= C. on q 1 to the left. Using Coulomb s law, on q 2 to the right, and the charge q 2 exerts a force F 2 on 1 ( )

Size: px

Start display at page:

Download "= C. on q 1 to the left. Using Coulomb s law, on q 2 to the right, and the charge q 2 exerts a force F 2 on 1 ( )"

Asher Hensley
5 years ago
Views:

Phsics Solutions to Chapter 5 5.. Model: Use the charge model.

1 Phsics Solutions to Chapter Model: Use the charge model. Solve: (a) In the process of charging b rubbing, electrons are removed from one material and transferred to the other because the are relativel free to move. Protons, on the other hand, are tightl bound in nuclei. So, electrons have been removed from the glass rod to make it positivel charged. 9 (b) Because each electron has a charge of.60 0 C, the number of electrons removed is C C 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 b 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 ver small weight of the balloon. (b) Model: Model the charged masses as point charges. Solve: (a) The charge q eerts a force on on q to the right, and the charge q eerts a force on on q to the left. Using Coulomb s law, 9 ( N m /C )(.0 C)(.0 C) (.0 m) K q q N 9 on on r (b) Newton s second law on either q or q is m a a on N m/s.0 kg Assess: Coulomb is a prett big unit. That is wh ( or ) is such a large force. on on 5.0. Model: The gravitational field of an object depends on its mass and etends through all of space. Solve: (a) The gravitational field strength due to a at the radius of its satellite s orbit is N/kg. That is, g G m, toward N/kg, toward orbit rorbit When the radius of the orbit is doubled,

2 g G m G m, toward, toward N/kg, toward new orbit ( rorbit ) rorbit (b) When the s densit is doubled, then m new ρ new V ρv m. Thus, assuming that V remains the same, (c) orbit g Gm, toward ( N/kg, toward ) g N/kg, toward. orbit rorbit g does not depend on the satellite s mass. Thus, 5.6. Model: A field is the agent that eerts an electric force on a charge. orbit Solve: (a) To balance the weight of a proton Σ w net on p 0 N. This means ( kg)( 9.8 N/kg) mg on p w q E mg E 9 q.60 0 C N/C Because on p must be upward and the proton charge is positive, the electric field at the location of the proton 7 E.0 0 N/C, downward. must also be pointing upward. Thus (b) In the case of the electron, mg ( 9. 0 kg)( 9.8 N/kg) E N/C 9 q.60 0 C Because on e must be upward and the electron has a negative charge, the electric field at the location of the E N/C, downward. electron must be pointing downward. Thus 5.5. Model: The charges are point charges. We must first identif the region of space where the third charge q is located. You can see from the figure that the forces can t possibl add to zero if q is above or below the ais or outside the charges. However, at some point on the -ais between the two charges the forces from the two charges will be oppositel directed. Solve: The mathematical problem is to find the position for which the forces on and on are equal in magnitude. If q is the distance from q, it is the distance L from q. The magnitudes of the forces are

3 Equating the two forces, K q q Kq q on r ( ) ( ) ( ) ( ) K q q K q q r L on Kq q K q q L L and L L The solution L is not allowed as ou can see from the figure. To find the magnitude of the charge q, we appl the equilibrium condition to charge q : K q q K q q on on q 9 q q q L 9 ( L) We are now able to check the static equilibrium condition for the charge q (or q ): q q K q q q q q K L L L 9 on on ( L ) ( L) The sign of the third charge q must be negative. A positive sign on q will not have a net force of zero either on the charge q or the charge q. In summar, a charge of q placed 9 L from the charge q will cause the - charge sstem to be in static equilibrium Model: The charged plastic beads are point charges and the spring is an ideal spring that obes 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, Kq r k r K k q The spring constant k is obtained b noting that the weight of a.0 g mass stretches the spring.0 cm. Thus mg k(.0 0 m) ( 0.98 N/m.5 0 m.0 0 m)(.5 0 m) (.0 0 kg)( 9.8 N/kg) k 0.98 N/m.0 0 m q. nc N m /C Model: The charged spheres are point charges. Each sphere is in static equilibrium and the string makes an angle θ 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 net T + w + e 0. In component form,

