Chapter 20: Electric Fields and Forces Solutions

Transcription

1 Chapter 0: lectric Fields and Forces Solutions Questions: 5, 9, 13, 14, 0 xercises & Problems: 8, 14, 15, 6, 3, 43, 45, 58 Q0.5: When you take clothes out of the drier right after it stops, the clothes often stick to your hands and arms. Is your body charged? If so, how did it acquire a charge? If not, why does this happen? Q0.5. Reason: The clothes are charged by rubbing in the drier. Your body is neutral but the clothes can polarize you a little and attract the opposite charge in your body. Assess: This is the same way a balloon sticks to the wall after you rub it on your hair. Q0.9: A plastic balloon that has been rubbed with wool will stick to a wall. a) Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from? b) Draw a charge diagram showing how the balloon is held to the wall. Q0.9. Reason: (a) No, we cannot conclude that the wall is charged. Attractive electric forces occur between (i) two opposite charges, or (ii) a charge and a neutral object that is polarized by the charge. Rubbing the balloon does charge the balloon. Since the balloon is rubber, its charge is negative. As the balloon is brought near the wall, the wall becomes polarized. The positive side of the wall is closer to the balloon than the negative side, so there is a net attractive electric force between the wall and the balloon. This causes the balloon to stick to the wall, with a normal force balancing the attractive force and an upward friction force balancing the very small weight of the balloon. (b) Assess: You have probably wondered at birthday parties how the balloons stick. Now you know. Tell everyone at the next party, and explain why they eventually fall off, too.

2 Q0.13: ach part of the figure shows two points near two charges. Compare the electric field strengths 1 and at these two points. Is 1 >, 1 =, or 1 <? Q0.13. Reason: (a) < 1, because the contributions from the two charges cancel at ( = 0), point 1 1 but the field contribution from each charge points to the right at point. (b) > 1, because the contributions from each charge point in the same direction (to the right) at point 1 but point in opposite directions and somewhat cancel at point (the cancellation is not complete since point is closer to the negative charge than the positive one). (c) = 1. The contributions from the different charges are in opposite directions at each of the two points, but in each case the nearest charge to the point makes the stronger field to the left. The contribution from the other charge reduces the field a bit, but there is still a net field to the left in each case. The magnitudes of the field strength are equal. (d) < 1, ( = 0), because the contributions from the two charges cancel at point 1 1 but at point both field contributions have an upward component that doesn t cancel. (e) < 1. At point 1 the contributions are in opposite directions and partially cancel. While point is farther from the charge on the left, both contributions are in the same direction, so the field is stronger there. (f) > 1. At point 1 the contributions are in the same direction (to the right), whereas at point they partially cancel because they are in opposite directions (not to mention that point is farther away from the positive charge). Assess: It is worth spending a few minutes to get comfortable with all these cases. There are various physics software packages that allow you to map the fields around various charge distributions; they would be good to play with also. Q0.14: Iontophoresis is a noninvasive process that transports drugs through the skin without needles. In the photo, the red electrode is positive and the black electrode is negative. The electric field between the electrodes will drive the negatively charged molecules of the anesthetic through the skin. Should the drug be placed at the red or the black electrode? xplain. Q0.14. Reason: Since like charges repel, negatively charged drug molecules should be placed near the negative (black) electrode to be pushed through the skin.

