solution1_pdf - huang (jwh2377) Homework 2 lang (58650)...

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huang (jwh2377) – Homework 2 – lang – (58650) 1 This print-out should have 55 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Calculate the net charge on a substance con- sisting oF a combination oF 3 . 8 × 10 13 protons and 5 . 0 × 10 13 electrons. Correct answer: 1 . 92 × 10 6 C. Explanation: Let: N p = 3 . 8 × 10 13 charges , q p = 1 . 60 × 10 19 C , N e = 5 . 0 × 10 13 charges and q e = 1 . 60 × 10 19 C . Q net = N p q p + N e q e Q net = (3 . 8 × 10 13 )(1 . 6 × 10 19 C) + (5 × 10 13 )( 1 . 6 × 10 19 C) = 1 . 92 × 10 6 C 002 10.0 points You have 2 . 2 kg oF water. One mole oF water has a mass oF 18 g / mol and each molecule oF water contains 10 electrons since water is H 2 O. What is the total electron charge contained in this volume oF water? Correct answer: 1 . 17766 × 10 8 C. Explanation: Let : N A = 6 . 02214 × 10 23 molec / mol , q e = 1 . 6 × 10 19 C / electron , m = 2 . 2 kg , M = 18 g / mol = 0 . 018 kg / mol , and Z = 10 electrons / molec . The mass is proportional to the number oF molecules, so For m grams in n molecules and M grams in N A molecules, m M = n N A n = m M N A Since 10 electrons are in each molecule oF water, then the total number oF electrons n e in the coin is n e = Z n = Z m M N A and the total charge q For the n e electrons is q = n e q e = Z m N A q e M = (10 electrons / molec) 2 . 2 kg 0 . 018 kg / mol × ( 6 . 02214 × 10 23 molec / mol ) × ( 1 . 6 × 10 19 C / electron ) = 1 . 17766 × 10 8 C . 003 10.0 points A particle with charge 2 μ C is located on the x -axis at the point 8 cm , and a second particle with charge 6 μ C is placed on the x -axis at 10 cm . The Coulomb constant is 8 . 9875 × 10 9 N · m 2 / C 2 . 2 4 6 8 10 2 4 6 8 10 2 μ C 6 8 x (cm) What is the magnitude oF the total elec- trostatic Force on a third particle with charge 8 μ C placed on the x -axis at 2 cm ? Correct answer: 81 . 7863 N. Explanation: Let : q 1 = 2 μ C = 2 × 10 6 C , q 2 = 6 μ C = 6 × 10 6 C , q 3 = 8 μ C = 8 × 10 6 C , x 1 = 8 cm = 0 . 08 m , x 2 = 10 cm = 0 . 1 m , and x 3 = 2 cm = 0 . 02 m .
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huang (jwh2377) – Homework 2 – lang – (58650) 2 Coulomb’s law (in vector form) for the elec- tric force exerted by a charge q 1 on a second charge q 3 , written v F 13 is v F 13 = k e q 1 q 3 r 2 ˆ r 13 , where ˆ r 13 is a unit vector directed from q 1 to q 3 ; i.e. , vr 13 = vr 3 vr 1 . x 13 = x 3 x 1 = ( 2 cm) (8 cm) = 0 . 1 m x 23 = x 3 x 2 = ( 2 cm) ( 10 cm) = 0 . 08 m ˆ x 13 = x 3 x 1 r ( x 3 x 1 ) 2 = ˆ ı ˆ x 23 = x 3 x 2 r ( x 3 x 2 ) 2 = +ˆ ı Since the forces are collinear, the force on the third particle is the algebraic sum of the forces between the Frst and third and the second and third particles.
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This note was uploaded on 02/18/2010 for the course PHYS 317L taught by Professor Lang during the Spring '10 term at University of Texas at Austin.

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solution1_pdf - huang (jwh2377) Homework 2 lang (58650)...

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