Physics Solution Manual for 1100 and 2101

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Unformatted text preview: rners. 958 ELECTRIC FORCES AND ELECTRIC FIELDS SOLUTION As pointed out in the REASONING, the magnitude of any +y 2 individual force vector is F = k q / L2 . With this in mind, we apply the component method for vector addition to the forces at corner A, which are shown in the drawing at the right, together with the appropriate components. The x component Σ Fx and the FAB FAB sin 60.0º FAC 60.0 +x A y component Σ Fy of the net force are FAB cos 60.0 ( Σ Fx )A = FAB cos 60.0° + FAC = F ( cos 60.0° + 1) ( Σ Fy )A = FAB sin 60.0° = F sin 60.0° where we have used the fact that FAB = FAC = F . The Pythagorean theorem indicates that the magnitude of the net force at corner A is ( Σ F )A = ( Σ Fx )A + ( Σ Fy )A 2 2 =F = F 2 ( cos 60.0° + 1) + ( F sin 60.0° ) 2 ( cos 60.0° + 1) + ( sin 60.0° ) = k ( 2 = 8.99 ×10 N ⋅ m / C 9 2 2 2 ) (5.0 ×10−6 C) ( 0.030 m ) 2 2 q 2 L 2 ( cos 60.0° + 1)2 + ( sin 60.0° )2 2 ( cos 60.0° + 1)2 + ( sin 60.0° )2 = 430 N We now apply the component method for vector addition to the forces at corner B. These forces, together with the appropriate components are shown in the drawing at the right. We note immediately that the two vertical components cancel, since they have opposite directions. The two horizontal components, in contrast, reinforce since they have the same direction. Thus, we have the following components for the net force at corner B: FBC +y 60.0º FBC cos 60.0º +x B FBA cos 60.0 FBA Chapter 18 Problems 959 ( Σ Fx )B = − FBC cos 60.0° − FBA cos 60.0° = −2 F cos 60.0° ( Σ Fy )B = 0 where we have used the fact that FBC = FBA = F . The Pythagorean theorem indicates that the magnitude of the net force at corner B is ( Σ F )B = ( Σ Fx )B + ( Σ Fy )B = ( −2 F cos 60.0° )2 + ( 0 )2 2 = 2k 2 q 2 L 2 ( cos 60.0° = 2 8.99 × 109 N ⋅ m 2 / C 2 ) = 2 F cos 60.0° (5.0 ×10−6 C ) ( 0.030 m ) 2 2 cos 60.0° = 250 N As discussed in the REASONING, the magnitude of the net force acting on the charge at corner C is the same as that acting on the charge at corner B, so ( ΣF )C = 250 N . 21. 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. q2 +y SOLUTION F13 F12 a. The magnitude F12 of the force exerted on q1 1.30 m 23.0° 23.0° by q2 is given by Coulomb’s law, Equation 18.1, +x where the distance is specified in the drawing: q1 1.30 m q3 F12 = k q1 q2 2 r12 (8.99 × 109 N ⋅ m2 /C2 ) (8.00 × 10−6 C) (5.00 × 10−6 C) = 0.213 N = (1.30 m )2 960 ELECTRIC FORCES AND ELECTRIC FIELDS 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: +0.166 N F a= = = +111 m /s 2 −3 m 1.50 × 10 kg where the plus sign indicates that the acceleration is along the +y axis . ______________________________________________________________________________ 22. REASONING We will use Coulomb’s law to calculate the force that any one charge exerts on another charge. Note that in such calculations there are three separations to consider. Some of the charges are a distance d apart, some a distance 2d, and some a distance 3d. The greater the distance, the smaller the force. The net force acting on any one charge is the vector sum of three forces. In the following drawing we represent each of those forces by an arrow. These arrows are not drawn to scale and are meant only to “symbolize” the three different force magnitudes that result from the three different distances used in Coulomb’s law. In the drawing the directions are determined by the facts that like charges repel and unlike charges attract. By examining the drawing we will be able to identify the greatest and the smallest net force. + A d + B d + C d − D The greatest net force occurs for charge C, because all three force contributions point in the same direction and two of the three have the greatest magnitude, while the third has the next Chapter 18 Problems 961 greatest magnitude. The smallest net force occurs for charge B, because two of the three force contrib...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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