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# Fundamental of physics solution manual pdf

Request PDF on ResearchGate | Fundamentals of Physics, Student's Solutions Manual | No other book on the market today can match the success of Halliday. Chapter 1 – Student Solutions Manual 3. Using the given conversion factors, we find (a) the distance d in rods to be d. Get instant access to our step-by-step Fundamentals Of Physics solutions manual . better than downloaded Fundamentals of Physics PDF solution manuals?. Author: LAKEESHA MACIVER Language: English, Spanish, Portuguese Country: South Sudan Genre: Religion Pages: 618 Published (Last): 10.11.2015 ISBN: 578-3-16376-654-2 ePub File Size: 29.53 MB PDF File Size: 12.67 MB Distribution: Free* [*Regsitration Required] Downloads: 42111 Uploaded by: FELICA

Fundamentals of Physics 7th Edition: Instructor's Manual. Pages·· FUNDAMENTALS OF PHYSICS 10TH EDITION SOLUTION MANUAL. Instructor's Solution Manual for Fundamentals of Physics, 6/E by Halliday, Resnick, and Wa. Fundamentals of Physics 7th Edition: Instructor's Manual. Pages·· Strength Hibbeler Solution manual, 8th edition- caite.info Authors: Halliday, Resnick, Walker. Find your textbook below for step-by-step solutions to every problem. Fundamentals of Physics, 9th Edition.

This is in the visible range. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. Applying Newton s second law to the x and y directions of both blocks A and B, we arrive at four equations: Since the force and acceleration are constant, we use the equations from Table This is then the solution:

Therefore, 3. We adopt the positive direction choices used in the textbook so that equations such as Eq. The coordinate origin is at ground level directly below the release point. We note that the initial speed of the projectile is the plane s speed at the moment of release.

The coordinate origin is at ground level directly below impact point between bat and ball. The Hint given in the problem is important, since it provides us with enough information to find v0 directly from Eq. The coordinate origin is at the point where the ball is kicked. Writing the kinematic equations for projectile motion: A little manipulation, however, will give an algebraic solution: If kicked at any angle between these two, the ball will travel above the cross bar on the goalposts.

We denote h as the height of a step and w as the width. To hit step n, the ball must fall a distance nh and travel horizontally a distance between n 1 w and nw.

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We take the origin of a coordinate system to be at the point where the ball leaves the top of the stairway, and we choose the y axis to be positive in the upward direction. We equate y to nh and solve for the time to reach the level of step n: Now, this is less than n and greater than n 1, so the ball hits the third step. To calculate the centripetal acceleration of the stone, we need to know its speed during its circular motion this is also its initial speed when it flies off.

Relative to the car the velocity of the snowflakes has a vertical component of 8. Since the horizontal component of velocity in a projectile-motion problem is constant neglecting air friction , we find the original x-component from Since the x and y components of the acceleration are constants, we can use Table for the motion along both axes.

Lengths are in kilometers and times are in hours. Therefore, using Eq. Chapter 5 Student Solutions Manual f f 5. We denote the two forces F1 and F2.

Consequently, the angle is either The scale reads the magnitude of either of these forces. In each case the tension force of the cord attached to the salami must be the same in magnitude as the weight of the salami because the salami is not accelerating. Thus the scale reading is mg, where m is the mass of the salami. Its value is The negative sign indicates the acceleration is down the plane.

The magnitude of the acceleration is 4. The magnitude of the acceleration can be found using constant acceleration kinematics Table The acceleration of the electron is vertical and for all practical purposes the only force acting on it is the electric force. The force of gravity is negligible. Since the force and acceleration are constant, we use the equations from Table The free-body diagram is shown next. FN is the normal force of the plane on the f block and mg is the force of gravity on the block.

Take the positive direction to be upward. According to Newton s third f f law F1on2 has the same magnitude as F2 on1. The free-body diagrams for part a are shown below.

F is the applied force and f f is the force exerted by block 1 on block 2. Since the blocks move together they have the same acceleration and the same symbol is used in both equations. Substituting mpg for F in the equation for the monkey, we solve for am: The forces on the balloon are the force of gravity mg down and the force of the air f Fa up. After the ballast is thrown out, the mass is M m where m is the mass of the ballast and the acceleration is upward. Another way to analyze this is to examine the forces on m3; one of the downward forces on it is T4.

The free-body diagrams is shown on the right. Note that Fm, ry and Fm, rx , respectively, and thought f of as the y and x components of the force Fm, r exerted by the motorcycle on the rider.

