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Unformatted text preview: lim (kl9356) – 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 935 V and the magnitude of the electric field is 152 N / C. Determine the value of q . Correct answer: 6 . 39939 × 10 − 7 C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , V = 935 V , and e = 152 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 = (935 V) 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) (152 N / C) = 6 . 39939 × 10 − 7 C . 002 10.0 points Four identical particles each have charge . 3 μ C and mass 0 . 02 kg. They are released from rest at the vertices of a square of side . 27 m. How fast is each charge moving when their distances from the center of the square dou bles? Correct answer: 0 . 45028 m / s. Explanation: Let : m = 0 . 02 kg , q = 0 . 3 μ C , and L = 0 . 27 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 . 02 kg)(0 . 27 m) × (3 × 10 − 7 C) 2 = 0 . 149793 m 2 / s 2 , v = radicalBigg (0 . 149793 m 2 / s 2 ) parenleftbigg 1 + 1 2 √ 2 parenrightbigg = . 45028 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 = 1 V / m 3 , b = 5 V / m 2 , and c = 5 V / m 2 . Find the electric potential at the point ( x,y,z ) = (9 m , 9 m , 7 m). Correct answer: 9 V. Explanation: lim (kl9356) – Homework 6 – Weathers – (22202) 2 Let : a = 1 V / m 3 , b = 5 V / m 2 , c = 5 V / m 2 , and ( x,y,z ) = (9 m , 9 m , 7 m) . The electric potential is V = ax 2 y bxz cy 2 = ( 1 V / m 3 ) (9 m) 2 (9 m) ( 5 V / m 2 ) (9 m) (7 m) ( 5 V / m 2 ) (9 m) 2 = 9 V . 004 (part 2 of 4) 10.0 points Find the xcomponent of the electric field at the same point. Correct answer: 127 V / m. Explanation: The electric field is the gradient of the po tential, so E x = ∂ V ∂x = 2 axy + bz = 2 ( 1 V / m 3 ) (9 m) (9 m) + ( 5 V / m 2 ) (7 m) = 127 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 note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Charge, Magnetism, Work

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