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Unformatted text preview: Patel, Kinal – Homework 6 – Due: Feb 8 2008, 7:00 pm – Inst: Weathers 1 This printout should have 13 questions. Multiplechoice 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 858 V and the magnitude of the electric field is 203 N / C. Determine the value of q . Correct answer: 4 . 03494 × 10 7 C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , V = 858 V , and e = 203 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 = (858 V) 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) (203 N / C) = 4 . 03494 × 10 7 C . keywords: 002 (part 1 of 1) 10 points Four identical particles each have charge . 2 μ C and mass 0 . 03 kg. They are released from rest at the vertices of a square of side . 29 m. How fast is each charge moving when their distances from the center of the square dou bles? Correct answer: 0 . 236499 m / s. Explanation: Let : m = 0 . 03 kg , q = 0 . 2 μ C , and L = 0 . 29 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 µ 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 mv 2 + 4 k e q 2 2 L + 2 k e q 2 2 L √ 2 k e q 2 L µ 2 + 1 √ 2 ¶ = 2 mv 2 v = s k e q 2 mL µ 1 + 1 2 √ 2 ¶ Since k e q 2 mL = (8 . 98755 × 10 9 N · m 2 / C 2 ) (0 . 03 kg)(0 . 29 m) × (2 × 10 7 C) 2 = 0 . 0413221 m 2 / s 2 , v = s (0 . 0413221 m 2 / s 2 ) µ 1 + 1 2 √ 2 ¶ = . 236499 m / s . keywords: 003 (part 1 of 4) 10 points The electric potential over a certain region of space is given by V = ax 2 y bxz cy 2 , where a = 4 V / m 3 , b = 6 V / m 2 , and c = 3 V / m 2 . Patel, Kinal – Homework 6 – Due: Feb 8 2008, 7:00 pm – Inst: Weathers 2 Find the electric potential at the point ( x,y,z ) = (8 m , 0 m , 8 m). Correct answer: 384 V. Explanation: Let : a = 4 V / m 3 , b = 6 V / m 2 , c = 3 V / m 2 , and ( x,y,z ) = (8 m , 0 m , 8 m) . The electric potential is V = ax 2 y bxz cy 2 = ( 4 V / m 3 ) (8 m) 2 (0 m) ( 6 V / m 2 ) (8 m)(8 m) ( 3 V / m 2 ) (0 m) 2 = 384 V . 004 (part 2 of 4) 10 points Find the xcomponent of the electric field at the same point....
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This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.
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