spring_07_exam4_phy2054

spring_07_exam4_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 Apr. 28, 2007 Exam 4 Solutions 1. A charge of 5 microC is placed at the origin and a +10 microC charge is located at x = 1.0 m. Where, on the x-axis, is the net electrostatic field equal to zero, and is the electrostatic poten- tial at this point also zero? (1) x = +2.41 m; no (2) x = 2.41 m; no (3) x = +1.00 m; yes (4) x = 0.62 m; no (5) none of these The electric field must be zero for points to the right of the smaller charge (x > 0) because in that region the field from the larger charge is compensated by the greater distance. If +2q is the charge at x = 1.0 and q is the charge at the origin, then the equation for E x = 0 can be written () 2 2 /2 / 1 0 kq x kq x −+ + = . Solving yields 2 1 2.41 x =++ ± m. 2. The sketch depicts a 50-turn rectangular loop of dimensions a = 55 cm and b = 130 cm, and rotated by an angle θ = 65 ° from the y axis. It is hinged along the z axis and immersed in a uni- form magnetic field B = 0.27 T pointing in the + z direction. The current in the loop is 4.5 A. The torque on the loop (in N m) and its direction of rotation, viewed from above, is (CW = clockwise and CCW = counterclockwise) (1) 0.0 (2) 0.79, CW (3) 0.37, CCW (4) 0.87, CW (5) 2.92, CCW Only the currents along the top and bottom of the loop contribute to the torque since the currents on the side generate zero force apiece. However, the torques from the top and bottom parts of the loop cancel because the forces are equal and opposite and the lever arm is the same. x z y ± a b B
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PHY2054 Spring 2007 2 3. Three electrons are placed at the corners of an equilateral triangle of side 1 nm. Find the elec- trical potential energy in J of this configuration. (1) +6.9 × 10 19 (2) 6.9 × 10 19 (3) +2.3 × 10 19 (4) 2.3 × 10 19 (5) zero The total electrical potential energy is found by adding the potential energy of all charge pairs. For three electrons arranged in an equilateral triangle of side d, the potential energy is 2 tot 3/ Uk e d = . This yields an energy of +6.9 × 10 19 J . 4. A parallel-plate capacitor has plates of area A , separated by distance d . A charge Q is placed on it so that the potential difference between the plates is V and the stored energy is U . With the device isolated, the plates are now moved to a separation distance of d / 2. What is now their potential difference, and the stored energy? (1) V / 2, U / 2 (2) 2 V , 2 U (3) 4 V , 4 U (4) 8 V , U / 4 (5) V , U (unchanged) The charge is unchanged and the E field is also unchanged since it is proportional to the charge density of the plate. The voltage = E × distance is 1/2 of its former value or V / 2, while the ca- pacitance becomes twice as large. Thus () ( ) 22 11 1 2 2/ 2 / 2 / 2 UC V C V C V U ′′ == = = 5. Two conductors are made of the same material and have the same length. Conductor a is a solid wire of diameter 2 mm. Conductor b is a hollow tube of inside diameter 2mm and outside diameter 4mm.
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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_exam4_phy2054 - PHY2054 Spring 2007 Prof. Eugene...

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