PHY 309 L. Solutions for Problem set # 1. Question Q.4 at the end of chapter 12: (a) When a piece of glass is rubbed with a nylon cloth, some electrons move from the glass to the nylon. Since the electrons carry negative charges, this means that the nylon becomes negatively charged while the glass becomes positively charged. Subsequently, when the glass rod touches the first pith ball, the rod recovers some of its missing electrons by stealing them from the ball. In this way, some of the rod s positive electric charge gets transferred to the ball, so the first pith ball becomes positively charged. Likewise, when the nylon cloth touches the second pith ball, some of the nylon s extra electrons move to the ball. Thus, some of the nylon s negative charge is transferred to the ball, so the second ball becomes negatively charged. (b) The two pithballs have opposite electric charges positive for the first ball and negative for the second so they attract each other. Question Q.14 at the end of chapter 12: Initially, the pith ball has no net electric charge. Although the molecules of cellulose, lignin, etc., in the pith naturally contain both positively and negatively charged atoms, their charges cancel each other and the net charge is zero. Moreover, both types of charge are distributed uniformly throughout the pith, so every little piece of the ball is neutral and not just the whole ball. But when a charged rod is brought near the ball, the pith becomes polarized. Suppose the rod is positively charged as in figure 12.9 on page 242 of the textbook. This charge attract the negatively charged atoms in the pith so they move a tiny distance closer to the rod. On the other hand, the positively charged atoms in the pith are repelled by the rod s charge so they move a tiny distance away from the rod. Consequently, the two types of charges cancel each other only in the middle of the pith ball but not on its surface! On the side of the ball facing the rod the surface becomes negatively charged because the negatively charged atoms move in while the positively charged atoms move out. Similarly, on the other 1
side of the ball the surface becomes positively charged because the positive atoms move in while the negative atoms move out. Altogether, the net charge of the pith ball remains zero, but the side of the ball facing the rod acquires a negative charge while the opposite side gets an opposite positive charge. Consequently, the side of the ball facing the positively charged rod is attracted to it while the other side of the ball is repelled by the rod. But since the ball s side attracted to the rod is closer to it than the repelled side, the attractive force is stronger than the repulsive! Thus, the net force on the neutral pith ball is towards the charged rod. Now suppose the rod is negatively charged. In this case, the polarization of the pith works in the opposite direction: the positively charged atoms move a tiny distance towards the rod while the negatively charged atoms move a tiny distance away from it. Consequently, the surface of the pith ball facing the rod becomes positively charged which is attracted to the negatively charged rod while the surface on the other side of the ball becomes negatively charged and is repelled by the rod. The two charges are equal and opposite so the net charge of the ball is zero, but the forces between the two sides of the ball and the rod do not cancel. Instead, because of un-equal distances from the rod, the attractive force on the near side of the ball is stronger than the repulsive force on the far side, so the net force of the neutral pith ball is towards the charged rod, regardless of rod s charge being positive or negative! Thus, until the rod touches the pith ball, they attract each other because of the ball s polarization. But once they touch, some of the rod s charge is transferred to the ball. After that, the ball and the rod have like-sign charges both positive or both negative and they repel each other. Once the ball picks up enough net charge, this repulsive force becomes stronger than the attraction due to polarization, so the ball suddenly flies away from the rod. 2
