Exam I - Version 047/AACDD – midterm 01 – Turner –...

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Unformatted text preview: Version 047/AACDD – midterm 01 – Turner – (60230) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Three identical point charges hang from three strings, as shown. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 , and the acceleration of gravity is 9 . 81 m / s 2 . 45 ◦ 45 ◦ F g 20.0 cm 20.0 cm + + + +q +q +q 0.10 kg 0.10 kg 0.10 kg What is the value of q ? 1. 2.44481e-06 2. 1.98228e-06 3. 1.58582e-06 4. 1.9162e-06 5. 2.31266e-06 6. 1.78405e-06 7. 8.58988e-07 8. 1.85013e-06 9. 1.6519e-06 10. 1.32152e-06 Correct answer: 1 . 32152 × 10 − 6 C. Explanation: Let : m = 0 . 10 kg , L = 20 . 0 cm , θ = 45 ◦ , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . r = 2 L sin θ = 2 L sin 45 ◦ = 2 L √ 2 2 = L √ 2 F T,x = F T sin θ F T,y = F T cos θ Each sphere is in equilibrium horizontally F electric − F T,x = 0 F electric − F T sin θ = 0 and vertically F T,y − F g = 0 F T cos θ − F g = 0 F T = F g cos θ . From the horizontal equilibrium, F electric = parenleftbigg F g cos θ parenrightbigg sin θ F electric = F g tan θ = F g (tan 45 ◦ ) = F g . For either of the end charges, F electric = k e q 2 r 2 + k e q 2 ( r 2 ) 2 = k e q 2 r 2 + 4 k e q 2 r 2 = 5 k e q 2 r 2 5 k e q 2 r 2 = mg . Thus | q | = radicalBigg r 2 mg 5 k e = radicalBigg ( L √ 2) 2 mg 5 k e = L · radicalbigg 2 mg 5 k e = (20 cm) parenleftbigg 1 m 100 cm parenrightbigg × radicalBigg 2(0 . 1 kg)(9 . 81 m / s 2 ) 5(8 . 98755 × 10 9 N · m 2 / C 2 ) = 1 . 32152 × 10 − 6 C . Version 047/AACDD – midterm 01 – Turner – (60230) 2 002 10.0 points An insulating sphere of radius 15 cm has a uniform charge density throughout its vol- ume. 15 cm 6 . 7517 cm 29 . 0129 cm p If the magnitude of the electric field at a distance of 6 . 7517 cm from the center is 22769 . 2 N / C, what is the magnitude of the electric field at 29 . 0129 cm from the center? 1. 2835.34 2. 17185.7 3. 12383.8 4. 13521.5 5. 131170.0 6. 27046.9 7. 42930.6 8. 60133.9 9. 82917.4 10. 62637.4 Correct answer: 13521 . 5 N / C. Explanation: Let : R = 15 cm , E 1 = 22769 . 2 N / C , r 1 = 6 . 7517 cm , r 2 = 29 . 0129 cm , V = 4 3 π R 3 , and ρ = Q V . R r 1 r 2 p Method 1: We know the magnitude of the electric field at a radius r 1 = 6 . 7517 cm (corresponding to a smaller sphere with sur- face area A 1 and volume V 1 ): the magnitude is E 1 = 22769 . 2 N / C. We want to find the magnitude E 2 at a radius r 2 = 29 . 0129 cm, corresponding to a sphere with surface area A 2 and volume V 2 that is larger than the insulating sphere. From Gauss’s Law, we know that since the flux is constant over the sphere, E 1 A 1 = Φ 1 = Q 1 ǫ relating the flux through the Gaussian sphere of radius r 1 to the charge enclosed, Q 1 . We also know Q 1 = ρ V 1 . For the outer sphere (radius r 2 = 29 . 0129 cm), E 2 A 2 = Φ 2 = Q ǫ with Q = ρ V (Not ρ V 2 , as the Gaussian sur- face is larger than the actual physical sphere,...
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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Exam I - Version 047/AACDD – midterm 01 – Turner –...

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