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Unformatted text preview: Reilly (mar3978) – oldmidterm 01 – Turner – (56725) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two spheres, fastened to “pucks”, are rid ing on a frictionless airtrack. Sphere 1 is charged with 2 nC, and sphere 2 is charged with 10 nC. Both objects have the same mass. 1 nC is equal to 1 × 10 − 9 C. As they repel, 1. sphere 1 accelerates 5 times as fast as sphere 2. 2. sphere 2 accelerates 25 times as fast as sphere 1. 3. sphere 1 accelerates 25 times as fast as sphere 2. 4. sphere 2 accelerates 5 times as fast as sphere 1. 5. they have the same magnitude of acceler ation. correct 6. they do not accelerate at all, but rather separate at constant velocity. Explanation: The force of repulsion exerted on each mass is determined by F = 1 4 π ǫ Q 1 Q 2 r 2 = ma where r is the distance between the centers of the two spheres: bardbl vector F 12 bardbl = bardbl vector F 21 bardbl . Since both spheres have the same mass and are subject to the same force, they have the same acceleration. 002 10.0 points If the potential in a region is given by the function V = 2 x − y 2 − cos( z ) , what is the ycomponent of the electric field at the point P = ( x ′ , y ′ , z ′ )? 1. E y = y ′ 2 2. E y = cos( z ′ ) 3. E y = sin( z ′ ) 4. E y = − 2 y ′ 5. E y = 2 y ′ correct 6. E y = y ′ 3 3 7. E y = x ′ 2 4 8. E y = 2 x ′ 9. E y = − sin( z ′ ) 10. E y = 2 Explanation: The electric field is the gradient of the po tential, so the ycomponent of the Efield eval uated at P is E y = − ∂ V ∂y = − ∂ ∂y bracketleftbig 2 x − y 2 − cos( x ) bracketrightbig = − [ − 2 y ] = 2 y . 003 10.0 points The two charges Q are fixed at the vertices of an equilateral triangle with sides of length a as shown. a a a Q Q q Reilly (mar3978) – oldmidterm 01 – Turner – (56725) 2 The work required to move a charge q from the other vertex to the center of the line join ing the fixed charges is 1. W = 0 2. W = 6 k Q q a 3. W = √ 2 k Q q a 4. W = 2 k Q q a correct 5. W = 4 k Q q a 6. W = k Q q a Explanation: U 1 = + k Q q a + k Q q a = +2 k Q q a U 2 = + k Q q a 2 + k Q q a 2 = +4 k Q q a W = U 2 − U 1 = 2 k Q q a . 004 10.0 points Five charges are placed in a closed box. Each charge (except the first) has a magnitude which is twice that of the previous one placed in the box. All charges have the same sign and (after all the charges have been placed in the box) the net electric flux through the box is 6 . 5 × 10 7 N · m 2 / C. What is the magnitude of the smallest charge in the box? Correct answer: 18 . 5652 μ C....
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 Spring '10
 TURNER,J
 Electric charge, Reilly

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