3 Assess: The field lines point from the red electrode (positive) to the black electrode (negative), but negative particles feel a force in the opposite direction from the local field lines. Q0.0: A typical commercial airplane is struck by lightning about once per year. When this happens, the external metal skin of the airplane might be burned, but the people and equipment inside the aircraft experience no ill effects. xplain why this is so. Q0.0. Reason: When lightning strikes, there is a tremendous transfer of charge from the cloud to the object struck. When lightning strikes the metal plane, this charge is distributed over the surface of the plane. There will initially be movement of charges over the plane (over a very short time interval) but a situation of static equilibrium will quickly be established. Since we have established that there is no electric field inside a conductor, the passengers will experience no change. Assess: In like manner, if you are inside a car struck by lighting, you will be safe. The tires will most likely be damaged as this tremendous amount of charge moves through them to ground. P0.8: If two identical conducting spheres are in contact, any excess charge will be evenly distributed between the two. Three identical metal spheres are labeled A, B, and C. Initially, A has charge q, B has charge q/, and C is uncharged. a) What is the final charge on each sphere if C is touched to B, removed, and then touched to A? b) Starting again from the initial conditions, what is the charge on each sphere if C is touched to A, removed, and then touched to B? P0.8. Prepare: When two identical conducting spheres are in contact the charge is evenly distributed between them. Solve: (a) The following grid shows the initial charge on each of the spheres and the charge after each event. (b) The following grid shows the initial charge on each of the spheres and the charge after each event. Assess: The end result depends on the order in which the various events occur.

4 P0.14: What are the magnitude and direction of the electric force on Charge A in the figure? P0.14. Prepare: Please refer to Figure P0.14. Charges A, B, and C are point F charges. Charge A experiences an electric force B on A F C on A F due to charge C. The force B on A to the left. Solve: Coulomb s law yields: F F (9 10 N m /C )( C)( C) B on A = = ( m) (9 10 N m /C )( C)( C) = = (.0 10 m) C on A The net force on A is due to charge B and an electric force F is directed to the right, and the force C on A N N F = F + F = + x + x = 5 5 on A B on A C on A ( N, -direction) ( N, -direction) 0 N is directed Assess: The force on A by C is the same (but in the opposite direction) as that of B on C because C has four times the charge and is twice the distance away compared to B. Check this statement against Coulomb s law! P0.15: In the figure, charge q experiences no net electric force. What is q1? P0.15. Prepare: Please refer to Figure P0.15. The charged particles are point charges. The charge q is in static equilibrium, so the net force on q is zero. If q is positive, q1 will have to be positive to make the net force zero on q. And, if q is negative, q1 will still have to be positive for q to be in equilibrium. We will assume that the charge q is positive. For this situation, the force on q by the nc charge is to the left and by q1 is to the right. Solve: We have 9 1 q1 q 1 ( 10 C) q Fnet on q = F q1 on q + F nc on q =, + x-direction, x-direction 0 N/C + = 4 πε0 (0. m) 4 πε0 (0.10 m) Thus, q1 (0. m) (0.10 m) 9 10 C = 0 N/C q1 = 8.0 nc Assess: If the charge q is assumed negative, the force on q by the nc charge is to the right and by q1 is to the left. The magnitude of q1 remains unchanged. P0.6: What are the strength and direction of an electric field that will balance the weight of a 1.0 g plastic sphere that has been charged to 3.0 nc?

5 P0.6. Prepare: A field is the agent that exerts an electric force on a charge. Because the weight of the plastic ball acts downward, the electric force must act upward. Σ ( F ) = F w. Solve: Newton s second law on the plastic ball is net y on with the electric force, 3 mg ( kg)(9.8 N/kg) 6 on q = = = = = N/C 9 F w q mg q C q To balance the weight Because F on q must be upward and the charge is negative, the electric field at the location of the plastic ball must be pointing downward. Thus F = q 6 = ( N/C, downward). Assess: means the sign of the charge q determines the direction of F or positive q, and F are pointing in the same direction. But and F point in opposite directions when q is negative.. For P0.3: Storm clouds may build up large negative charges near their bottom edges. The earth is a good conductor, so the charge on the cloud attracts an equal and opposite charge on the earth under the cloud. The -field strength near the earth depends on the shape of the earth's surface, as we can explain with a simple model. The top metal plate has uniformly distributed negative charge. The bottom metal plate, which has a high point, has an equal and opposite charge that is free to move. a) Sketch the two plates and the region between them, showing the distribution of positive charge on the bottom plate. b) Complete your diagram by sketching -field lines between the two plates. Be sure to note the direction of the field. Where is the field strongest? c) xplain why it is more dangerous to be on top of a hill or mountain during a lightning storm than on level ground. P0.3. Prepare: The high part of the bottom conductor is closest to the upper conductor, so the positive charges like to congregate there. Solve: (c) We see from the figures that the field is strongest at the high point of the bottom conductor (the earth); this is where the air is mostly likely to break down and become