## Solutions- Fundamentals of physics, 8th edition

Now, the magnitude of the force exerted on the rider by the motorcycle is the same magnitude of force exerted by the rider on the motorcycle, so the answer is 6. With the acceleration 5. Chapter 6 Student Solutions Manual 1. To bring the bureau into a state of motion, the person should push with any force greater than this value. T is the tension force of the rope on the crate, f f FN is the normal force of the floor on the crate, mg is the f force of gravity, and f is the force of friction.

We assume the crate is motionless. The equations for the x and the y components of the force according to Newton s second law are: We solve for the tension: Thef free-body diagrams for block B and for the knot just above block A are shown next. Applying Newton s second law in the x and y directions for block B and then doing the same for the knot results in four equations: The free-body diagrams for the two blocks are shown next.

T is the magnitude of the f f tension force of the string, FNA is the normal force on block A the leading block , FNB is f f the normal force on block B, f A is kinetic friction force on block A, f B is kinetic friction force on block B. Applying Newton s second law to the x and y directions of both blocks A and B, we arrive at four equations: We take the direction of the boat s motion to be positive.

The free-body diagram for the ball is shown below. Note that the tension in the upper string is greater than the tension in the lower string. It must balance the downward pull of gravity and the force of the lower string. The second equation gives the tension in the lower string: Thus, f We choose inward horizontally towards the center of the circular path as the positive direction. We see that the static friction found in part a is less than this, so the car rolls no skidding and successfully negotiates the curve.

For fmax and fk we use Eq. Replace fs with fk in Fig. This amounts to replacing the static coefficient with the kinetic coefficient in Eq. Now for the second part of the problem, with the block projected uphill the friction direction is reversed from what is shown in Fig. Newton s second law for the uphill motion and Eq. Sample Problem treats the angle of repose the minimum angle necessary for a stationary block to start sliding downhill: Consequently, when the block comes to rest, the incline is not steep enough to cause it to start slipping down the incline again.

Probably the most appropriate picture in the textbook to represent the situation in this problem is in the previous chapter: Chapter 7 Student Solutions Manual 3. The force of the cable is upward and the force of gravity is mg downward. Thus, using Eq. This force is upward, opposite to the force of gravity which has magnitude Mg.

The displacement is downward, so the work done by the cord s force is, using Eq. Since the block starts from rest, we use Eq.

According to the work-kinetic energy theorem, this gives the change in the kinetic energy: The work-kinetic energy theorem is used to solve for xf. The total work is the sum of the work done by gravity on the elevator, the work done by gravity on the counterweight, and the work done by the motor on the system: Since the elevator moves at constant velocity, its kinetic energy does not change and according to the work-kinetic energy theorem the total work done is zero.

We note that 10 min is equivalent to s. Designating the relaxed position as read by that scale as xo we look again at the first picture: A convenient approach is provided by Eq. Note that we have set the applied force equal to the weight in order to maintain constant velocity zero acceleration. This can also be inferred from the answers for parts a and b.

Chapter 8 Student Solutions Manual 5. We neglect any work done by friction. We work with SI units, so the speed is converted: Therefore, the ramp must be about 2. They remain the same if the mass is reduced. Since the problem asks for the speed at the 1 bottom, we write mv 2 for Kf. Since Ki is greater than before, Kf is greater. This means the final speed of the flake is greater. Point C is our reference point for computing gravitational potential energy.

Elastic potential energy of the spring is zero when the spring is relaxed. Information given in the second sentence allows us to compute the spring constant. Thus, Eq. Place the x axis along the path of the block and the y axis normal to the floor. The x and the y component of Newton's second law are x: Thus, f 7. FN There is the same potential energy change in both circumstances, so we can equate the kinetic energy changes as well: We use energy conservation in the form of Eq.

Again, there is some uncertainty in reading the graph which makes the last digit not very significant. At that level 5. Therefore, the particle remains in the region 1.

In our present problem, the height change is equal to the rod length L. Chapter 9 Student Solutions Manual We need to find the coordinates of the point where the shell explodes and the velocity of the fragment that does not fall straight down.

Since one fragment has a velocity of zero after the explosion, the momentum of the other equals the momentum of the shell before the explosion. Let M be the mass of the shell and let V0 be the velocity of the fragment. This information is used in the form of initial conditions for a projectile motion problem to determine where the fragment lands.

Using the linear momentum-impulse theorem stated in Eq. Let m be the mass of the ball and v its speed as it leaves the foot.

