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Unformatted text preview: Physics 204, morning summer 2008 Recitation ClassWork 2 solutions Chapter 18 Conceptual Questions #1,10,12,14,15 Concepts & calculations #69,72,73 1. REASONING AND SOLUTION In Figure 18.8, the grounding wire is removed first, followed by the rod, and the sphere is left with a positive charge. If the rod were removed first, followed by the grounding wire, the sphere would not be left with a charge. Once the rod is removed, the repulsive force caused by the presence of the rubber rod is no longer present. Since the wire is still attached, free electrons will enter the sphere from the ground until the sphere is once again neutral. 10. REASONING AND SOLUTION From Equation 18.2, we see that the electric field E that exists at a point in space is the force per unit charge, F/q0, experienced by a small test charge q0 placed at that point. When a positive or negative charge is placed at a point where there is an electric field, the charge will experience a force given by F = q 0 E . If the charge q0 is positive, then F and E are in the same direction; that is, a positive charge experiences a force in the direction of the electric field. Similarly, if the charge q0 is negative, then F and E are in the opposite direction; a negative charge experiences a force that is opposite to the direction of the electric field. If two different charges experience forces that differ in both magnitude and direction when they are separately placed at the same point in space, we can conclude that the two charges differ in both magnitude and polarity. 12. REASONING AND SOLUTION The charges at corners 1 and 3 are both +q while the charges at corners 2 and 4 are both –q. Let’s concentrate first on the electric field produced at the center C of the rectangle by the positive charges. The positive charge at corner 1 produces an electric field at C that points toward corner 3. In contrast, the positive charge at corner 3 produces an electric field at C that points toward corner 1. Thus, the two fields have opposite directions. The magnitudes of the fields are identical because the charges have the same magnitude and are equally far from the center. Therefore, the fields from the two positive charges cancel. In a similar manner, the fields due to the negative charges on corners 2 and 4 also cancel. Consequently, the net electric field at the center C due to all four charges is zero. 14. SSM REASONING AND SOLUTION In general, electric field lines are always directed away from positive charges and toward negative charges. At a point in space, the direction of the electric field is tangent to the electric field line that passes through that point. The magnitude of the electric field is strongest in regions where the field lines are closest together. Furthermore, the text states that electric field lines created by a positive point charge are directed radially outward, while those created by a negative point charge are directed radially inward. a. In both drawings I and II the electric field is the same everywhere. This statement is false. By inspection, we see that the field lines in II get closer together as we proceed from left to right in the drawing; hence, the magnitude of the electric field is getting stronger. The direction of the electric field at any point along a field line can be found by constructing a line tangent to the field line. Clearly, the direction of the electric field in II changes as we proceed from left to right in the drawing. b. As you move from left to right in each case, the electric field becomes stronger. This statement is false. While this is true for the field in II by reasons that are discussed in part (a) above, we can see by inspection that the field lines in I remain equally spaced and point in the same direction as we go from left to right. Thus, the field in I remains constant in both magnitude and direction. c. The electric field in I is the same everywhere but becomes stronger in II as you move from left to right. This statement is true for the reasons discussed in parts (a) and (b) above. d. The electric fields in both I and II could be created by negative charges located somewhere on the left and positive charges located somewhere on the right. Since electric field lines are always directed away from positive charges and toward negative charges, this statement is false. e. Both I and II arise from a single point charge located somewhere on the left. Since the electric field lines are radial in the vicinity of a single point charge, this statement is false. 15. REASONING AND SOLUTION The effects of air resistance and gravity are ignored throughout this discussion. As shown in Figure 18.25, the electric field lines between the plates of a parallel plate capacitor are parallel and equally spaced, except near the edges. Ignoring edge effects, then, we can treat the electric field between the plates as a constant electric field. Since the force on the charge is given by Equation 18.2 as F = q0E , we can deduce that the force on the positive charge, and therefore its acceleration, is constant in magnitude and is directed toward the negative plate. Since the charge enters the region with a horizontal velocity and the acceleration is downward, we can conclude that the horizontal component of the velocity will remain constant while the vertical component of the velocity will increase in the downward direction as the particle moves from left to right. This situation is analogous to that of a projectile fired horizontally near the surface of the earth. We can deduce, therefore, that the charged particle will follow a parabolic trajectory. a. Two possible trajectories for the particle are shown in the following drawings. For case 1, the particle hits the bottom plate and never emerges from the capacitor. The point at which the particle hits the bottom plate (when indeed it does) depends on the initial speed v of the particle when it enters the capacitor and the magnitude E of the electric field between the plates. For certain values of v and E, the particle may exit from the right side of the capacitor without striking the bottom plate. This is shown in case 2. Note that after the particle leaves the capacitor, it will move in a straight line at constant speed according to Newton's first law (assuming that no other forces are present). Case 1 + + + + + + + + + + + + + + parabolic path
