Resnick hallidy krane volume 2 5th edition pdf download

Getting Resnick Halliday physics volume 1 pdf download should be easier now. So, relax and get all the physics books you need here on collegelearners. HRKphysics 5th edition volume 1 pdf free download by Michael Resnick and David Halliday is the most popular introductory physics textbook in the world. I found this book to be a little on the complex end of the fundamental physics forum but it was certainly manageable and I think it is apt at the task of boosting a diligent students confidence level.

I recommend this book to any program advisor who is in the market for a challenging but realistic text. Written for the full year or three term Calculus-based University Physics course for science and engineering majors, the publication of the first edition of Physics in launched the modern era of Physics textbooks. It was a new paradigm at the time and continues to be the dominant model for all texts.

Halliday Resnick Krane physics volume 2 5th edition pdf download is the most realistic option for schools looking to teach a more demanding course. This means X is Carbon, C. The fraction is then 1.

The number of water molecules is E The total positive charge in 0. The mass of protons needed is then 3. Ignoring the mass of the electron why not?

P Assume that the spheres initially have charges q1 and q2. The charges must actually have opposite charge. P a The third charge, q3 , will be between the first two.

Now that we have found the position of the third charge we need to find the magnitude. The negative sign is because the force between the first and second charge must be in the opposite direction to the force between the first and third charge. If the charge 3 is moved toward charge 1 then the force of attraction with charge 1 will increase.

But moving charge 3 closer to charge 1 means moving charge 3 away from charge 2, so the force of attraction between charge 3 and charge 2 will decrease. So charge 3 experiences more attraction to ward the charge that it moves toward, and less attraction to the charge it moves away from.

Sounds unstable to me. They would then repel again. P Displace the charge q a distance y. This means take the derivative, and set it equal to zero. E The answers to a and b are the same! Since the gravitational force is down, the electric force, and consequently the electric field, must be directed up.

This is a dipole, and the field is given by Eq. The same is true for 2 and 8, 3 and 9, on up to 6 and By symmetry we expect the field to point along a line so that three charges are above and three below. That would mean E If both charges are positive then Eq. Point P is on the perpendicular bisector of both dipoles, so we can use Eq.

E Look at Eq. We are ignoring the sign of the charge here. If each surface atom can have at most one excess electron, then the fraction of atoms which are charged is 6. E Imagine switching the positive and negative charges.

The electric field would also need to switch directions. By symmetry, then, the electric field can only point vertically down. E We want to fit the data to Eq. Finding the radius will take a little more work.

We can choose one point, and make that the reference point, and then solve for R. E A sketch of the field looks like this. E Consider a view of the disk on edge. Combining these expressions, mv i 9. E a The average speed between the plates is 1.

The speed with which the electron hits the plate is twice this, or 2. E If each value of q measured by Millikan was a multiple of e, then the difference between any two values of q must also be a multiple of q. The smallest difference would be the smallest multiple, and this multiple might be unity.

The differences are 1. This is a pretty clear indication that the fundamental charge is on the order of 1. If so, the likely number of fundamental charges on each of the drops is shown below in a table arranged like the one in the book: 4 8 12 5 10 14 7 11 16 The total number of charges is 87, while the total charge is Combining, 2yvx t0 2yd 2 6.

The only thing we care about is the angle. E c Equal and opposite, or 5. In short, along the axis the electric field is directed in the same direction as the dipole moment. All that is left is to add the expressions. Assume this charge difference is evenly distributed on the top half of the ring.

We are only interested then in the vertical component as we integrate around the top half of the ring. First we need an octopole which is constructed from a quadrupole. We want to keep things as simple as possible, so the construction steps are 1. The octopole will be two stacked, offset quadrupoles. The horizontal component is given by a variation of the work required to derive Eq.

Since the vertical and horizontal components are equal then E P a Swap all positive and negative charges in the problem and the electric field must reverse direction. But this is the same as flipping the problem over; consequently, the electric field must point parallel to the rod. This means that as we go farther from the point charge we need more and more field lines or fewer and fewer.

This is a real answer, so this means the electron either hits the top plate, or it misses both plates. This is clearly on the upper plate. The two charge are positive, but since we will eventually focus on the area between the charges we must subtract the two field contributions, since they point in opposite directions. The faces, however, have the same size but are organized in pairs with opposite directions. These will cancel, so the total flux is zero in all three cases. E a The flat base is easy enough, since according to Eq.

The last steps are just substituting the area of a circle for the flat side of the hemisphere. E By Eq. The charge inside the die is 8. E If the electric field is uniform then there are no free charges near or inside the net.

The flux through the netting must be equal to, but opposite in sign, from the flux through the opening.

E There is no flux through the sides of the cube. E a There is only a flux through the right and left faces. Note that this is the flux through faces opposite the charge; for faces which touch the charge the electric field is parallel to the surface, so the flux would be zero.

