Electric Field 1-solutions

Electric Field 1-solutions - gilvin(jg47854 3 Electric...

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gilvin (jg47854) - 3. Electric Field 1 - meyers - (21235) 1 This print-out should have 26 questions. Multiple-choice questions may continue on + the next column or page - find all choices before answering. This homework is due Tuesday, February 1, at midnight Tucson time. 001 10.0 points Two small spheres carry equal amounts of electric charge. There are equally spaced points (a , b , and c) which lie along the same a 9. a 10. a b c + b c + b c line. + a b c What is the direction of the net electric field at each point due to these charges? Explanation: Since the field originates from positive charges and terminates on the negative charges, + 1. a 2. a 3. a 4. a + b c + b c + b c + b c a b c 002 10.0 points A droplet of ink in an industrial ink-jet printer carries a charge of 2 × 10 −10 C and is deflected onto paper by a force of 0.0003 N. Find the strength of the electric field to produce this force. Correct answer: 1.5 × 10 6 V/m. Explanation: Let : F e = 0.0003 N and q = 2 × 10 −10 C . The electrical force is F e = E q 0.0003 N 5. a correct 6. a 7. a 8. + b c + b c + b c E = F e = q 2 × 10 −10 C = 1.5 × 10 6 V/m . 003 (part 1 of 2) 10.0 points In 1909 Robert Millikan was the first to find the charge of an electron in his now-famous oil drop experiment. In the experiment tiny oil drops are sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops are observed with a magnifying eyepiece, and the electric field is adjusted so that the upward force q E on some negatively charged oil drops is just sufficient to balance the downward force m g of gravity. Millikan accurately measured the charges on many oil drops and found the values to be
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E = (4.99592 * 10 ? 13 kg) (9.8 m/s 2 ) e = 5.44 * 10 ? 18 C ? gilvin (jg47854) - 3. Electric Field 1 - meyers - (21235) 2 whole-number multiples of 1.6 × 10 −19 C — Explanation: the charge of the electron. For this he won the Nobel Prize. Let : F = 0.227 N and If a drop of mass 4.99592×10 −13 kg remains E = 4.27 × 10 5 N/C . stationary in an electric field of 9 × 10 5 N/C, what is the charge on this drop? The acceler- ation of gravity is 9.8 m/s 2 . Correct answer: 5.44 × 10 −18 C. Explanation: The electric field is E = F q q = F E = 0.227 N 4.27 × 10 5 N/C = 5.31616 × 10 −7 . Let : m = 4.99592 × 10 −13 kg , E = 9 × 10 5 N/C , and g = 9.8 m/s 2 . When suspended, m g = E q q = m g 006 (part 1 of 3) 10.0 points A charge of 7 µC is at the origin. What is the magnitude of the electric field on the x axis at x = 6 m? The Coulomb constant is 8.98755 × 10 9 N m 2 /C 2 . Correct answer: 1747.58 N/C. 9 × 10 5 N/C = 5.44 × 10 −18 C . 004 (part 2 of 2) 10.0 points How many extra electrons are on this particu- lar oil drop (given the presently known charge of the electron)? Correct answer: 34. Explanation: Explanation: Let : k = 8.98755 × 10 9 N m 2 /C 2 , q = 7 µC = 7 × 10 −6 C , and x = 6 m . The electric field at a point P located a dis- tance x from a charge q is E (x) = k q (1) x 2 r p,o .
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