Question Q.23 at the end of chapter 12: C A B D On this diagram, the red arrows are the electric fields of the positive charge +q only, the blue arrows are the electric fields of the negative charge q only, and the purple arrows are the net electric fields due to both charges. Problem E.8 at the end of chapter 12: The proton has positive charge +e = +1.602176 10 19 C while the electron has exactly opposite charge e. Like any other pair of opposite-sign electric charges, the electron and the proton attract each other with a Coulomb force F = k( e)(+e) r 2 = ke2 r 2 (1) where the indicates attractive rather than repulsive force, r is the distance between the two particles, and k = 1 4πε 0 = 8.98755 10 9 N m 2 /C 2. (2) In a hydrogen atom, the average distance between the electron and the proton is about r = 0.5 Å = 5 10 11 m. At this distance, the numerical value of the attractive force is F = ke2 r 2 = (9 109 Nm 2 /C 2 )(1.6 10 19 C) 2 (5 10 11 m) 2 9 10 8 N. (3) This may not look like a big force, but please remember that the electron has has a very tiny mass m e 9.1 10 31 kg. So the Coulomb force (3) gives it a tremendous acceleration a e = F/m e 1 10 23 m/s 2, or 10,000 billion billion gees! 3
Problem E.10 at the end of chapter 12: A charge q in electric field E experiences an electrostatic force F = qe. (4) This is a vector equation, so for a positive charge q the force F has the same direction as the electric field E, while for a negative charge q the direction of the force F is opposite from the direction of the field E. The test charge in question q = +4µC is positive, so the electrostatic force acting on it has the same direction as the electric field. Since we know the direction of the force is downward, it follows that the direction of the electric field E is also downward. As to the magnitude E of the electric field, eq. (4) implies F = q E = E = F q, (5) hence E = 12 N 4 10 6 C = 3 106 N/C 3,000,000 V/m. (6) Problem E.12 at the end of chapter 12: The force on an electric charge q due to electric field E is F = qe. (4) This is a vector equation, so for a positive charge q the force F has the same direction as the electric field E, while for a negative charge q the direction of the force F is opposite from the direction of the field E. In this problem, the electric field E points to the right while the charge q is negative. Thus, the direction of the force F is opposite from the direction of E; in other words, the force on the charge is directed to the left. As to the magnitude of this force, F = q E = (3.0 10 6 C) (8.5 10 +4 N/C) = 25.5 10 2 N 0.25 N. (7) 4
Problem SP.1 at the end of chapter 12: Note: I am not sure the authors of this problem paid attention to significant figures, so I am calculating all the answers to 2 significant figures and then rounding them off to just 1 significant figure warranted by the problem data. (a) F AB = kq A q B r 2 AB = (9.0 109 Nm 2 /C 2 )(+0.10 C)(+0.02 C) (2 m) 2 = 4.5 10 6 N 5 10 6 N. (8) (b) F CB = kq C q B r 2 CB = (9.0 109 Nm 2 /C 2 )(+0.04 C)(+0.02 C) (1 m) 2 = 7.2 10 6 N 7 10 6 N. (9) (c) Since all 3 charges are positive, all forces between them are repulsive. Charge A sits left from charge B, so its force F AB pushes charge B to the right. On the other hand, charge C sits right from charge B, so its force F CB pushes charge B to the left. The net force on charge B is therefore F net onb = F AB F CB = 4.5 10 6 N 7.2 10 6 N = 2.7 10 6 N 3 10 6 N. (10) The magnitude of this force is 2.7 3 mega Newtons, and its direction is to the left. (d) The net force on the test charge B is FonB net = q B E(B) (11) where E(B) is the electric field at point B created by all the charges except B itself. Given the net force obtained in part (c), this electric field is E(B) = Fnet onb q B = 2.7 106 N +0.02 C = 1.35 10 8 V/m 1 10 8 V/m. (12) In other words, the electric field created by charges A and C at point B has magnitude 135 100 mega-volt per meter and direction to the left. 5
Note units: 1 V/m (Volt per meter) is the same as 1 N/C (Newton of force per Coulomb of charge). (e) In this part we change the value of the test charge B from q B = +0.02 C to q B = 0.06 C but we don t change the charges A or C or the locations of any charges. Consequently, the electric field E(B) on the test charge B from the other charges A and C remains exactly as in part (d). Therefore, the force on the new test charge is simply FonnewB net = (new q B ) (old E(B)) = ( 0.06 C) ( 1.35 10 8 V/m) = +8.1 10 6 N +8 10 6 N. (13) Note that the direction of this 8 mega Newton force is to the right rather than to the left. The non-textbook problem I is postponed to the next homework. 6