homework 1 - montes (dam2772) – Homework1 – shih –...

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Unformatted text preview: montes (dam2772) – Homework1 – shih – (56310) 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 The diagram above shows an isolated, pos- itive charge Q. Point B is twice as far away from Q as point A. + Q A B What is the ratio of the electric field strength at point A to the electric field strength at point B? 1. 2 to 1 2. 4 to 1 correct 3. 1 to 2 4. 8 to 1 5. 1 to 1 Explanation: The electric field drops as 1 r 2 from a point charge. The electric field at point A would be Q d 2 where d is the distance between the charge and point A. The electric field at point B would be Q (2 d ) 2 . The ratio of E A to E B is then 1 1 / 4 or 4 to 1. 002 10.0 points When combing your hair, you scuff electrons from your hair onto the comb. Is your hair then positively or negatively charged? What about the comb? 1. Neither is charged. 2. Both are negatively charged. 3. Both are positively charged. 4. positively charged; negatively charged correct 5. negatively charged; positively charged Explanation: Excess electrons rubbed from your hair leave it positively charged. Excess electrons on the comb give it a negative charge. 003 10.0 points A particle with charge 5 μ C is located on the x-axis at the point − 4 cm , and a second particle with charge − 6 μ C is placed on the x-axis at 8 cm . 2 4 6 8 10 − 2 − 4 − 6 − 8 − 10 5 μ C − 6 μ C 2 μ C x → (cm) What is the magnitude of the total elec- trostatic force on a third particle with charge 2 μ C placed on the x-axis at − 2 cm ? The Coulomb constant is 8 . 9875 × 10 9 N · m 2 / C 2 . Correct answer: 235 . 473 N. Explanation: Let : q 1 = 5 μ C = 5 × 10 − 6 C , q 2 = − 6 μ C = − 6 × 10 − 6 C , q 3 = 2 μ C = 2 × 10 − 6 C , x 1 = − 4 cm = − . 04 m , x 2 = 8 cm = 0 . 08 m , and x 3 = − 2 cm = − . 02 m . Coulomb’s law (in vector form) for the elec- tric force exerted by a charge q 1 on a second charge q 3 , written vector F 13 is vector F 13 = k e q 1 q 3 r 2 ˆ r 13 , where ˆ r 13 is a unit vector directed from q 1 to q 3 ; i.e. , vectorr 13 = vectorr 3 − vectorr 1 . x 13 = x 3 − x 1 = ( − 2 cm) − ( − 4 cm) = 0 . 02 m x 23 = x 3 − x 2 montes (dam2772) – Homework1 – shih – (56310) 2 = ( − 2 cm) − (8 cm) = − . 1 m ˆ x 13 = x 3 − x 1 radicalbig ( x 3 − x 1 ) 2 = +ˆ ı ˆ x 23 = x 3 − x 2 radicalbig ( x 3 − x 2 ) 2 = − ˆ ı Since the forces are collinear, the force on the third particle is the algebraic sum of the forces between the first and third and the second and third particles. vector F = vector F 13 + vector F 23 = k e bracketleftbigg q 1 r 2 13 ˆ r 13 + q 2 r 2 23 ˆ r 23 bracketrightbigg q 3 = 8 . 9875 × 10 9 N · m 2 / C 2 × bracketleftbigg (5 × 10 − 6 C) (0 . 02 m) 2 (+ˆ ı ) + ( − 6 × 10 − 6 C) ( − . 1 m) 2 ( − ˆ ı ) bracketrightbigg × (2 × 10 − 6 C) = 224 . 688 N + (10 . 785 N) = 235 . 473 N...
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This note was uploaded on 03/22/2012 for the course PHYS 302L taught by Professor Tsoi during the Spring '10 term at University of Texas.

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homework 1 - montes (dam2772) – Homework1 – shih –...

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