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Homework_06

# Homework_06 - Nguyen Thanh Homework 6 Due Feb 8 2008 7:00...

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Nguyen, Thanh – Homework 6 – Due: Feb 8 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points At distance r from a point charge q , the elec- tric potential is 277 V and the magnitude of the electric field is 233 N / C. Determine the value of q . Correct answer: 3 . 66406 × 10 - 8 C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , V = 277 V , and e = 233 N / C . E = k e q r 2 and V = k e q r , so that V E = r . The potential is V = k e q r = k e q V E = k e q E V q = V 2 k e E = (277 V) 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) (233 N / C) = 3 . 66406 × 10 - 8 C . keywords: 002 (part 1 of 1) 10 points Four identical particles each have charge 0 . 2 μ C and mass 0 . 01 kg. They are released from rest at the vertices of a square of side 0 . 47 m. How fast is each charge moving when their distances from the center of the square dou- bles? Correct answer: 0 . 321765 m / s. Explanation: Let : m = 0 . 01 kg , q = 0 . 2 μ C , and L = 0 . 47 m . Each charge moves off on its diagonal line, and by symmetry each has the same speed. The initial potential energy of the system of charges is U i = k e q 2 4 L + 2 L 2 . The final potential energy is obtained by dou- bling the distances. By conservation of energy ( K + U ) i = ( K + U ) f 4 k e q 2 L + 2 k e q 2 L 2 = 4 2 m v 2 + 4 k e q 2 2 L + 2 k e q 2 2 L 2 k e q 2 L 2 + 1 2 = 2 m v 2 v = s k e q 2 m L 1 + 1 2 2 Since k e q 2 m L = (8 . 98755 × 10 9 N · m 2 / C 2 ) (0 . 01 kg)(0 . 47 m) × (2 × 10 - 7 C) 2 = 0 . 0764898 m 2 / s 2 , v = s (0 . 0764898 m 2 / s 2 ) 1 + 1 2 2 = 0 . 321765 m / s . keywords: 003 (part 1 of 4) 10 points The electric potential over a certain region of space is given by V = a x 2 y - b x z - c y 2 , where a = 8 V / m 3 , b = 5 V / m 2 , and c = 8 V / m 2 .

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Nguyen, Thanh – Homework 6 – Due: Feb 8 2008, 7:00 pm – Inst: Weathers 2 Find the electric potential at the point ( x, y, z ) = (8 m , 1 m , 1 m). Correct answer: 464 V. Explanation: Let : a = 8 V / m 3 , b = 5 V / m 2 , c = 8 V / m 2 , and ( x, y, z ) = (8 m , 1 m , 1 m) . The electric potential is V = a x 2 y - b x z - c y 2 = ( 8 V / m 3 ) (8 m) 2 (1 m) - ( 5 V / m 2 ) (8 m) (1 m) - ( 8 V / m 2 ) (1 m) 2 = 464 V . 004 (part 2 of 4) 10 points Find the x -component of the electric field at the same point. Correct answer: - 123 V / m. Explanation: The electric field is the gradient of the po- tential, so E x = - ∂ V ∂x = - 2 a x y + b z = - 2 ( 8 V / m 3 ) (8 m) (1 m) + ( 5 V / m 2 ) (1 m) = - 123 V / m . 005 (part 3 of 4) 10 points Find the y -component of the electric field at the same point.
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Homework_06 - Nguyen Thanh Homework 6 Due Feb 8 2008 7:00...

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