Unformatted text preview: PHY2054 Spring 2007 Prof. Eugene Dunnam
Prof. Paul Avery
Feb. 6, 2007 Exam 1 Solutions ‘ “ . 'l. A charge Q, and a charge Q2 = 1000Q1 are located 25 cm apart. The ratio of the electrostatic
force on 91 to that on Q; is: ’ (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 tn'pled.
The ratio of the new electrostatic force to the old force is: (i0 4/9 (2) 2/3
(3) 9/4
(4) 4/3
(5) l 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 electronEl‘cE 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 Eentripetal force with the coulomb ﬂn'ce, WE = k(6e)e/r2, where 6e is] the charge of the carbon nucleus, r is the orbital radius and m is the mass of the electron. Solv.
ingfor vgives v = \léke2 lmr = 6.5t5x106 m/s = 6560 Ian/s. ...
View
Full Document
 Spring '09
 DUNNAM

Click to edit the document details