**tipler - volume - 2**

Resolução dos exercicios do Tipler volume 2

(Parte **1** de 11)

Chapter 21 The Electric Field 1: Discrete Charge Distributions

Conceptual Problems

*1 ••

Similarities: Differences:

The force between charges and masses varies as 1/r2.

There are positive and negative charges but only positive masses.

The force is directly proportional to the product of the charges or masses.

Like charges repel; like masses attract.

The gravitational constant G is many orders of magnitude smaller than the Coulomb constant k.

2 • Determine the Concept No. In order to charge a body by induction, it must have charges that are free to move about on the body. An insulator does not have such charges.

3 •• Determine the Concept During this sequence of events, negative charges are attracted from ground to the rectangular metal plate B. When S is opened, these charges are trapped on B and remain there when the charged body is removed. Hence B is negatively charged and correct. is )(c

4 •• (a) Connect the metal sphere to ground; bring the insulating rod near the metal sphere and disconnect the sphere from ground; then remove the insulating rod. The sphere will be negatively charged.

(b) Bring the insulating rod in contact with the metal sphere; some of the positive charge on the rod will be transferred to the metal sphere.

(c) Yes. First charge one metal sphere negatively by induction as in (a). Then use that negatively charged sphere to charge the second metal sphere positively by induction.

Chapter 21 2

*5 •• Determine the Concept Because the spheres are conductors, there are free electrons on them that will reposition themselves when the positively charged rod is brought nearby.

(a) On the sphere near the positively charged rod, the induced charge is negative and near the rod. On the other sphere, the net charge is positive and on the side far from the rod. This is shown in the diagram.

(b) When the spheres are separated and far apart and the rod has been removed, the induced charges are distributed uniformly over each sphere. The charge distributions are shown in the diagram.

6 • Determine the Concept The forces acting on +q are shown in the diagram. The force acting on +q due to −Q is along the line joining them and directed toward −Q. The force acting on +q due to +Q is along the line joining them and directed away from

Because charges +Q and −Q are equal in magnitude, the forces due to these charges are equal and their sum (the net force on +q) will be to the right and socorrect. is )(e Note that the vertical components of these forces add up to zero.

*7 • Determine the Concept The acceleration of the positive charge is given by

.0 EFa r r mqm ==Because q0 and m are both positive, the acceleration is in the same direction as the electric field. correct. is )(d

Determine the Concept Er is zero wherever the net force acting on a test charge is zero. At the center of the square the two positive charges alone would produce a net electric field of zero, and the two negative charges alone would also produce a net electric field of zero. Thus, the net force acting on a test charge at the midpoint of the

The Electric Field 1: Discrete Charge Distributions 3 square will be zero. correct. is )(b

9 •• (a) The zero net force acting on Q could be the consequence of equal collinear charges being equidistant from and on opposite sides of Q.

(b) The charges described in (a) could be either positive or negative and the net force on Q would still be zero.

(c) Suppose Q is positive. Imagine a negative charge situated to its right and a larger positive charge on the same line and the right of the negative charge. Such an arrangement of charges, with the distances properly chosen, would result in a net force of zero acting on Q.

(d) Because none of the above are correct, correct. is )(d

10 • Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the sketch to the right we’ve assigned 2 field lines to each charge q.

*1 • Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right we’ve assigned 2 field lines to each charge q.

Chapter 21 4

*12 • Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right we’ve assigned 7 field lines to each charge q.

13 • Determine the Concept A positive charge will induce a charge of the opposite sign on the near surface of the nearby neutral conductor. The positive charge and the induced charge on the neutral conductor, being of opposite sign, will always attract one another.

correct. is )(a

*14 • Determine the Concept Electric field lines around an electric dipole originate at the positive charge and terminate at the negative charge. Only the lines shown in (d) satisfy this requirement. correct. is )(d

*15 •• Determine the Concept Because θ ≠ 0, a dipole in a uniform electric field will experience a restoring torque whose magnitude isθsinxpE. Hence it will oscillate about its equilibrium orientation, θ = 0. If θ << 1, sinθ ≈ θ, and the motion will be simple harmonic motion. Because the field is nonuniform and is larger in the x direction, the force acting on the positive charge of the dipole (in the direction of increasing x) will be greater than the force acting on the negative charge of the dipole (in the direction of decreasing x) and thus there will be a net electric force on the dipole in the direction of increasing x. Hence, the dipole will accelerate in the x direction as it oscillates about θ = 0.

(c) False. Electric field lines diverge from any point in space occupied by a positive charge.

The Electric Field 1: Discrete Charge Distributions 5

(e) True

17 •• Determine the Concept The diagram shows the metal balls before they are placed in the water. In this situation, the net electric field at the location of the sphere on the left is due only to the charge –q on the sphere on the right. If the metal balls are placed in water, the water molecules around each ball tend to align themselves with the electric field. This is shown for the ball on the right with charge –q.

(a) The net electric field | r |

E net that produces a force on the ball on the left is the

field | r |

E due to the charge –q on the ball on the right plus the field due to the layer of positive charge that surrounds the ball on the right. This layer of positive charge is due to the aligning of the water molecules in the electric field, and the amount of positive charge in the layer surrounding the ball on the left will be less than +q. Thus, Enet < E. Because Enet < E, the force on the ball on the left is less than it would be if the balls had not been placed in water. Hence, the force will decrease when the balls are placed in the water.

(b) When a third uncharged metal ball is placed between the first two, the net electric field at the location of the sphere on the right is the field due to the charge +q on the sphere on the left, plus the field due to the charge –Q and +Q on the sphere in the middle. This electric field is directed to the right.

The field due to the charge –Q and +Q on the sphere in the middle at the location of the sphere on the right is to the right. It follows that the net electric field due to the charge +q on the sphere on the left, plus the field due to the charge –Q and +Q on the sphere in the middle is to the right and has a greater magnitude than the field due only to the charge

+q on the sphere on the left. Hence, the force on either sphere will increaseif a third uncharged metal ball is placed between them.

Remarks: The reduction of an electric field by the alignment of dipole moments with the field is discussed in further detail in Chapter 24.

Chapter 21 6

*18 •• Determine the Concept Yes. A positively charged ball will induce a dipole on the metal ball, and if the two are in close proximity, the net force can be attractive.

*19 •• Determine the Concept Assume that the wand has a negative charge. When the charged wand is brought near the tinfoil, the side nearer the wand becomes positively charged by induction, and so it swings toward the wand. When it touches the wand, some of the negative charge is transferred to the foil, which, as a result, acquires a net negative charge and is now repelled by the wand.

Estimation and Approximation

(Parte **1** de 11)