solucionario fisica resnick halliday 5ta ed vol 2 (2)

solucionario fisica resnick halliday 5ta ed vol 2 (2)

(Parte 1 de 8)

Instructor Solutions Manual for

Physics by Halliday, Resnick, and Krane

Paul Stanley Beloit College

A Note To The Instructor...

The solutions here are somewhat brief, as they are designed for the instructor, not for the student.

Check with the publishers before electronically posting any part of these solutions; website, ftp, or server access must be restricted to your students.

I have been somewhat casual about subscripts whenever it is obvious that a problem is one dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution. Although this does not change the validity of the answer, it will sometimes obfuscate the approach if viewed by a novice. There are some traditional formula, such as which are not used in the text. The worked solutions use only material from the text, so there may be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an easier approach existed. But if it was not in the text, I did not use it here.

I also tried to avoid reinventing the wheel. There are some exercises and problems in the text which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer you to the previous solution.

I adopt a different approach for rounding of significant figures than previous authors; in particular, I usually round intermediate answers. As such, some of my answers will differ from those in the back of the book.

Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual with considerably more detail and, when appropriate, include discussion on any physical implications of the answer. These student solutions carefully discuss the steps required for solving problems, point out the relevant equation numbers, or even specify where in the text additional information can be found. When two almost equivalent methods of solution exist, often both are presented. You are encouraged to refer students to the Student’s Solution Manual for these exercises and problems. However, the material from the Student’s Solution Manual must not be copied.

Paul Stanley Beloit College stanley@clunet.edu

E25-2 Use Eq. 25-4:

E25-3 Use Eq. 25-4:

E25-5 (a) Use Eq. 25-4,

We must assess the direction of the force of q3 on q1; it will be directed along the line which connects the two charges, and will be directed away from q3. The diagram below shows the directions.

F 12θ F net

From this diagram we want to find the magnitude of the net force on q1. The cosine law is appropriate here:

E25-6 Originally F0 = CQ20 = 0.088N, where C is a constant. When sphere 3 touches 1 the charge on both becomes Q0/2. When sphere 3 the touches sphere 2 the charge on each becomes

E25-7 The forces on q3 are ~F31 and ~F32. These forces are given by the vector form of Coulomb’s Law, Eq. 25-5,

These two forces are the only forces which act on q3, so in order to have q3 in equilibrium the forces must be equal in magnitude, but opposite in direction. In short,

Note that r31 and r32 both point in the same direction and are both of unit length. We then get q1 = −4q2.

E25-8 The horizontal and vertical contributions from the upper left charge and lower right charge are straightforward to find. The contributions from the upper left charge require slightly more work.

charge will contribute

(a) The horizontal component of the net force is then

E25-9 The magnitude of the force on the negative charge from each positive charge is

The force from each positive charge is directed along the side of the triangle; but from symmetry only the component along the bisector is of interest. This means that we need to weight the above answer by a factor of 2cos(30◦) = 1.73. The net force is then 24.5N.

E25-1 This problem is similar to Ex. 25-7. There are some additional issues, however. It is easy enough to write expressions for the forces on the third charge

Then

The only way to satisfy the vector nature of the above expression is to have r31 = ±r32; this means that q3 must be collinear with q1 and q2. q3 could be between q1 and q2, or it could be on either side. Let’s resolve this issue now by putting the values for q1 and q2 into the expression:

Since squared quantities are positive, we can only get this to work if r31 = r32, so q3 is not between q1 and q2. We are then left with so that q3 is closer to q1 than it is to q2. Then r32 = r31 + r12 = r31 + 0.618m, and if we take the square root of both sides of the above expression,

E25-12 The magnitude of the magnetic force between any two charges is kq2/a2, where a = 0.153m. The force between each charge is directed along the side of the triangle; but from symmetry only the component along the bisector is of interest. This means that we need to weight the above answer by a factor of 2cos(30◦) = 1.73. The net force on any charge is then 1.73kq2/a2.

The angle between the strings and the plane of the charges is θ, given by

The force of gravity on each ball is directed vertically and the electric force is directed horizontally. The two must then be related by tanθ = FE/FG,

E25-13 On any corner charge there are seven forces; one from each of the other seven charges. The net force will be the sum. Since all eight charges are the same all of the forces will be repulsive. We need to sketch a diagram to show how the charges are labeled.

The magnitude of the force of charge 2 on charge 1 is

where r12 = a, the length of a side. Since both charges are the same we wrote q2. By symmetry we expect that the magnitudes of F12, F13, and F14 will all be the same and they will all be at right angles to each other directed along the edges of the cube. Written in terms of vectors the forces would be

The force from charge 5 is

and is directed along the side diagonal away from charge 5. The distance r15 is also the side diagonal distance, and can be found from

By symmetry we expect that the magnitudes of F15, F16, and F17 will all be the same and they will all be directed along the diagonals of the faces of the cube. In terms of components we would have

The last force is the force from charge 8 on charge 1, and is given by

and is directed along the cube diagonal away from charge 8. The distance r18 is also the cube diagonal distance, and can be found from

We can add the components together. By symmetry we expect the same answer for each components, so we’l just do one. How about i. This component has contributions from charge 2, 6, 7,

or 1

The three components add according to Pythagoras to pick up a final factor of √ 3, so

E25-14 (a) Yes. Changing the sign of y will change the sign of Fy; since this is equivalent to putting the charge q0 on the “other” side, we would expect the force to also push in the “other” direction.

(b) The equation should look Eq. 25-15, except all y’s should be replaced by x’s. Then

(c) Setting the particle a distance d away should give a force with the same magnitude as

(d) Let the distance be d = √ x2 + y2, and then use the fact that Fx/F = cosθ = x/d. Then

Fx = F x xq0 q

and

(Parte 1 de 8)

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