RCW1%20solutions

RCW1%20solutions - Physics 204, morning summer 2008...

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Unformatted text preview: Physics 204, morning summer 2008 Recitation ClassWork 1 solutions Chapter 18 Conceptual Question #3,4,6,8 Concepts & calculations #70,71,75 3. REASONING AND SOLUTION When the charged insulating rod is brought near to (but not touching) the sphere, the free electrons in the sphere will move. If the rod is negatively charged, the free electrons will move to the side of the sphere that is opposite to the side where the rod is; if the rod is positively charged, the free electrons will migrate to the side of the sphere where the rod is. In either case, the region of the sphere near the vicinity of the rod will acquire a charge that has the opposite sign as the charge on the rod. a. Since oppositely charged objects always attract each other, the rod and sphere will always experience a mutual attraction. b. Since the side of the sphere in the vicinity of the rod will always have charge that is opposite in sign to the charge on the rod, the rod and the sphere will always attract each other. They never repel each other. 4. REASONING AND SOLUTION On a dry day, just after washing your hair to remove natural oils and drying it thoroughly, you run a plastic comb through it. As the surface of the comb rubs against your hair, the comb becomes electrically charged. If the comb is brought near small bits of paper, the charge on the comb causes a separation of charge on the bits of paper. Since the paper is neutral, it contains equal amounts of positive and negative charge. The charge on the comb causes the regions of the bits of paper that are closest to the comb to become oppositely charged; therefore, the bits of paper are attracted to the comb. This situation is analogous to that in Figure 18.9. 6. REASONING AND SOLUTION A proton and electron are held in place on the x axis. The proton is at x = −d, while the electron is at x = + d . They are released simultaneously, and the only force that affects their motions is the electrostatic force of attraction that each applies to the other. According to Newton's third law, the force F pe exerted on the proton by the electron is equal in magnitude and opposite in direction to the force F exerted on ep the electron by the proton. In other words, F = −Fep . According to Newton's second law, this equation can be pe written mp ap = − me ae (1) where m p and me are the respective masses and ap and ae are the respective accelerations of the proton and the electron. Since the mass of the electron is considerably smaller than the mass of the proton, the acceleration of the electron at any instant must be considerably greater than the acceleration of the proton at that instant in order for Equation (1) to hold. Since the electron has a much greater acceleration than the proton, it will attain greater velocities than the proton and, therefore, reach the origin first. 8. 1 SSM REASONING AND SOLUTION Identical point 2 charges are fixed to opposite corners of a square, as shown in the figure at the right. There exists an electric field at each point in space in the vicinity of this configuration. The electric field at a given point is the resultant of the electric field at that point due to the charge at corner 1 and the charge at corner 3. C 4 The magnitude of the electric field due to a single point 3 charge q is given by Equation 18.3: E = k q / r 2 , where r is the distance from the charge to the point in question. Since the point charges at corners 1 and 3 are identical, they each have the same value of q. Furthermore, they are equidistant from the center point C. Therefore, at the center of the square, each charge gives rise to an electric field that is equal in magnitude and opposite in direction. The resultant electric field at the center of the square is, therefore, zero. Note that this result is independent of the polarity of the charges. The distance from either empty corner, 2 or 4, to either of the charges is the same. Since the charges are equidistant from either empty corner, each charge gives rise to an electric field that is equal in magnitude. Since the direction of the electric field due to a point charge is radial (radially inward for negative charges and radially outward for positive charges), we see that the electric fields due to each of the two point charges will be mutually perpendicular. Their resultant can be found by using the Pythagorean theorem. The magnitude of the force experienced by a third point charge placed in this system is, from Equation 18.2: F = q0 E , where q0 is the magnitude of the charge and E is the magnitude of the electric field at the location of the charge. Since the electric field is zero at the center of the square, a third point charge will experience no force there. Thus, a third point charge will experience the greater force at one of the empty corners of the square. 