acting in the vertical direction. (Friction is horizontal, and thus totally irrelevant for this question!) QE = Mg means E = Mg/Q. The next two questions refer to this situation: Two positively charged particles, labeled 1 and 2, are placed a distance R apart in empty space and are released from rest. Particle 1 has mass m and charge + Q ; particle 2 has mass 2 m and charge + Q/2 . Each particle feels only the static Coulomb force due to the other particle. (There is no gravity, no friction, nor any other forces in this problem.) 2. How do the magnitudes of the initial accelerations of the two particles compare? [a 1 = magnitude of initial acceleration of particle 1, a 2 = magnitude of initial acceleration of particle 2] A) a 1 = 2 a 2 B) a 1 = (1/2)a 2 C) a 1 = 4 a 2 D) a 1 = a 2 E) None of these. The forces on each must be equal (and opposite) by Newton III. But F=ma, so the smaller mass has the larger acceleration, by the ratio of masses. 3. As the particles continue to move apart after their release, the speed of each particle... A) decreases as time goes by B) increases as time goes by C) stays the same. The force is repulsive, and although it gets smaller and smaller, it is still always repulsive. A positive force means a positive acceleration, forever. Acceleration means increasing speed in this case. SPEED just keeps on getting bigger and bigger! Phys 11120, page 2 The following two questions refer to this situation: Two negative charges are each located a distance r from the origin, as shown. Note that the upper charge is twice as strong as the one to the right. (-2Q compared to -Q) 4. At the origin, the direction of electric field is ... A) Up and right at exactly a 45 degree angle to the +x axis. B) Down and left at exactly a 45 degree angle to the -x axis. C) Straight up the page D) The field has no direction, because the field is zero E) None of these: the field points in some other direction than the choices given above. There are two Field vectors at the origin , one pointing straight right, one pointing straight up with twice the magnitude. The sum of those two arrows is indeed "up and right", but it is NOT at a 45 degree angle! (The two forces would have to be equal in magnitude to get that angle) 5. ( Assume, as usual, that Voltage is zero at infinity.) At the origin, given the charge configuration above, the VOLTAGE is .... configuration above, the VOLTAGE is ....
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