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Physics 122 Solutions to Chapter 25 25.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So, electrons have been removed from the glass rod to make it positively charged. (b) Because each electron has a charge of 1.60 1019 C , the number of electrons removed is 5 109 C 3.13 1010 1.60 10 19 C 25.6. Model: Use the charge model. Solve: (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 electric force and an upward frictional force balancing the very small weight of the balloon. (b) 25.11. Model: Model the charged masses as point charges. Visualize: Solve: (a) The charge q1 exerts a force F1 on 2 on q2 to the right, and the charge q2 exerts a force F2 on 1 on q1 to the left. Using Coulomb’s law, F1 on 2 F2 on 1 K q1 q2 2 12 r 9.0 10 9 N m2 /C2 1.0 C 1.0 C 1.0 m 2 9.0 109 N (b) Newton’s second law on either q1 or q2 is F1 on 2 m1a1 a1 9.0 109 N 9.0 109 m/s2 1.0 kg Assess: Coulomb is a pretty “big” unit. That is why F1 on 2 or F2 on 1 is such a large force. 25.20. Model: The gravitational field of an object depends on its mass and extends through all of space. Solve: (a) The gravitational field strength due to a planet at the radius of its satellite’s orbit is 12 N/kg. That is, G mplanet gorbit , toward planet 12 N/kg, toward planet 2 rorbit When the radius of the orbit is doubled, G m 1Gm planet planet gnew orbit , toward planet , toward planet 3 N/kg, toward planet 2 2 2rorbit 4 rorbit (b) When the planet’s density is doubled, then mnew newV 2V 2mplanet. Thus, assuming that V remains the same, G2mplanet gorbit , toward planet 24 N/kg, toward planet 2 rorbit (c) gorbit does not depend on the satellite’s mass. Thus, gorbit 12 N/kg, toward planet . 25.26. Model: A field is the agent that exerts an electric force on a charge. Visualize: Solve: (a) To balance the weight of a proton Fnet y Fon p w 0 N. This means Fon p w q E mg E 1.67 1027 kg 9.8 N/kg mg 1.02 107 N/C q 1.60 1019 C Because Fon p must be upward and the proton charge is positive, the electric field at the location of the proton must also be pointing upward. Thus E 1.02 10 7 N/C, downward . (b) In the case of the electron, 9.11 1031 kg 9.8 N/kg mg E 5.58 1011 N/C q 1.60 1019 C Because Fon e must be upward and the electron has a negative charge, the electric field at the location of the electron must be pointing downward. Thus E 5.58 10 11 N/C, downward . 25.54. Model: The charges are point charges. Visualize: We must first identify the region of space where the third charge q3 is located. You can see from the figure that the forces can’t possibly add to zero if q3 is above or below the axis or outside the charges. However, at some point on the x-axis between the two charges the forces from the two charges will be oppositely directed. Solve: The mathematical problem is to find the position for which the forces F1 on 3 and F2 on 3 are equal in magnitude. If q3 is the distance x from q1, it is the distance L x from q2. The magnitudes of the forces are F1 on 3 K q1 q3 2 13 r Kq q3 x F2 on 3 2 K q2 q3 2 23 r K 4q q3 L x 2 Equating the two forces, Kq q3 x 2 K 4q q3 L x 2 L x 4x 2 x 2 L and L 3 The solution x L is not allowed as you can see from the figure. To find the magnitude of the charge q3, we apply the equilibrium condition to charge q1: F2 on 1 F3 on 1 K q2 q1 L2 K q3 q1 L 1 3 2 4q 9 q3 q3 4 q 9 We are now able to check the static equilibrium condition for the charge 4q (or q2): F1 on 2 F3 on 2 K q1 q2 2 L K q3 q2 L x 2 q L2 4 9 q L 2 3 2 q L2 The sign of the third charge q3 must be negative. A positive sign on q3 will not have a net force of zero either on the charge q or the charge 4q. In summary, a charge of 94 q placed x 13 L from the charge q will cause the 3charge system to be in static equilibrium. 