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Unformatted text preview: Physics 1B: Electricity & Magnetism Summer Session I, 2010 Final Exam, July 30, 2010 This is version A ! Useful coefficients: Mass of proton, m p = 1 . 67 10 27 kg; Mass of electron, m e = 9 . 11 10 31 kg Unit of elementary charge e = 1 . 6 10 19 C Coulombs constant, k e = 8 . 99 10 9 N m 2 C 2 Permittivity of free space = 1 4 k e = 8 . 85 10 12 C 2 N 1 m 2 1 eV = 1.6 10 19 J g = 9 . 8 m s 2 Permeability of free space o = 4 10 7 T m A 1 Speed of light c = 3 10 8 m/s 1. Which of the following characteristics are held in common by both gravitational and electrostatic forces when dealing with either point masses or charges? a. inverse square distance law applies b. forces are conservative c. potential energy is a function of distance of separation d. all of the above choices are valid Solution: d. 2. An electrical device with a resistance of 1 is operated at a voltage of 2.0 V. How many electrons pass through a point inside the device in a time t = 0.8 seconds? a. 16 electrons b. 6.4 10 19 electrons c. 1.56 10 18 electrons d. 1.0 10 19 electrons e. 7.8 10 17 electrons We must calculate the current via I = V/R = 2V/1.0 = 2.0 A. From the definition of current, I = Q/t we can calculate the total charge which passes through the device: Q = I t = (2.0 C/s)(0.8 s) = 1.6 C Finally, we divide by the unit of elementary charge: N = Q/(1.6 10 19 C/e ) = 1.6 C / (1.6 10 19 C/e ) = 1 . 10 +19 electrons 1 3. In the core of a supergiant star, nuclear fusion is occurring: in one reaction, three helium nuclei ( 4 He; 2 protons, 2 neutrons each) fuse together to form a single carbon nucleus ( 12 C; six protons and six neutrons in each). Suppose three 4 He nuclei have just fused to create a 12 C nucleus, and theres an extra 4 He nucleus sitting 4 nm away (1 nm = 10 9 m). What is the resulting electrostatic force between the 12 C and 4 He nuclei? (Note: temperatures here are high enough that all atoms are ionized, i.e., all electrons have been stripped, so assume electrons are not relevant here.) a. 1 . 8 10 8 N b. 9 . 6 10 8 N c. 3 . 10 11 N d. 9 . 6 10 10 N e. 1 . 7 10 10 N Solution: q 1 = +6e; q 2 = +2e. F e = k e q 1 q 2 r 2 = 8 . 99 10 9 Nm 2 C 2 2 6 (1 . 6 10 19 C ) 2 (4 10 9 m ) 2 = 1 . 7 10 10 N 4. Two protons are initially 2 10 9 m apart. What is the CHANGE in potential energy (P.E. final P.E. initial ) if theyre brought in to a separation distance of 1 10 9 m? a. 0.18 eV b. 0.36 eV c. 0.72 eV d. 1.44 eV e. 2.88 eV The initial potential energy is PE(initial)= ( k e q 1 q 2 )/r = (8 . 99 10 9 N m 2 C 2 (1 . 6 10 19 C ) 2 / 2 e 9 m ) = 1 . 15 10 19 J....
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This note was uploaded on 09/13/2010 for the course PHYS PHYS 1B taught by Professor Berman during the Summer '09 term at UCSD.
 Summer '09
 Berman
 Charge, Magnetism, Mass

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