HW3-solutions - HW3 Tsoi (60250) 1 This print-out should...

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Unformatted text preview: HW3 Tsoi (60250) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Through what potential difference would an electron need to be accelerated for it to achieve a speed of 6 . 1 % of the speed of light (2 . 99792 10 8 m / s), starting from rest? Correct answer: 950 . 714 V. Explanation: Let : s = 6 . 1% = 0 . 061 , c = 2 . 99792 10 8 m / s , m e = 9 . 10939 10 31 kg , and q e = 1 . 60218 10 19 C . The speed of the electron is v = 0 . 061 c = 0 . 061 ( 2 . 99792 10 8 m / s ) = 1 . 82873 10 7 m / s , By conservation of energy 1 2 m e v 2 = ( q e ) V V = m e v 2 2 q e = ( 9 . 10939 10 31 kg ) ( 1 . 82873 10 7 m / s ) 2 2 (1 . 60218 10 19 C) = 950 . 714 V . 002 10.0 points (a) A test charge + q is brought to a point A a distance r from the center of a sphere having a net charge + Q . (b) Next, a test charge +5 q is brought to a point B a distance 5 r from the center of the sphere. Compared with the electrostatic potential energy of configuration (a), the potential en- ergy in configuration (b) is 1. smaller. 2. the same. correct 3. greater. Explanation: U A = k Qq r and U B = k Q (5 q ) 5 r = U A 003 (part 1 of 2) 10.0 points An electron gun fires electrons at the screen of a television tube. The electrons start from rest and are accelerated through a potential difference of 13500 V. The charge and mass on an electron are 1 . 6 10 19 C and 9 . 11 10 31 kg, respectively. What is the energy of the electrons when they hit the screen? Correct answer: 2 . 16 10 15 J. Explanation: Let : V = 13500 V . W = K = K f = e V = ( e ) (13500 V) 1 . 6 10 19 J 1 eV = 2 . 16 10 15 J . 004 (part 2 of 2) 10.0 points What is the speed of impact of electrons with the screen of the picture tube? Correct answer: 6 . 88625 10 7 m / s. Explanation: Let : V = 13500 V , e = 1 . 6 10 19 C , and m e = 9 . 11 10 31 kg . HW3 Tsoi (60250) 2 e V = 1 2 mv 2 v = radicalbigg 2 e V m e = radicalBigg 2 (1 . 6 10 19 C) (13500 V) 9 . 11 10 31 kg = 6 . 88625 10 7 m / s . 005 10.0 points Four positive charges of magnitude q are ar- ranged at the corners of a square, as shown. At the center C of the square, the potential due to one charge alone is V , and the electric field due to one charge alone has magnitude E . b b b b b C + q + q + q + q Which of the following correctly gives the electric potential and the magnitude of the electric field at the center of the square due to all four charges? Electric Electric Potential Field 1. 4 V 2 E 2. 2 V 4 E 3....
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HW3-solutions - HW3 Tsoi (60250) 1 This print-out should...

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