4 - Copy (2)

4 - Copy (2) - Version 027/AABCD midterm 01a turner(56705...

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Version 027/AABCD – midterm 01a – turner – (56705) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points P . x y ++++ −−−− L . x y ++++++ −−−−−− M . x y +++++ +++++ + + + + + G . x y −−−− ++++ S . x y ++++ ++++ ±or which confguration(s) does the total electric feld vector at the origin have non- zero components in the x direction as well as the y direction ( i.e. , both x and y components are non-zero)? 1. Confgurations P , G and L only 2. Confgurations P and S only 3. Confgurations S and L only 4. Confguration P only 5. Confgurations S , M and L only 6. Confguration S only 7. Confguration M only 8. Confguration G only correct 9. Confgurations P , S and L only 10. Confgurations P and L only Explanation: Basic Concepts: Δ E = k Δ q r 2 ˆ r and E = s Δ E . Symmetry oF the confguration will cause some component oF the electric feld to be zero. Solution: Confguration P : It is anti- symmetric about the y -axis (opposite sign oF charges), so the electric feld has no y - component. x y P Confguration S : It is symmetric by a ro- tation oF 180 , so the electric felds generated by these two pieces have opposite directions; thereFore the total feld is zero.
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Version 027/AABCD – midterm 01a – turner – (56705) 2 x y ++++ ++++ S Confguration G : It is anti-symmetric by rotation oF a 180 , so the total feld has non- zero components in both x and y directions, just like the feld generated by just one piece oF charge. x y −−−− ++++ G Confguration M : It is symmetric about the x -axis, so the y component oF the total feld must vanish. x y +++++ +++++ + + + + + M Confguration L : It is symmetric about the y -axis, so the x component oF the total feld must vanish. x y ++++++ −−−−−− L 002 10.0 points Three identical point charges, each oF mass 150 g and charge + q , hang From three strings, as in the fgure. 9 . 8 m / s 2 10 . 7 cm 150 g + q 150 g + q 38 150 g + q IF the lengths oF the leFt and right strings are each 10 . 7 cm, and each Forms an an- gle oF 38 with the vertical, determine the value oF q . The acceleration oF gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1. 0.415744 2. 0.66606 3. 0.517167 4. 1.1998 5. 1.07128 6. 0.750419 7. 1.14671 8. 0.825952 9. 1.04714 10. 0.479555 Correct answer: 0 . 66606 μ C. Explanation: Let : θ = 38 , m = 150 g = 0 . 15 kg , L = 10 . 7 cm = 0 . 107 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . Consider the Forces acting on the charge on the right. There must be an electrostatic Force F acting on this charge, keeping it balanced against the Force oF gravity mg . The electro- static Force is due to the other two charges and is thereFore horizontal.
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4 - Copy (2) - Version 027/AABCD midterm 01a turner(56705...

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