# 11 - fryer(vdf96 – Electric Fields Electric Potential and...

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Unformatted text preview: fryer (vdf96) – Electric Fields, Electric Potential and Energy – graves – (6) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Usually the force of gravity on electrons is neglected. To see why, we can compare the force of the Earth’s gravity on an electron with the force exerted on the electron by an electric field of magnitude of 50000 V / m (a relatively small field). The acceleration of gravity is 9 . 8 m / s 2 , the mass of an electron is 9 . 10939 × 10 − 31 kg, and the charge on an electron is − 1 . 602 × 10 − 19 C. What is the force exerted on the electron by an electric field of magnitude of 50000 V / m? Correct answer: − 8 . 01 × 10 − 15 N. Explanation: Let : E = 50000 V / m and e = − 1 . 602 × 10 − 19 C . The electrical force is F e = E e = (50000 V / m) ( − 1 . 602 × 10 − 19 C) = − 8 . 01 × 10 − 15 N . 002 (part 2 of 2) 10.0 points What is the force of the Earth’s gravity on the electron? Correct answer: 8 . 9272 × 10 − 30 N. Explanation: Let : g = 9 . 8 m / s 2 and m e = 9 . 10939 × 10 − 31 kg . The gravitational force is F g = g m e = (9 . 8 m / s 2 ) (9 . 10939 × 10 − 31 kg) = 8 . 9272 × 10 − 30 N . We can see F e ≫ F g . 003 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 E = F e q = . 0003 N 2 × 10 − 10 C = 1 . 5 × 10 6 V / m . 004 10.0 points The diagram shows an isolated, positive charge Q , where point B is twice as far away from Q as point A . + Q A B 10 cm 20 cm The ratio of the electric field strength at point A to the electric field strength at point B is 1. E A E B = 2 1 . 2. E A E B = 1 2 . 3. E A E B = 1 1 . 4. E A E B = 8 1 . 5. E A E B = 4 1 . correct Explanation: Let : r B = 2 r A . fryer (vdf96) – Electric Fields, Electric Potential and Energy – graves – (6) 2 The electric field strength E ∝ 1 r 2 , so E A E B = 1 r 2 A 1 r 2 B = r 2 B r 2 A = (2 r ) 2 r 2 = 4 . 005 (part 1 of 5) 10.0 points A wall has a negative charge distribution pro- ducing a uniform horizontal electric field. A small plastic ball of mass 0 . 0198 kg, carrying charge of − 77 μ C, is suspended by an un- charged, nonconducting thread 0 . 354 m long. The thread is attached to the wall and the ball hangs in equilibrium in the electric and gravitational fields, as shown....
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## This note was uploaded on 12/11/2011 for the course PHYSICS 101 taught by Professor Graves during the Fall '10 term at San Jose City College.

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11 - fryer(vdf96 – Electric Fields Electric Potential and...

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