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Unformatted text preview: roofner (bar784) – Homework 6 – Weathers – (22202) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points At distance r from a point charge q , the elec tric potential is 431 V and the magnitude of the electric field is 310 N / C. Determine the value of q . Correct answer: 6 . 66732 × 10 − 8 C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , V = 431 V , and e = 310 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 = (431 V) 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) (310 N / C) = 6 . 66732 × 10 − 8 C . 002 10.0 points Four identical particles each have charge . 5 μ C and mass 0 . 03 kg. They are released from rest at the vertices of a square of side . 34 m. How fast is each charge moving when their distances from the center of the square dou bles? Correct answer: 0 . 546045 m / s. Explanation: Let : m = 0 . 03 kg , q = 0 . 5 μ C , and L = 0 . 34 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 parenleftbigg 4 L + 2 L √ 2 parenrightbigg . 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 mv 2 + 4 k e q 2 2 L + 2 k e q 2 2 L √ 2 k e q 2 L parenleftbigg 2 + 1 √ 2 parenrightbigg = 2 mv 2 v = radicalBigg k e q 2 mL parenleftbigg 1 + 1 2 √ 2 parenrightbigg Since k e q 2 mL = (8 . 98755 × 10 9 N · m 2 / C 2 ) (0 . 03 kg)(0 . 34 m) × (5 × 10 − 7 C) 2 = 0 . 220283 m 2 / s 2 , v = radicalBigg (0 . 220283 m 2 / s 2 ) parenleftbigg 1 + 1 2 √ 2 parenrightbigg = . 546045 m / s . 003 (part 1 of 4) 10.0 points The electric potential over a certain region of space is given by V = ax 2 y bxz cy 2 , where a = 3 V / m 3 , b = 8 V / m 2 , and c = 1 V / m 2 . Find the electric potential at the point ( x,y,z ) = (1 m , 4 m , 8 m). Correct answer: 68 V. Explanation: roofner (bar784) – Homework 6 – Weathers – (22202) 2 Let : a = 3 V / m 3 , b = 8 V / m 2 , c = 1 V / m 2 , and ( x,y,z ) = (1 m , 4 m , 8 m) . The electric potential is V = ax 2 y bxz cy 2 = ( 3 V / m 3 ) (1 m) 2 (4 m) ( 8 V / m 2 ) (1 m) (8 m) ( 1 V / m 2 ) (4 m) 2 = 68 V . 004 (part 2 of 4) 10.0 points Find the xcomponent of the electric field at the same point. Correct answer: 40 V / m. Explanation: The electric field is the gradient of the po tential, so E x = ∂ V ∂x = 2 axy + bz = 2 ( 3 V / m 3 ) (1 m) (4 m) + ( 8 V / m 2 ) (8 m) = 40 V / m . 005 (part 3 of 4) 10.0 points Find the ycomponent of the electric field at the same point....
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This document was uploaded on 10/26/2011 for the course PHYS 2220 at North Texas.
 Fall '00
 Littler
 Charge, Magnetism, Work

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