6 conducting and allow the opposite charges to rush toward each other (we call this lightning). Assess: This is why we are told to stay away from high and tall things in a lightning storm. Stay away from tall things and crouch down low and hug your knees if you must remain outside. Inside metal cars is usually a safe place to be. P0.43: Objects A and B are both positively charged. Both have a mass of 100 g, but A has twice the charge of B. When A and B are placed with 10 cm between their centers, B experiences an electric force of 0.45 N. a) How large is the force on A? b) What are the charges q A and qb? P0.43. Prepare: Positively charged objects A and B are point charges. The repulsive force by A on B and by B on A are equal and given by Coulomb s law. Solve: (a) It is given that F A on B = 0.45 N. By Newton s third law, FB on A = FA on B = 0.45 N. (b) Coulomb s law is F Kq q K( q )( q ) = F = 0.45 N = = r r A B B B B on A A on B (0.45 N) r (0.45 N)(10 10 m) 7 6 qb = = = C q 9 A= qb= C K ( N m /C ) Assess: A relatively large force of 0.45 N between charges separated by 10 cm must mean significant charge on the objects. P0.45: What are the strength and direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above or below the horizontal?

7 P0.45. Prepare: The electric field is that of the two 1 nc charges located on the y- axis. Please refer to Figure P0.45. We denote the top 1 nc charge by q1 and the bottom ( and ) 1 nc charge by q. The electric fields 1 of both the positive charges are directed away from their respective charges. With vector addition, they yield the net electric field net at the point P indicated by the dot. Solve: The electric fields from q 1 and q are 9 9 q1 ( N m /C )(1 10 C) 1 = K, along + x-axis, along x-axis = + r1 (0.05 m) = (3600 N/C, along + x-axis) 1 q =, θ above + x -axis = (70 N/C, θ above + x -axis) 4πε 0 r 1 tan 10 cm/5 cm, tan () θ = θ = = Because We will now calculate the components of these electric fields. The electric field due to q 1 is away from q1 along + x and that due to q is away from q in the first quadrant. Their components are 1 x 1 1 y = = 0 = cos63.45 x = sin63.45 y The x and y components of the net electric field are: ( ) = + = + cos63.45 = 39 N/C net x 1x x 1 ( ) = + = 0 + sin63.45 = 644 N/C net y 1y y Thus, the strength of the electric field at P is net = (39 N/C) + (644 N/C) = 3975 N/C which will be reported as 4000 N/C. To find the angle this net vector makes with the x-axis, we calculate 644 N/C tanφ = 39 N/C φ = 9.3 Assess: Because of the inverse square dependence on distance, <. 1 Additionally, because the point P has no special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis. P0.58: A 0.10 g honeybee acquires a charge of +3 pc while flying. a) The electric field near the surface of the earth is typically 100 N/C, directed downward. What is the ratio of the electric force on the bee to the bee s weight? b) What electric field strength and direction would allow the bee to hang suspended in the air? P0.58. Prepare: We have sufficient information to determine the electric force on the charged bee and the weight of the bee. Knowing these two forces we can determine their ratio. When these two forces are equal in magnitude and opposite in direction, the bee will hang suspended in air. Solve: (a) The ratio of the electric force to the weight is determined by:

8 F q = =.3 10 w mg F (b) The bee will hang suspended when the electric force is equal to the weight: q = mg. = mg q = 7 / N/C. This gives an electric field of Assess: Table 0. informs us that an electric field of 10 6 N/C will create a spark in air. Note that the required electric field is greater than the air breakdown electric field. As a result we don t expect the bees to just hang in the air due to the charge acquired while flying; they will have provide some of the lift. 6 = w or