70. CONCEPT QUESTIONS a. The electrical force that each charge exerts on the middle charge is shown in the drawing below. F21 is the force exerted on 2 by 1, and F23 is the force exerted on 2 by 3. Each force has the same magnitude, because the charges have the same magnitude and the distances are equal. −q F21 +q 1 F23 2 (a) +q +q F23 +q 3 1 2 F21 +q +q 3 1 F21 −q 2 F23 (b) 3 +q (c) b. The net electric force F that acts on 2 is shown in the diagrams below. F21 F23 F21 F23 F23 (a) F21 F F = 0 N F (b) (c) It can be seen from the diagrams that the largest electric force occurs in (a), followed by (c), and then by (b). SOLUTION The magnitude F21 of the force exerted on 2 by 1 is the same as the magnitude F23 of the force exerted on 2 by 3, since the magnitudes of the charges are the same and the distances are the same. Coulomb’s law gives the magnitudes as F21 = F23 = kq q r2 (8.99 × 10 9 = )(8.6 × 10 C)(8.6 × 10 2 (3.8 × 10−3 m ) N ⋅ m /C 2 −6 2 −6 C ) = 4.6 × 104 N In part (a) of the drawing, both F21 and F23 point to the left, so the net force has a magnitude of ( ) F = 2 F12 = 2 4.6 × 104 N = 9.2 × 104 N In part (b) of the drawing, F21 and F23 point in opposite directions, so the net force has a magnitude of 0 N . In part (c) the magnitude can be obtained from the Pythagorean theorem: 2 2 F = F21 + F23 = ( 4.6 × 104 N ) + ( 4.6 × 104 N ) 2 2 = 6.5 × 104 N 71. CONCEPT QUESTIONS a. The gravitational force is an attractive force. To neutralize this force, the electrical force must be a repulsive force. Therefore, the charges must both be positive or both negative. b. Newton’s law of gravitation, Equation 4.3, states that the gravitational force depends inversely on the square of the distance between the earth and the moon. Coulomb’s law, Equation 18.1 states that the electrical force also depends inversely on the square of the distance. When these two forces are added together to give a zero net force, the distance can be algebraically eliminated. Thus, we do not need to know the distance between the two bodies. SOLUTION Since the repulsive electrical force neutralizes the attractive gravitational force, the magnitudes of the two forces are equal: kq q r 2 = Electrical force, Equation 18.1 GM e M m r2 Gravitational force, Equation 4.3 Solving this equation for the magnitude q of the charge on either body, we find q = GM e M m k ⎛ N ⋅ m2 ⎞ 6.67 × 10−11 5.98 × 1024 kg 7.35 × 1022 kg ⎜ 2 ⎟ kg ⎠ = ⎝ = 5.71 × 1013 C 2 N⋅m 8.99 × 109 C2 ( )( ) 75. CONCEPT QUESTIONS a. The drawing at the right shows the forces that act on the charges at each corner. For example, FAB is the force exerted on the charge at corner A by the charge at corner B. The directions of the forces are consistent with the fact that like charges repel and unlike charges attract. Coulomb’s law indicates that all of the forces shown have the same magnitude, namely, 2 FBC + FBA FAB F = k q / L , where q is the magnitude of each of 2 the charges and L is the length of each side of the equilateral triangle. The magnitude is the same for each force, because q and L are the same for each pair of charges. B + A − FAC FCA C FCB b. The net force acting at each corner is the sum of the two force vectors shown in the drawing, and the net force is greatest at corner A. This is because the angle between the two vectors at A is 60º. With the angle less than 90º, the two vectors partially reinforce one another. In comparison, the angles between the vectors at corners B and C are both 120º, which means that the vectors at those corners partially offset one another. c. The net forces acting at corners B and C have the same magnitude, since the magnitudes of the individual vectors are the same and the angles between the vectors at both B and C are the same (120º). Thus, vector addition by either the tail‐to‐head method (see Section 1.6) or the component method (see Section 1.8) will give resultant vectors that have different directions but the same magnitude. The magnitude of the net force is the smallest at these two corners. SOLUTION As pointed out in the answer to Concept Question (a), the magnitude of any +y 2 individual force vector is F = k q / L2 . With FAB 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 y component Σ Fy of the net force are FAB sin 60.0º FAC 60.0 A +x FAB cos 60.0º ( 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 ( cos 60.0° + 1) ( 2 = F 2 ( cos 60.0° + 1) + ( F sin 60.0° ) + ( sin 60.0° ) = k = 8.99 ×10 N ⋅ m / C = 430 N 9 2 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 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: +y FBC 60.0 FBC cos 60.0º B +x FBA cos 60.0º FBA ( 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 = 2k 2 q 2 L 2 = ( ( −2 F cos 60.0° )2 + ( 0 )2 cos 60.0° = 2 8.99 × 109 N ⋅ m 2 / C2 ) = 2 F cos 60.0° (5.0 ×10−6 C ) ( 0.030 m )2 2 cos 60.0° = 250 N As discussed in the answer to Concept Question (c), 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 . These values of 430 and 250 N for the magnitudes of the net forces at corners A and B, respectively, are consistent with our answers to the Concept Questions. ...
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