Physics Solution Manual for 1100 and 2101

85 10 12 c 2 n m 2 020 m 2 cos 00o 79

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Unformatted text preview: auses the magnitude of the net field at P to be twice as great as it is when only the first charge is present. Since both fields have the same direction, the magnitude of E2 must, then, be the same as the magnitude of E1. But the second charge is further away from point P than is the first charge, and more distant charges create weaker fields. To offset the weakness that comes from the greater distance, the second charge must have a greater magnitude than that of the first charge. b. The drawing at the right shows the q2 +q 1 E2 P E1 electric fields at point P due to the two − d d charges in the case that the second charge is negative. The presence of the second charge causes the magnitude of the net field at P to be twice as great as it is when only the first charge is present. Since the fields now have opposite directions, the magnitude of E2 must be greater than the magnitude of E1. This is necessary so that E2 can offset E1 and still lead to a net field with twice the magnitude as E1. To create this greater field E2, the second charge must now have a greater magnitude than it did in question (a). SOLUTION a. The magnitudes of the field contributions of each charge are given according to kq Equation 18.3 as E = 2 . With q2 present, the magnitude of the net field at P is twice what r it is when only q1 is present. Using Equation 18.3, we can express this fact as follows: k q1 d2 + k q2 ( 2d ) 2 =2 k q1 d2 or k q2 ( 2d )2 = k q1 d2 Solving for q2 gives q2 = 4 q1 = 4 ( 0.50 µ C ) = 2.0 µ C Thus, the second charge is q2 = +2.0 µC . b. Now that the second charge is negative, we have k q2 ( 2d ) 2 − k q1 d2 =2 k q1 d2 or k q2 ( 2d )2 =3 Solving for q2 gives q2 = 12 q1 = 12 ( 0.50 µ C ) = 6.0 µ C Thus, the second charge is q2 = −6.0 µC . k q1 d2 984 ELECTRIC FORCES AND ELECTRIC FIELDS 49. REASONING Since we know the initial velocity and time, we can determine the particle’s displacement from an equation of kinematics, provided its acceleration can be determined. The acceleration is given by Newton’s second law as the net force acting on the particle divided by its mass. The net force is the electrostatic force, since the particle is moving in an electric field. The electrostatic force depends on the particle’s charge and the electric field, both of which are known. SOLUTION To obtain the displacement x of the particle we employ Equation 3.5a from the equations of kinematics: x = v0 x t + 1 a x t 2 . We use this equation because two of the 2 variables, the initial velocity v0x and the time t, are known. The initial velocity is zero, since the particle is released from rest. The acceleration ax can be found from Newton’s second law, as given by Equation 4.2a, as the net force ΣFx acting on the particle divided by its mass m: ax = ΣFx / m . Only the electrostatic force Fx acts on the proton, so it is the net force. Setting ΣFx = Fx in Newton’s second law gives ax = Fx / m . Substituting this result into Equation 3.5a, we have that F (1) x = v0 xt + 1 axt 2 = v0 xt + 1 x t 2 2 2m Since the particle is moving in a uniform electric field Ex, it experiences an electrostatic force Fx given by Fx = q0 Ex (Equation 18.2), where q0 is the charge. Substituting this expression for Fx into Equation (1) gives F q E x = v0 x t + 1 x t 2 = v0 xt + 1 0 x t 2 2m 2 m ( +12 ×10−6 C ) ( +480 N/C ) 2 −2 −2 = ( 0 m/s ) (1.6 ×10−2 s ) + 1 (1.6 ×10 s ) = +1.9 ×10 m 2 −5 3.8 ×10 kg ______________________________________________________________________________ 50. REASONING The following drawing shows the two particles in the electric field Ex. They are separated by a distance d. If the particles are to stay at the same distance from each other after being released, they must have the same acceleration, so ax, 1 = ax, 2. According to Newton’s second law (Equation 4.2a), the acceleration ax of each particle is equal to the net force ΣFx acting on it divided by its mass m, or ax = ΣFx / m . Chapter 18 Problems 985 d Ex q1 = −7.0 µ C m1 = 1.4 × 10 −5 q2 = +18 µ C −5 m2 = 2.6 × 10 kg +x kg SOLUTION The net force acting on each particle and its resulting acceleration are: q1: The charge q1 experiences a force q1Ex due to the electric field (see Equation 18.2). The charge also experiences an attractive force in the +x direction due to the presence of q2. This force is given by Coulomb’s law as + k q1 q2 / d 2 (see Equation 18.1). The net force acting on q1 is ΣFx, 1 = q1Ex + k q1 q2 d2 The acceleration of q1 is ΣFx, 1 ax,1 = m1 = q1Ex + k q1 q2 d2 m1 q2: The charge q2 experiences a force q2 E x due to the electric field. It also experiences an attractive force in the −x direction due to the presence of q1. This force is given by Coulomb’s law as − k q1 q2 / d 2 . The net force acting on q2 is ΣFx, 2 = q2 Ex − k q1 q2 d2 The acceleration of q2 is ax, 2 = ΣFx, 2 m2 = q2 E x − k q1 q2 d2 m2 Setting ax,1 = ax,2 gives q1Ex + k m1 q1 q2 d2 = q2 Ex − k m2 q1 q2 d2 986 ELECTRIC FORCES AND ELECTRIC FIELDS Solving this expression for d,...
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