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Homework Solutions 3

# Homework Solutions 3 - Ash(bka258 HW3 Tsoi(60250 This...

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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 Q q 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 .

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Ash (bka258) – HW3 – Tsoi – (60250) 2 e Δ V = 1 2 m v 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 0 , and the electric field due to one charge alone has magnitude E 0 . 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 0 0 correct 2. 2 V 0 4 E 0 3. 0 0 4. 0 2 E 0 5. 4 V 0 2 E 0 Explanation: The electric potential is scalar, so the elec- tric potential at the center is 4 V 0 . The electric field is a vector with both magnitude and di- rection in the space. All of the electric field components due to the four charges have the same magnitudes but different directions. In the configuration, the sum of the four compo- nent vectors is zero, so the total field at C is zero.
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