4 Dividing the two equations, ( ) T + w + ( ) ( ) T w ( ) net e 0 N net + + e 0 N Kq T sinθ + 0 N + 0 N T cosθ mg + 0 N 0 N d Kq Kq T sinθ T cosθ + mg d Lsinθ 9 9 ( N m /C )( 00 0 C) Kq θ θ sin tan.59 0 L mg.0 m kg 9.8 N/kg ( ) or small-angles, tanθ sinθ. With this approimation we obtain sinθ rad and θ Model: The electric field is that of a positive point charge located at the origin. Please refer to igure P5.6. Place the 5 nc charge at the origin. Solve: The electric field is ( N m / C )( C) q E, awa from q, awa from q πε r r N m /C, awa from q r At each of the three points, E q i j (.0 0 m) + (.0 0 m) ˆ ˆ i j ˆ ˆ ( i j) ( θ θ ) 5.0 N m /C, awa from N/C cos ˆ + sin ˆ N/C N/C N m /C 5 E, awa from q.5 0 iˆ N/C (.0 0 m) 5.0 N m /C ˆ E ˆ, awa from q (.0 0 i j) N/C (.0 0 m) + (.0 0 m) Model: The electric field is that of three point charges. Solve: (a) In the figure, the distances are r r tan / 8.. Using the equation for the field of a point charge, cm + cm.6 cm and the angle is θ

5 9 9 ( N m /C )(.0 0 C) K q E E 9000 N/C r ( 0.06 m) We now use the angle θ to find the components of the field vectors: E E cosθ iˆ E sinθ ˆj 850iˆ 80 ˆj N/C E E cosθ iˆ + E sinθ ˆj 850iˆ + 80 ˆj N/C E is easier since it has onl an -component. Its magnitude is 9 9 ( N m /C )(.0 0 C) r ( m) K q E 0,000 N/C E ˆ Eˆ i 0,000 i N/C (b) The electric field is defined in terms of an electric force acting on charge q: E q. Since forces obe a principle of superposition ( net + + ) it follows that the electric field due to several charges also obes a principle of superposition. (c) The net electric field at a point cm to the right of q is E ˆ net E + E + E 7,00 i N/C. The -components of E and E cancel, giving a net field pointing along the -ais Model: The charged ball attached to the string is a point charge. The ball is in static equilibrium in the eternal electric field when the string makes an angle θ 0 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 net T + w + e 0. In component form, The above two equations simplif to Dividing both equations, we get net T 0 N qe 0 N + + T sinθ qe T cosθ mg net T mg + 0 N 0 N qe mg tanθ kg 9.8 N/kg tan0 mg E 00,000 N/C 7 tanθ q.78 0 C 78 nc Solve: (a) Kinetic energ is K mv, so the velocit squared is v K/m. rom kinematics, a particle moving through distance with acceleration a, starting from rest, finishes with v a. To gain 8 K 0 J of kinetic energ in.0 µm requires an acceleration 8 v K / m K.0 0 J a.0 0 m/s m 9. 0 kg.0 0 m 6 8

6 (b) The force that produces this acceleration is ma kg.0 0 m/s.0 0 N (c) The electric field is.0 0 N 6 E N/C e.6 0 C (d) The force on an electron due to charge q is K q e r. To have a breakdown, the force on the electron must be at least.0 0 N. The minimum charge that could cause a breakdown will be the charge that causes eactl a force of.0 0 N: ( 0.0 m) (.0 0 N) ( N m / C )(.6 0 C) K q e r N q C 68 nc 9 9 r Ke

5 10 C C

5 10 C C Chapter solutions Q.. Reason: (a) Yes, the field would be zero at a point on the line between the two charges, closer to the 1 nc charge. (b) In this case the contributions from the two charges are in