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

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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 – find 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 μ C 8 μ C 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 vector F 13 is vector 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. , vectorr 13 = vectorr 3 vectorr 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 radicalbig ( x 3 x 1 ) 2 = ˆ ı ˆ x 23 = x 3 x 2 radicalbig ( 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 first and third and the second and third particles. vector F = vector F 13 + vector F 23 = k e bracketleftbigg q 1 r 2 13 ˆ r 13 + q 2 r 2 23 ˆ r 23 bracketrightbigg q 3 = 8 . 9875 × 10 9 N · m 2 / C 2 × bracketleftbigg (2 × 10 6 C) ( 0 . 1 m) 2 ( ˆ ı ) + ( 6 × 10 6 C) (0 . 08 m) 2 (+ˆ ı ) bracketrightbigg × (8 × 10 6 C) = ( 14 . 38 N) + ( 67 . 4063 N) = 81 . 7863 N bardbl vector F bardbl = 81 . 7863 N .
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