25.58. Model: The charged plastic beads are point charges and the spring is an ideal spring that obeys Hooke’s law. Solve: Let q be the charge on each plastic bead. The repulsive force between the beads pushes the beads apart. The spring is stretched until the restoring spring force on either bead is equal to the repulsive Coulomb force. That is, k x r 2 K Kq2 k x q r2 The spring constant k is obtained by noting that the weight of a 1.0 g mass stretches the spring 1.0 cm. Thus mg k(1.0 102 m) k q 1.0 10 3 1.0 10 2 m 0.98 N/m 4.5 102 m 4.0 102 m 4.5 102 m 9.0 109 N m2 /C2 kg 9.8 N/kg 0.98 N/m 2 33.2 nC 25.59. Model: The charged spheres are point charges. Visualize: Each sphere is in static equilibrium and the string makes an angle with the vertical. The three forces acting on each sphere are the electric force, the weight of the sphere, and the tension force. Solve: In static equilibrium, Newton’s first law is Fnet T w Fe 0 . In component form, Fnet x Tx wx Fe x 0 N T sin 0 N T sin Fnet y Ty wy Fe y 0 N Kq2 0N d2 T cos mg 0 N 0 N Kq2 Kq2 2 2 d 2L sin T cos mg Dividing the two equations, 2 9.0 10 9 N m 2 /C2 100 10 9 C Kq2 sin tan 2 4.59 10 4 4L mg 4 1.0 m 2 5.0 10 3 kg 9.8 N/kg 2 For small-angles, tan sin . With this approximation we obtain sin 0.07714 rad and 4.42. 25.63. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P25.63. Place the 5 nC charge at the origin. Solve: The electric field is 9 2 2 9 1 q 9 10 N m / C 5 10 C E , away from q , away from q 2 2 r 4 0 r 45.0 N m 2 /C , away from q 2 r At each of the three points, 45.0 N m 2 /C 9.0 10 4 N/C cos iˆ sin ˆj E1 , away from q 2 2 2.0 10 2 m 1.0 10 2 m 1 ˆ 2 9.0 10 4 N/C i 5 5 ˆj 4.02 10 4 iˆ 8.05 10 4 ˆj N/C 45.0 N m2 /C 4.5 105iˆ N/C E2 , away from q 2 2 1.0 10 m 45.0 N m2 /C 4.02 104 iˆ 8.05 104 ˆj N/C E3 , away from q 2 2 2.0 10 2 m 1.0 10 2 m 25.66. Model: The electric field is that of three point charges. Visualize: Solve: (a) In the figure, the distances are r1 r3 tan 1 1 cm 3 cm 2 1/ 3 18.43 . Using the equation for the field of a point charge, 2 3.162 cm and the angle is E1 E3 K q1 2 1 r 9.0 10 9 N m2 /C2 1.0 10 9 C 0.03162 m 2 9000 N/C We now use the angle to find the components of the field vectors: N/C E1 E1 cos iˆ E1 sin ˆj 8540iˆ 2840 ˆj E3 E3 cos iˆ E3 sin ˆj 8540iˆ 2840 ˆj N/C E2 is easier since it has only an x-component. Its magnitude is E2 K q2 r22 9.0 10 9 N m2 /C2 1.0 10 9 C 0.0300 m 10,000 N/C E 2 2 E2iˆ 10,000iˆ N/C (b) The electric field is defined in terms of an electric force acting on charge q: E F q . Since forces obey a principle of superposition ( Fnet F1 F2 principle of superposition. ) it follows that the electric field due to several charges also obeys a (c) The net electric field at a point 3 cm to the right of q2 is Enet E1 E2 E3 27,100iˆ N/C . The y-components of E1 and E2 cancel, giving a net field pointing along the x-axis. 25.68. Model: The charged ball attached to the string is a point charge. Visualize: The ball is in static equilibrium in the external electric field when the string makes an angle 20 with the vertical. The three forces acting on the charged ball are the electric force due to the field, the weight of the ball, and the tension force. Solve: In static equilibrium, Newton’s second law for the ball is Fnet T w Fe 0 . In component form, Fnet x Tx 0 N qE 0 N Fnet y Ty mg 0 N 0 N The above two equations simplify to T cos mg T sin qE Dividing both equations, we get tan 5.0 103 kg 9.8 N/kg tan20 qE mg tan q 1.78 107 C 178 nC mg E 100,000 N/C 25.76. Solve: (a) Kinetic energy is K 21 mv2 , so the velocity squared is v2 2K/m. From kinematics, a particle moving through distance x with acceleration a, starting from rest, finishes with v2 2ax. To gain K 2 1018 J of kinetic energy in x 2.0 m requires an acceleration a v2 2K / m K 2.0 10 18 J 1.10 1018 m/s2 2x 2x mx 9.11 10 31 kg 2.0 10 6 m (b) The force that produces this acceleration is F ma 9.11 10 31 kg 1.10 1018 m/s2 1.0 10 12 N (c) The electric field is E F 1.0 1012 N 6.25 106 N/C e 1.6 1019 C (d) The force on an electron due to charge q is F K q e r 2 . To have a breakdown, the force on the electron must be at least 1.0 1012 N. The minimum charge that could cause a breakdown will be the charge that causes exactly a force of 1.0 1012 N: F Kqe r2 1.0 10 0.01 m 1.0 10 12 N r2F N q 6.9 10 8 C 68 nC 9 2 2 19 Ke 9.0 10 N m / C 1.6 10 C 2 12