The magnitude of the electric field from the sheet on the right is the same, but it points directly away from the sheet on the right. The magnitude of the electric field from the plate on the right is the same, but it points directly away from the plate on the right. This centripetal force is from the electrostatic attraction with the sphere; so long as the proton is outside the sphere the electric field is equivalent to that of a point charge Q Eq.

E The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. Then 4. Then 5. Then E The electric field must do work on the electron to stop it.

Choose the orientation of the axis so that the infinite line of charge is along the z axis. Now for the q enc part. E a The near field is given by Eq. P a The flux through all surfaces except the right and left faces will be zero.

P The net force on the small sphere is zero; this force is the vector sum of the force of gravity W , the electric force FE , and the tension T. The electric field inside any conductor is zero.

This is no charge on the outside surface. The negative sign indicates that the field is pointing inward toward the axis of the cylinder. The positive sign indicates that the field is pointing outward from the axis of the cylinder. P Subtract Eq. P a Put a spherical Gaussian surface inside the shell centered on the point charge. Keep it centered on the charge. With all the symmetry here it appears as if the shell has no effect; the field just looks like a point charge field.

So the shell does make a difference. So the electric field outside the shell looks like a point charge field originating from the center of the shell regardless of where inside the shell the point charge is placed! The conductor acts as a shield. The force is between the shell and the outside charge. P The repulsive electrostatic forces must exactly balance the attractive gravitational forces. Numerically, 1. P If a point is an equilibrium point then the electric field at that point should be zero.

If it is a stable point then moving the test charge assumed positive a small distance from the equilibrium point should result in a restoring force directed back toward the equilibrium point. In other words, there will be a point where the electric field is zero, and around this point there will be an electric field pointing inward.

But there should be nothing inside the surface except an empty point P , so we have a contradiction. P a Follow the example on Page Then the potential in the vicinity of a point charge will be given by Eq.

The total electric potential energy is the sum of these three terms, or zero. E There are six interaction terms, one for every charge pair. Number the charges clockwise from the upper left hand corner. Moving the first part from infinity to the location where we want to construct the electron is easy and takes no work at all. Bringing in the third part requires work against the force of repulsion between the third charge and both of the other two charges.

Potential energy then exists in the form U13 and U23 , where all three charges are the same, and all three separations are the same.

Then s 2 1. The speed of the car is then r s 2K 2 3. Then Q 3. E a The electric field between two parallel plates is uniform and perpendicular to the plates. E a This is an energy conservation problem: 1 2 79 e2 2 79 e 1.

I expect it the difference to go to zero as the two points A and B get closer together. The numerator will go to zero as d gets smaller. The denominator, however, stays finite, which is a good thing. E a Since both charges are positive the electric potential from both charges will be positive.

E The potential is given by Eq. That is between d and e. E The potential on the positive plate is 2 5. E The dotted lines are equipotential lines, the solid arrows are electric field lines. Note that there are twice as many electric field lines from the larger charge! E This can easily be done with a spreadsheet. The following is a sketch; the electric field is the bold curve, the potential is the thin curve.

Then 1 q1 1 That is 0. But if the radius were decreased by a factor of , so would the charge 1. Consequently, smaller metal balls can be raised to higher potentials with less charge. The power required by the drive belt is the product 3. One of these pairs does not change during the process, so it can be ignored when finding the change in potential energy. P a First apply Eq. Then q Pick a point, call it x, y. Equipotential surfaces are also equi-Stanley surfaces.

P a We follow the work done in Section for a uniform line of charge, starting with Eq. P The spheres are small compared to the separation distance.

The potential from the sphere at a distance of This problem is effectively the same as Exercise ; we have a total charge that is divided between two unequal size spheres which are at the same potential on the surface. The extra factor of two is because only half of the decays result in an increase in charge.

P a None. Since the ions have positive charge then the current density is in the same direction as the velocity. Then s 4 0. E The drift velocity is given by Eq. E The resistance of an object with constant cross section is given by Eq. The material is possibly platinum. E Use the equation from Exercise The diameter is 0. We can now find the conductivity, j 1. E The resistance for potential differences less than 1. Then m 9. Then 1. There are two electrons for every alpha particle, and each alpha particle has a charge equal in magnitude to two electrons.

The electrons will move 1 cm in 1. It takes 0. P The resistance when on is 2. P Originally we have a resistance R1 made out of a wire of length l1 and cross sectional area A1. This is possibly silver. Since this term is so much larger than the other two it is the only significant effect. The disks will be effectively in series, so we will add the resistances to get the total.

The current density is given by Eq. The above expression can be inverted to give the electric field as a function of radial distance, since the current is a constant in the above expression. E Eq. The potential energy has changed by E a From Eq.

If there are N capacitors, then the total charge will be N q, and we want this total charge to be 1. C eq This is the equivalent capacitance for the entire arrangement. There are three capaci- tors, so the total charge to pass through the ammeter is 0.

