20075ee1_1_HW9 - CHAPTER 9 9.1 A point charge Q =-0.3 C and...

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CHAPTER 9 9.1. A point charge, Q = 0 . 3 µ C and m = 3 × 10 16 kg, is moving through the field E = 30 a z V/m. Use Eq. (1) and Newton’s laws to develop the appropriate differential equations and solve them, subject to the initial conditions at t = 0: v = 3 × 10 5 a x m/s at the origin. At t = 3 µ s, find: a) the position P ( x, y, z ) of the charge: The force on the charge is given by F = q E , and Newton’s second law becomes: F = m a = m d 2 z dt 2 = q E = ( 0 . 3 × 10 6 )(30 a z ) describing motion of the charge in the z direction. The initial velocity in x is constant, and so no force is applied in that direction. We integrate once: dz dt = v z = qE m t + C 1 The initial velocity along z , v z (0) is zero, and so C 1 = 0. Integrating a second time yields the z coordinate: z = qE 2 m t 2 + C 2 The charge lies at the origin at t = 0, and so C 2 = 0. Introducing the given values, we find z = ( 0 . 3 × 10 6 )(30) 2 × 3 × 10 16 t 2 = 1 . 5 × 10 10 t 2 m At t = 3 µ s, z = (1 . 5 × 10 10 )(3 × 10 6 ) 2 = . 135cm. Now, considering the initial constant velocity in x , the charge in 3 µ s attains an x coordinate of x = vt = (3 × 10 5 )(3 × 10 6 ) = . 90 m. In summary, at t = 3 µ s we have P ( x, y, z ) = ( . 90 , 0 , . 135) . b) the velocity, v : After the first integration in part a , we find v z = qE m t = (3 × 10 10 )(3 × 10 6 ) = 9 × 10 4 m / s Including the intial x -directed velocity, we finally obtain v = 3 × 10 5 a x 9 × 10 4 a z m / s . c) the kinetic energy of the charge: Have K . E . = 1 2 m | v | 2 = 1 2 (3 × 10 16 )(1 . 13 × 10 5 ) 2 = 1 . 5 × 10 5 J 9.2. A point charge, Q = 0 . 3 µ C and m = 3 × 10 16 kg, is moving through the field B = 30 a z mT. Make use of Eq. (2) and Newton’s laws to develop the appropriate differential equations, and solve them, subject to the initial condition at t = 0, v = 3 × 10 5 m/s at the origin. Solve these equations (perhaps with the help of an example given in Section 7.5) to evaluate at t = 3 µ s: a) the position P ( x, y, z ) of the charge; b) its velocity; c) and its kinetic energy: We begin by visualizing the problem. Using F = q v × B , we find that a positive charge moving along positive a x , would encounter the z -directed B field and be deflected into the negative y direction. 1
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9.2 (continued) Motion along negative y through the field would cause further deflection into the negative x direction. We can construct the differential equations for the forces in x and in y as follows: F x a x = m dv x dt a x = qv y a y × B a z = qBv y a x F y a y = m dv y dt a y = qv x a x × B a z = qBv x a y or dv x dt = qB m v y (1) and dv y dt = qB m v x (2) To solve these equations, we first differentiate (2) with time and substitute (1), obtaining: d 2 v y dt 2 = qB m dv x dt = qB m 2 v y Therefore, v y = A sin( qBt/m ) + A cos( qBt/m ). However, at t = 0, v y = 0, and so A = 0, leaving v y = A sin( qBt/m ). Then, using (2), v x = m qB dv y dt = A cos qBt m Now at t = 0, v x = v x 0 = 3 × 10 5 . Therefore A = v x 0 , and so v x = v x 0 cos( qBt/m ), and v y = v x 0 sin( qBt/m ). The positions are then found by integrating v x and v y over time: x ( t ) = v x 0 cos qBt m dt + C = mv x 0 qB sin qBt m +
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