HW1%20solutions

# HW1%20solutions - Physics 204 morning summer 2008 Homework...

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Unformatted text preview: Physics 204, morning summer 2008 Homework 1 solutions Chapter 18 Problems 3, 6, 56, 8, 17 3. REASONING AND SOLUTION The total charge of the electrons is q = N(–e) = (6.0 1013)(–1.60 10–19 C) q = – 9.6 10–6 C = –9.6 C The net charge on the sphere is, therefore, qnet = +8.0 μC – 9.6 μC = −1.6 μC 6. REASONING a. The number N of electrons is 10 times the number of water molecules in 1 liter of water. The number of water molecules is equal to the number n of moles of water molecules times Avogadro’s number NA: N = 10 n NA . b. The net charge of all the electrons is equal to the number of electrons times the change on one electron. SOLUTION a. The number N of water molecules is equal to 10 n NA , where n is the number of moles of water molecules and NA is Avogadro’s number. The number of moles is equal to the mass m of 1 liter of water divided by the mass per mole of water. The mass of water is equal to its density Thus, the number of electrons is times the volume, as expressed by Equation 11.1. ⎛ ⎞ ⎛ ⎞ ρV m N = 10 n NA = 10 ⎜ ⎟ NA = 10 ⎜ ⎟ NA ⎝ 18.0 g/mol ⎠ ⎝ 18.0 g/mol ⎠ ⎡ −3 3 3 ⎛ 1000 g ⎞ ⎤ ⎢ 1000 kg/m 1.00 × 10 m ⎜ −1 23 ⎟⎥ = 10 ⎢ ⎝ 1 kg ⎠ ⎥ 6.022 × 10 mol ⎢ ⎥ 18.0 g/mol ⎣ ⎦ ( )( ) ( ) = 3.35 × 1026 electrons b. ( The net charge Q of all the electrons is equal to the number of electrons times the change on one electron: Q = 3.35 × 1026 )( −1.60 × 10 −19 ) C = −5.36 × 107 C . 56. REASONING AND SOLUTION The electrostatic forces decreases with the square of the distance separating the charges. If this distance is increased by a factor of 5 then the force will decrease by a factor of 25. The new force is, then, F= 3.5 N = 0.14 N 25 8. REASONING The number N of excess electrons on one of the objects is equal to the charge q on it divided by the charge of an electron (−e), or N = q/(−e). Since the charge on the object is negative, we can write q = − q , where q is the magnitude of the charge. The magnitude of the charge can be found from Coulomb’s law (Equation 18.1), which states that the magnitude F of the electrostatic force exerted on each object is given by F = k q q / r 2 , where r is the distance between them. SOLUTION The number N of excess electrons on one of the objects is N= q q −q = = (1) −e −e e To find the magnitude of the charge, we solve Coulomb’s law, F = k q q / r 2 , for q : q = F r2 k Substituting this result into Equation (1) gives N= q = e F r2 k = e ( 4.55 ×10−21 N )(1.80 ×10−3 m )2 8.99 ×109 N ⋅ m 2 / C2 1.60 × 10−19 C =8 17. REASONING a. There are two electrostatic forces that act on q1; that due to q2 and that due to q3. The magnitudes of these forces can be found by using Coulomb’s law. The magnitude and direction of the net force that acts on q1 can be determined by using the method of vector components. b. According to Newton’s second law, Equation 4.2b, the acceleration of q1 is equal to the net force divided by its mass. However, there is only one force acting on it, so this force is the net force. SOLUTION a. The magnitude F12 of the force exerted F13 2 r12 1.30 m 23.0° q1 where the distance is specified in the drawing: F12 = F12 23.0° on q1 by q2 is given by Coulomb’s law, Equation 18.1, k q1 q2 q2 +y 1.30 m +x q3 (8.99 × 109 N ⋅ m2 /C2 )(8.00 × 10−6 C )(5.00 × 10−6 C) = 0.213 N = (1.30 m )2 Since the magnitudes of the charges and the distances are the same, the magnitude of F13 is the same as the magnitude of F12, or F13 = 0.213 N. From the drawing it can be seen that the x‐components of the two forces cancel, so we need only to calculate the y components of the forces. Force y component F12 +F12 sin 23.0° = +(0.213 N) sin 23.0° = +0.0832 N F13 +F13 sin 23.0° = +(0.213 N) sin 23.0° = +0.0832 N F Fy = +0.166 N Thus, the net force is F = + 0.166 N (directed along the +y axis) . b. According to Newton’s second law, Equation 4.2b, the acceleration of q1 is equal to the net force divided by its mass. However, there is only one force acting on it, so this force is the net force: a= F +0.166 N = = +111 m /s 2 m 1.50 × 10−3 kg where the plus sign indicates that the acceleration is along the +y axis . ...
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