Homework Solutions 3 - Ash (bka258) – HW3 – Tsoi –...

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Unformatted text preview: Ash (bka258) – 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 3 . 4 % of the speed of light (2 . 99792 × 10 8 m / s), starting from rest? Correct answer: 295 . 357 V. Explanation: Let : s = 3 . 4% = 0 . 034 , 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 . 034 c = 0 . 034 ( 2 . 99792 × 10 8 m / s ) = 1 . 01929 × 10 7 m / s , By conservation of energy 1 2 m e v 2 = − ( − q e ) Δ V Δ V = m e v 2 2 q e = ( 9 . 10939 × 10 − 31 kg ) × ( 1 . 01929 × 10 7 m / s ) 2 2 (1 . 60218 × 10 − 19 C) = 295 . 357 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 +6 q is brought to a point B a distance 6 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. the same. correct 2. greater. 3. smaller. Explanation: U A = k Qq r and U B = k Q (6 q ) 6 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 11500 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: 1 . 84 × 10 − 15 J. Explanation: Let : Δ V = 11500 V . W = Δ K = K f = e Δ V = ( e ) (11500 V) · 1 . 6 × 10 − 19 J 1 eV = 1 . 84 × 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 . 35572 × 10 7 m / s. Explanation: Let : Δ V = 11500 V , e = 1 . 6 × 10 − 19 C , and m e = 9 . 11 × 10 − 31 kg . Ash (bka258) – HW3 – Tsoi – (60250) 2 e Δ V = 1 2 mv 2 v = radicalbigg 2 e Δ V m e = radicalBigg 2 (1 . 6 × 10 − 19 C) (11500 V) 9 . 11 × 10 − 31 kg = 6 . 35572 × 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?...
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This note was uploaded on 09/29/2008 for the course PHY 303 taught by Professor Erskine/tsoi during the Spring '08 term at University of Texas at Austin.

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Homework Solutions 3 - Ash (bka258) – HW3 – Tsoi –...

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