spring_07_exam1_phy2054

spring_07_exam1_phy2054 - PHY2054 Spring 2007 Prof Eugene...

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PHY2054 Spring 2007 1 Prof. Eugene Dunnam P r o f . P a u l A v e r y Feb. 6, 2007 Exam 1 Solutions 1. A charge Q 1 and a charge Q 2 = 1000 Q 1 are located 25 cm apart. The ratio of the electrostatic force on Q 1 to that on Q 2 is: (1) none of these (2) 1000 (3) 0.0032 (4) 0.001 (5) 316 The ratio of forces is 1.0, as can be seen from Coulomb’s law or Newton’s third law. 2. The charge on each of two tiny spheres is doubled while their separation distance is tripled. The ratio of the new electrostatic force to the old force is: (1) 4/9 (2) 2/3 (3) 9/4 (4) 4/3 (5) 1 Doubling the charge makes the numerator 4 times larger, while tripling the distance makes the denominator 9 times larger. Together they yield a force that is 4/9 of the original force. 3. An electron circles a carbon nucleus (6 protons, 6 neutrons) at a distance of 0.0353 nanome- ters. What is its velocity in km/s? (1) 6560 (2) 9280 (3) 2680 (4) 153 (5) 1950 Here we balance centripetal force with the coulomb force, or () 22 /6 / mv r k e e r = , where 6e is the charge of the carbon nucleus, r is the orbital radius and m is the mass of the electron. Solv- ing for v gives 26 6 / 6.56 10 / 6560 vk e m r m s == × = km/s.
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PHY2054 Spring 2007 2 4. The electric field required to suspend a proton against the force of gravity is: (1) 102 nanovolts/m directed upward (2) 102 nanovolts/m directed downward (3) 557 picovolts/m directed upward (4) 557 picovolts/m directed downward (5) none of these The E field required to balance the proton is found from p Ee m g = , where m p is the mass of the proton and e is its charge. Solving yields E = 102 nanovolts/m, directed upwards to balance the gravitational force. 5. Negative charge Q is located on the y -axis at y = a and charge 2 Q is located at y = a . The direction of the electric field at the point ( a , 0) due to these charges is toward (1) First quadrant (2) + y (3) x (4) + x (5) y The net y component of E is positive and the net x component is positive, putting the direction of E in the first quadrant. 6. A charge q is placed at the origin and a charge 4 q is placed on the x -axis at x = 2 cm. At what value of x on the x axis (in cm) does the electric field equal zero? (1) 2.00 (2) +4.00 (3) +0.67 (4) 4.00 (5) 0.67 The point where E = 0 must be to the left of the charge q (why?). The x components of E from the charge q and 4q then point in the negative and positive directions, respectively. They satisfy the relation () 2 2 /4 / 0 kq x kq x d −+ = , where d is the distance between the charges. Solving for x (the negative root is the one we want) yields x = d = 2 cm.
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This test prep was uploaded on 04/17/2008 for the course PHY 2054 taught by Professor Avery during the Spring '08 term at University of Florida.

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spring_07_exam1_phy2054 - PHY2054 Spring 2007 Prof Eugene...

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