– – – – – – – – – – – – – – – Case 2 + + + + + + + + + + + + + + + parabolic path – – – – – – – – – – – – – – – straight line path b. The direction of the particle's instantaneous displacement vector and velocity vector are in the instantaneous direction of motion of the particle. Furthermore, the direction of the linear momentum vector is always in the same direction as the particle's velocity, and, therefore, in the instantaneous direction of motion. Since the particle travels in a parabolic path as discussed in part (a), we know that the directions of the particle's displacement vector, velocity vector, and linear momentum vector are always tangent to the path shown in part (a). They are not, therefore, parallel to the electric field inside the capacitor. From Newton's second law, we know that the acceleration vector of the particle is in the direction of the net force that acts on the particle. Since the particle is positively charged, we know that the net force on the particle at any point is in the same direction as the direction of the electric field at that point. Therefore, the acceleration vector is parallel to the electric field in the capacitor. 69. CONCEPT QUESTIONS a. The conservation of electric charge states that, during any process, the net electric charge of an isolated system remains constant (is conserved). Therefore, the net charge (q1 + q2) on the two spheres before they touch is the same as the net charge after they touch. b. When the two identical spheres touch, the net charge will spread out equally over both of them. When the spheres are separated, the charge on each is the same. SOLUTION a. Since the final charge on each sphere is +5.0 C, the final net charge on both spheres is 2(+5.0 C) = +10.0 C. The initial net charge must also be +10.0 C. The only spheres whose net charge is +10.0 C are B (qB = − 2.0 μ C) and D (qD = +12.0 μ C) . b. Since the final charge on each sphere is +3.0 C, the final net charge on the three spheres is 3(+3.0 C) = +9.0 C. The initial net charge must also be +9.0 C. The only spheres whose net charge is +9.0 C are A (qA = − 8.0 μ C), C (qC = +5.0 μ C) and D (qD = +12.0 μ C) . c. Since the final charge on a given sphere in part (b) is +3.0 C, we would have to add −3.0 C to make it electrically neutral. Since the charge on an electron is −1.6 × 10−19 C, the number of electrons that would have to be added is Number of electrons = −3.0 × 10−6 C
= 1.9 × 1013 −19
−1.6 × 10 C 72. CONCEPT QUESTIONS a. The magnitude of the electric field is obtained by dividing the magnitude of the force (obtained from the meter) by the magnitude of the charge. Since the charge is positive, the direction of the electric field is the same as the direction of the force. b. As in part (a), the magnitude of the electric field is obtained by dividing the magnitude of the force by the magnitude of the charge. Since the charge is negative, however, the direction of the force (as indicated by the meter) is opposite to the direction of the electric field. Thus, the direction of the electric field is opposite to that of the force. SOLUTION a. According to Equation 18.2, the magnitude of the electric field is E= F 40.0 μ N
=
= 2.0 N /C q 20.0 μ C As mentioned in the answer to Concept Question (a), the direction of the electric field is the same as the direction of the force, or due east . b. The magnitude of the electric field is E= F 20.0 μ N
=
= 2.0 N /C q 10.0 μ C Since the charge is negative, the direction of the electric field is opposite to the direction of the force, or due east . Thus, the electric fields in parts (a) and (b) are the same. 73. CONCEPT QUESTION Part (a) of the drawing given in the text. The electric field produced by a charge points away from a positive charge and toward a negative charge. Therefore, the electric field E+2 produced by the +2.0 C charge points away from it, and the electric fields E−3 and E−5 produced by the −3.0 C and −5.0 C charges point toward them (see the drawing that follows). The magnitude of the electric field produced by a point charge is given by Equation 18.3 as E = k q /r2. Since the distance from each charge to the origin is the same, the magnitude of the electric field is proportional only to the magnitude q of the charge. Thus, the x component Ex of the net electric field is proportional to 5.0 C (2.0 C + 3.0 C). Since only one of the charges produces an electric field in the y direction, the y component Ey of the net electric field is proportional to the magnitude of this charge, or 5.0 C. Thus, the x and y components are equal, as indicated in the right drawing, where the net electric field E is also shown. −5.0 μ C E−5 Ey
E+2 +2.0 μ C O E −3.0 μ C Ex E−3 Part (b) of the drawing given in the text. Using the same arguments as earlier, we find that the electric fields produced by the four charges are shown in the left drawing. These fields also produce the same net electric field E as before, as indicated in the following drawing. E+6 +1.0 μ C Ey E−1
+4.0 μ C E −1.0 μ C Ex E+4
E+1
+6.0 μ C SOLUTION Part (a) of the drawing given in the text. The net electric field in the x direction is Ex (8.99 × 109 N ⋅ m2 /C2 )( 2.0 × 10−6 C ) + (8.99 × 109 N ⋅ m2 /C2 )(3.0 × 10−6 C )
=
( 0.061 m )2 ( 0.061 m )2 = 1.2 × 10 N /C
7 The net electric field in the y direction is Ey (8.99 × 109 N ⋅ m2 /C2 )(5.0 × 10−6 C ) = 1.2 × 107 N /C =
( 0.061 m )2 The magnitude of the net electric field is 2
2
E = Ex + E y = (1.2 × 107 N /C ) + (1.2 × 107 N /C )
2 2 = 1.7 × 107 N /C Part (b) of the drawing given in the text. The magnitude of the net electric field is the same as determined for part (a); E = 1.7 ×107 N/C . ...
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This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.
 Spring '09
 rollino
 Physics

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