This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full Document
 Fall '00
 Littler
 Charge, Electric Potential, Electrostatics, Magnetism, Work, Electric charge, Fundamental physics concepts, charge density

Click to edit the document details