exam1_sol - PHY2054 Spring 2008 Prof. Pradeep Kumar Prof....

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PHY2054 Spring 2008 1 P r o f . P r a d e e p K u m a r P r o f . P a u l A v e r y Feb. 5, 2008 Exam 1 Solutions 1. Charges 9 Q and 3 Q are held in place at positions x = 0 m and x = 4 m, respectively. At what position in x (in m) should a third charge be placed so that it is in equilibrium? (1) +2.5 (2) +9.5 (3) +1.5 (4) +4.7 (5) 5.5 The point where E = 0 must be between the charges (why?). The x components of E from the charge 9q and 3q then point in the negative and positive directions, respectively. They satisfy the relation () 2 2 9/ 3 / 0 kqx k q x d −+− = , where d is the distance between the charges. Solving for x (the positive root is the one we want) yields ( ) 3 3 / 2 2.54 xd =− = m . 2. A charge of +4 e is added to a small, uncharged beryllium sphere ( A = 9, Z = 4, density = 1850 kg/m 3 ) of radius 2.3 μ m. What is the magnitude of the electric field (in V/m) required to cause the beryllium sphere to float in the presence of gravity ( g = 9.8 m/s 2 )? (1) 1.4 × 10 6 (2) 5.2 × 10 3 (3) 2.5 × 10 4 (4) 7.7 × 10 6 (5) 9.0 × 10 5 The equation showing the balance of gravitational force and electrical force is mg = qE, where 3 4/ 3 mr π ρ = and q = 4e. Solving for E yields 36 /3 1.4 10 Er g e πρ == × V/m.
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PHY2054 Spring 2008 2 3. Refer to the previous problem. How much positive charge (in coulombs) is carried by the pro- tons in 0.34 gram of beryllium? (1) 14,600 (2) 3,600 (3) 130,000 (4) 390 (5) None of these The number of atoms in a mass m is given by 0 / mN A , where N 0 is Avogadro’s number, and m and A are expressed in grams. The total charge Q is therefore 0 / 14,600 Qm N Z eA = = C. 4. An electron moving horizontally at a velocity of 2.3 × 10 7 m/s enters a region with an electric field of magnitude E = 52,600 V/m in the same direction. How far in meters will the electron travel before it turns around? (1) 0.029 (2) 2.8 (3) 0.78 (4) 14 (5) 53 For constant acceleration, the velocity equation is 2 0 2 va d = , where v 0 is the initial velocity, d is the distance to stop it and a = eE/m e is the acceleration of the electron. Using v 0 = 2.3 × 10 7 m/s yields 2 0 / 2 0.029 e dm v e E == m. 5. Charges are arranged on a square of side d as shown in the diagram. In what direction is the electric field at the center of the square?
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This note was uploaded on 07/16/2011 for the course PHY 2005 taught by Professor Lee during the Summer '08 term at University of Florida.

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exam1_sol - PHY2054 Spring 2008 Prof. Pradeep Kumar Prof....

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