Chapter 22 Solutions

Chapter 22 Solutions - wright (daw2557) – Chapter 22 –...

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Unformatted text preview: wright (daw2557) – Chapter 22 – de – (14443) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An object having a net charge of 19 . 7 μ C is placed in a uniform electric field of 730 N / C directed vertically up. The acceleration of gravity is 9 . 8 m / s 2 . What is the mass of this object if it “floats” in the field? Correct answer: 1 . 46745 g. Explanation: Let : Q = 19 . 7 μ C = 1 . 97 × 10- 5 C , E = 730 N / C , and g = 9 . 8 m / s 2 . Call the vertical direction the y-direction, so the the unit vector ˆ points up. Then force equilibrium in the vertical direction for a charge Q of mass m yields summationdisplay F = Q E ˆ + mg (- ˆ ) = 0 For this to hold, Q E- mg = 0 m = Q E g = (1 . 97 × 10- 5 C) (730 N / C) 9 . 8 m / s 2 × 1000 g 1 kg = 1 . 46745 g ....
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This note was uploaded on 03/23/2011 for the course PHYS 1444 taught by Professor Sureshsharma during the Spring '09 term at UT Arlington.

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Chapter 22 Solutions - wright (daw2557) – Chapter 22 –...

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