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solution2 - fullwood (tf5349) Homework 1 Electrostatics...

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fullwood (tf5349) – Homework 1 Electrostatics – Catala – (12054) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 10.0 points ±ind the magnitude of the electric Feld at a point midway between two charges +17 . 2 × 10 9 C and +62 . 9 × 10 9 C separated by a distance of 77.2 cm. The value of the Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 . Correct answer: 2757 . 41 N / C. Explanation: Let : q 1 = 17 . 2 × 10 9 C , q 2 = 62 . 9 × 10 9 C , x = 77 . 2 cm , and k C = 8 . 99 × 10 9 N · m 2 / C 2 . r 1 = r 2 = x 2 = 77 . 2 cm 2 · p 1 m 100 cm P = 0 . 386 m The electric Feld is E = k C q r 2 . The magnitudes of the electric Felds are E 1 = 8 . 99 × 10 9 N · m 2 / C 2 × 1 . 72 × 10 8 C (0 . 386 m) 2 = 1037 . 8 N / C , and E 2 = 8 . 99 × 10 9 N · m 2 / C 2 × 6 . 29 × 10 8 C (0 . 386 m) 2 = 3795 . 21 N / C . If the 17 . 2 × 10 9 C charge is the left one and the positive direction is to the right, E net = E 1 E 2 = 1037 . 8 N / C 3795 . 21 N / C = 2757 . 41 N / C b v E net b = 2757 . 41 N / C . The electric Feld at midpoint between the charges +17 . 2 × 10 9 C and +62 . 9 × 10 9 C is 2757 . 41 N / C toward the +17 . 2 × 10 9 C charge. 002 10.0 points A charge of +2 . 20 × 10 9 C is placed at the origin, and another charge of +6 . 50 × 10 9 C is placed at x = 1 . 7 m. ±ind the point (coordinate) between these two charges where a charge of +3 . 80 × 10 9 C should be placed so that the net electric force on it is zero. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 0 . 625258 m. Explanation: Let : q 1 = 2 . 20 × 10 9 C at the origin , q 2 = 6 . 50 × 10 9 C , q 3 = 3 . 80 × 10 9 C , and x 2 = 1 . 7 m . q 3 is between q 1 and q 2 , so if r 1 , 3 = P , then r 2 , 3 = x 2 P . F electric = k C q 1 q 2 r 2 The net electric force is zero, so F 1 , 3 = F 2 , 3 k C q 1 q 3 r 2 1 , 3 = k C q 2 q 3 r 2 2 , 3 q 1 r 2 1 , 3 = q 2 r 2 2 , 3 q 1 P 2 = q 2 ( x 2 P ) 2 ( x 2 P ) 2 P 2 = q 2 q 1 x 2 P P = r q 2 q 1 x 2 P = P r q 2 q 1 P p 1 + r q 2 q 1 P = x 2
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fullwood (tf5349) – Homework 1 Electrostatics – Catala – (12054) 2 P = x 2 1 + r q 2 q 1 = 1 . 7 m 1 + r 6 . 5 × 10 9 C 2 . 2 × 10 9 C = 0 . 625258 m . 003 (part 1 of 2) 10.0 points One gram of platinum has 3 . 09 × 10 21 atoms, and each platinum atom has 78 elec- trons. How many electrons are contained in 6.00 g of platinum? Correct answer: 1 . 44612 × 10 24 electrons. Explanation:
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This note was uploaded on 06/10/2010 for the course PHY 100 taught by Professor Pro.kiroshima during the Spring '10 term at Adelphi.

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solution2 - fullwood (tf5349) Homework 1 Electrostatics...

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