E a The equivalent capacitance is given by Eq. For series capacitors, the charge on the equivalent capacitor is the same as the charge on each of the capacitors. This statement is wrong in the Student Solutions! Then for the 4. E Look back at the solution to Ex.

If C3 breaks down electrically then the circuit is effectively two capacitors in parallel. E Consider any junction other than A or B. Call this junction point 0; label the four nearest junctions to this as points 1, 2, 3, and 4. Similar expressions exist for the other three capacitors. For the junction 0 the net charge must be zero; there is no way for charge to cross the plates of the capacitors. E a The capacitance of an air filled parallel-plate capacitor is given by Eq.

The energy density as a function of radial distance is found from Eq. Try transformer oil. E Capacitance with dielectric media is given by Eq. Mica wins. The minimum plate area is then dC 2. E a The capacitance of a cylindrical capacitor is given by Eq. We can look back to Table to get the dielectric properties of Pyrex. E a Insert the slab so that it is a distance a above the lower plate. To get the slab back out we will need to do work on the slab equal to the energy difference.

To get the slab in we will need to do work on the slab equal to the energy difference. P The capacitance of the cylindrical capacitor is from Eq. P Insert the slab so that it is a distance d above the lower plate. Each interior plate has a charge q on each surface; the exterior plate one pink, one gray has a charge of q on the interior surface only. As far as point e is concerned point a looks like it is originally positively charged, and point d is originally negatively charged.

Note the negative sign reflecting the opposite polarity of C2. P a If terminal a is more positive than terminal b then current can flow that will charge the capacitor on the left, the current can flow through the diode on the top, and the current can charge the capacitor on the right.

Current will not flow through the diode on the left. The capacitors are effectively in series. Since the capacitors are identical and series capacitors have the same charge, we expect the capacitors to have the same potential difference across them. But the total potential difference across both capacitors is equal to V, so the potential difference across either capacitor is 50 V. The output pins are connected to the capacitor on the right, so the potential difference across the output is 50 V.

If we assume the diode is resistanceless in this configuration then the potential difference across it will be zero. The net result is that the potential difference across the output pins is 0 V. In real life the potential difference across the diode would not be zero, even if forward biased. It will be somewhere around 0. Suddenly the problem is very easy.

Current had to flow through the connecting wires to get the charge from one capacitor to the other. The slabs are I assume the same thickness. The capacitance of one of the slabs is then given by Eq. There would be a similar expression for the other slab. The equivalent series capacitance would be given by Eq. Find the equivalent capacitance of the series combination on the right, and then add on the parallel part on the left.

All have an area A. E a E Go all of the way around the circuit. It is a simple one loop circuit, and although it does not matter which way we go around, we will follow the direction of the larger emf. E 5. Then an additional potential difference of The total resistance of the circuit is 1. The internal resistance of the battery is then 0. E a Define the current i1 as moving to the left through r1 and the current i2 as moving to the left through r2.

Rearrange and solve for i, p 2. The battery provides an emf of 2. It is a consequence of the fact that power is related to the current squared, and with any quadratics we expect two solutions. Both are possible, but it might be that only one is stable, or even that neither is stable, and a small perturbation to the friction involved in turning the motor will cause the system to break down.

E In parallel connections of two resistors the effective resistance is less than the smaller resistance but larger than half the smaller resistance. In series connections of two resistors the effective resistance is greater than the larger resistance but less than twice the larger resistance. Since the effective resistance of the parallel combination is less than either single resistance and the effective resistance of the series combinations is larger than either single resistance we can conclude that 3.

The resistors are then 4. E Points B and C are effectively the same point! E Focus on the loop through the battery, the 3. The loop rule yields The potential difference across the 5.

That is a maximum of 3. Solving for R2 yields 2. If the resistors are arranged in a square array consisting of n parallel branches of n series resistors, then the effective resistance is R. Each will dissipate a power P , together they will dissipate n2 P. So we want nine resistors, since four would be too small. E a Work through the circuit one step at a time. This is also the current through R1 , since all the current through the battery must also go through R1.

The potential difference across each of the three remaining resistors is 6. We then combine this branch with the resistor which connects points F and H. The resistance of the first resistor is 9. E See Exercise The resistance ratio is r1 0.

We can choose any standard resistors we want, and we could use any tolerance, but then we will need to check our results. There are other choices. Keep setting up.. These are only numerical problems where are the other questions and mcqs solution.

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Unknown January 12, at PM. Unknown April 14, at AM. Unknown September 16, at PM. Unknown September 28, at PM. Unknown October 4, at AM. Unknown February 18, at PM. Unknown February 25, at PM. Unknown March 17, at PM. Student Solutions Manual to accompany Physics, 5th edition: Written for the complete year or three term Calculus-based University Physics course for science and engineering majors, the publication of the primary edition of Physics in launched the fashionable era of Physics textbooks.

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