HW03-solutions

# HW03-solutions - Munoz (gm7794) – HW03 – TSOI –...

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Unformatted text preview: Munoz (gm7794) – HW03 – TSOI – (92515) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two negatively charged spheres with different radii are shown in the figure below.- Q- Q The two conductors are now conneted by a wire. Which of the following occurs when the two spheres are connected with a conducting wire? 1. Negative charge flows from the larger sphere to the smaller sphere until the elec- tric field at the surface of each sphere is the same. 2. Negative charge flows from the smaller sphere to the larger sphere until the electric potential of each sphere is the same. correct 3. No charge flows. 4. Negative charge flows from the larger sphere to the smaller sphere until the elec- tric potential of each sphere is the same. 5. Negative charge flows from the smaller sphere to the larger sphere until the elec- tric field at the surface of each sphere is the same. Explanation: When the wire is connected, charge will flow until each surface is at the same potential. When disconnected the potential of each sphere is given by V = k e q r . The smaller sphere is at a more negative po- tential than the larger sphere, so negative charge will flow from the smaller sphere to the large one until they are at the same po- tential. 002 10.0 points Through what potential difference would an electron need to be accelerated for it to achieve a speed of 3 . 5% of the speed of light (2 . 99792 × 10 8 m / s), starting from rest? Correct answer: 312 . 987 V. Explanation: Let : s = 3 . 5% = 0 . 035 , 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 . 035 c = 0 . 035 ( 2 . 99792 × 10 8 m / s ) = 1 . 04927 × 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 . 04927 × 10 7 m / s ) 2 2 (1 . 60218 × 10- 19 C) = 312 . 987 V . 003 10.0 points Four charges are placed at the corners of a square of side a , with q 1 = q 2 =- q , q 3 = q 4 = + q , where q is positive. Initially there is no charge at the center of the square. Munoz (gm7794) – HW03 – TSOI – (92515) 2 q 1 =- q q 2 =- q q 4 = + q q 3 = + q q Find the work required to bring the charge q from infinity and place it at the center of the square. 1. W =- 2 k q 2 a 2 2. W = 0 correct 3. W = 2 k q 2 a 4. W = 2 k q 2 a 2 5. W =- 2 k q 2 a 6. W =- 4 k q 2 a 2 7. W = 4 k q 2 a 2 8. W = 8 k q 2 a 2 9. W =- 4 k q 2 a 10. W = 4 k q 2 a Explanation: Based on the superposition principle, the potential at the center due to the charges at the corners is V = V 1 + V 2 + V 3 + V 4 = k q r (- 1- 1 + 1 + 1) = 0 ....
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## This note was uploaded on 06/10/2011 for the course PHY 302l taught by Professor Morrison during the Summer '08 term at University of Texas.

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HW03-solutions - Munoz (gm7794) – HW03 – TSOI –...

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