Physics Solution Manual for 1100 and 2101

3 108 v and the charge on an electron is q0 160 1019

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Unformatted text preview: on 18.2, F = qE. Since the charge q is negative, the electric field E must point downward, as the product qE in the expression F = qE must be positive, since the force F points upward. The magnitudes of the two forces must be equal, so that mg = q E . This expression can be solved for E. SOLUTION The magnitude of the electric field E is E= mg (0.012 kg)(9.80 m/s 2 ) = = 6.5 × 103 N/C –6 q 18 ×10 C As discussed in the reasoning, this electric field points downward . ______________________________________________________________________________ 998 67. ELECTRIC FORCES AND ELECTRIC FIELDS +z SSM REASONING The drawing at the right shows the set-up. Here, the electric field E points along the +y –F –q axis and applies a force of +F to the +q charge and a force of –F to the –q charge, where q = 8.0 µC denotes +y the magnitude of each charge. Each force has the same magnitude of F = E q , according to Equation 18.2. E +q The torque is measured as discussed in Section 9.1. According to Equation 9.1, the torque produced by each +F force has a magnitude given by the magnitude of the +x force times the lever arm, which is the perpendicular distance between the point of application of the force and the axis of rotation. In the drawing the z axis is the axis of rotation and is midway between the ends of the rod. Thus, the lever arm for each force is half the length L of the rod or L/2, and the magnitude of the torque produced by each force is (E q )(L/2). SOLUTION The +F and the –F force each cause the rod to rotate in the same sense about the z axis. Therefore, the torques from these forces reinforce one another. Using the expression (E q )(L/2) for the magnitude of each torque, we find that the magnitude of the net torque is Magnitude of = E q L + E q L = E q L net torque 2 2 ( )( ) = 5.0 × 103 N/C 8.0 × 10 –6 C ( 4.0 m ) = 0.16 N ⋅ m ______________________________________________________________________________ 68. REASONING The magnitude of the electrostatic force that acts on particle 1 is given by Coulomb’s law as F = k q1 q2 / r 2 . This equation can be used to find the magnitude q2 of the charge. SOLUTION Solving Coulomb’s law for the magnitude q2 of the charge gives ( 3.4 N )( 0.26 m ) F r2 = = 7.3 × 10−6 C 9 2 2 −6 k q1 8.99 × 10 N ⋅ m /C 3.5 × 10 C 2 q2 = ( )( ) (18.1) Since q1 is positive and experiences an attractive force, the charge q2 must be negative . ______________________________________________________________________________ 69. SSM WWW REASONING Each particle will experience an electrostatic force due to the presence of the other charge. According to Coulomb's law (Equation 18.1), the Chapter 18 Problems 999 magnitude of the force felt by each particle can be calculated from F = k q1 q2 / r 2 , where q1 and q2 are the respective charges on particles 1 and 2 and r is the distance between them. According to Newton's second law, the magnitude of the force experienced by each particle is given by F = ma , where a is the acceleration of the particle and we have assumed that the electrostatic force is the only force acting. SOLUTION a. Since the two particles have identical positive charges, q1 = q2 = q , and we have, using the data for particle 1, kq r2 2 = m1a1 Solving for q , we find that q= m1a1r 2 k = (6.00 × 10 –6 kg) (4.60 × 103 m/s 2 ) (2.60 × 10 –2 m) 2 8.99 × 109 N ⋅ m 2 /C 2 = 4.56 × 10 –8 C b. Since each particle experiences a force of the same magnitude (From Newton's third law), we can write F1 = F2, or m1a1 = m2a2. Solving this expression for the mass m2 of particle 2, we have (6.00 ×10 –6 kg)(4.60 × 103 m/s 2 ) = 3.25 ×10 –6 kg a2 8.50 × 103 m/s 2 ______________________________________________________________________________ m2 = m1a1 = 70. REASONING AND SOLUTION The electric field is defined by Equation 18.2: E = F/q0. Thus, the magnitude of the force exerted on a charge q in an electric field of magnitude E is given by F=qE (1) The magnitude of the electric field can be determined from its x and y components by using the Pythagorean theorem: 2 2 E = Ex + E y = ( 6.00 ×103 N/C) + (8.00 ×103 N/C ) 2 2 = 1.00 × 104 N/C a. From Equation (1) above, the magnitude of the force on the charge is F = (7.50 × 10–6 C)(1.00 × 104 N/C) = 7.5 × 10 –2 N 1000 ELECTRIC FORCES AND ELECTRIC FIELDS b. From the defining equation for the electric field, it follows that the direction of the force on a charge is the same as the direction of the field, provided that the charge is positive. Thus, the angle that the force makes with the x axis is given by 3 Ey −1 8.00 × 10 N/C = tan 6.00 × 103 N/C = 53.1° Ex ______________________________________________________________________________ θ = tan −1 71. REASONING The two charges lying on the x axis produce no net electric field at the coordinate origin. This is because they have identical charges, are located the same distance from the origin, and produce electric fields that point in opposite directions. The electric field produced by q3 at the origin points away from the charge, or along the −y direction. The electric field produced by q4 at the origin points toward the charge, or along the +y direction. The net electric field is, then, E = –E3 + E4, where E3 and E4 can be determined by using Equation 18.3. SOLUTION The net electric field at th...
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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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