No external forces with horizontal components act on the man-stone system and the vertical forces sum to zero, so the total momentum of the system is conserved. Since the man and the stone are initially at rest, the total momentum is zero both before and after the stone is kicked. Let ms be the mass of the stone and vs be its velocity after it is kicked; let mm be the mass of the man and vm be his velocity after he kicks the stone.

We take the axis to be positive in the direction of motion of the stone. The negative sign indicates that the man moves in the direction opposite to the direction of motion of the stone. Our notation is as follows: Let m2 be the mass of the cart that is originally at rest and v2f be its velocity after the collision.

Then, according to Eq. Let m2 be the mass of the body that is originally at rest and v2f be its velocity after the collision. During the majority of this time, no pellet is in contact with the wall, so the average force in part c is much less than the average force in part d.

The impulse on it equals the change in f f f its momentum: Its x and y components have equal magnitudes. Let mF be the mass of the freight car and vF be its initial velocity. Let mC be the mass of the caboose and v be the common final velocity of the two when they are coupled.

We apply the conservation of linear momentum to the x and y axes respectively. These values are given in Appendix C. The numerical result is rcom 7. We solve for the angular velocity at the start of the interval: We use the parallel axis theorem: We use the parallel-axis theorem. Two forces act on the ball, the force of the rod and the force of gravity. No torque about the pivot point is associated with the force of the rod since that force is along the line from the pivot point to the ball.

For the position shown, the torque is counter-clockwise.

We take a torque that tends to cause a counterclockwise rotation from rest to be positive and a torque tending to cause a clockwise rotation tof be negative. Its initial kinetic energy is zero. We assume the sense of initial rotation is positive.

We employ energy methods in this solution; thus, considerations of positive versus negative sense regarding the rotation of the wheel are not relevant. Chapter 11 Student Solutions Manual 5. By Eq. Only the x and z components of the position and velocity f f vectors are nonzero, so Eq. In this case the negative sign indicates that the direction of the torque is opposite the direction of the initial angular momentum, implicitly taken to be positive.

An equally good method for finding W is Eq. Let Ii be the initial rotational inertia of the system and let If be the final rotational inertia. This energy came from the man s store of internal energy. No external torques act on the system consisting of the train and wheel, so the total angular momentum of the system which is initially zero remains zero.

The initial angular momentum is the angular momentum of the falling putty wad. If one of the balls is lowered a distance h, the other is raised the same distance and the sum of the potential energies of the balls does not change. We need consider only the potential energy of the putty wad. Take the lowest point on the path to be the zero of potential energy.

The initial kinetic energy is the sum of that of the balls and wad: Conservation of energy provides the relation: The origin is marked O and may be anywhere.

This result is independent of the location of the origin. This result depends on h, the distance from the origin to one of the lines of motion.

This must be equal to the total kinetic energy gained: The magnitude of the angular momentum of the child about the center of the merry-go-round is given by Eq.

This problem involves the vector cross product of vectors lying in the xy plane.

## Fundamentals of Physics Solutions Manual

In a similar argument to that given in the previous part, we have f b f fg b f fg b g eb2. We make the unconventional choice of clockwise sense as positive, so that the angular acceleration is positive as is the linear acceleration of the center of mass, since we take rightwards as positive.

Contradicting what we assumed in setting up our force equation, f the friction force is found to point rightward with magnitude 4. Chapter 12 Student Solutions Manual f 5. Three forces act on the sphere: Since the sphere is in equilibrium they sum to zero. Then Newton s second law gives vertical component: The following two equations result from setting the sum of forces equal to zero with upwards positive , and the sum of torques about x2 equal to zero: The x axis is along the meter stick, with the origin at the zero position on the scale.

The forces acting on it are shown on the diagram below. Since the meter stick is in equilibrium, the sum of the torques about x2 must vanish: We consider the wheel as it leaves the lower floor. The floor no longer exerts a force on the wheel, and the only forces acting are the force F applied horizontally at the axle, the force of gravity mg acting vertically at the center of the wheel, and the force of the step corner, shown as the two components fh and fv. If the minimum force is applied the wheel does not accelerate, so both the total force and the total torque acting on it are zero.

We calculate the torque around the step corner. The second diagram indicates that the distance from the line of F to the corner is r h, where r is the radius of the wheel and h is the height of the step. The solution for F is 2 6. We examine the box when it is about to tip. Since it will rotate about the lower right edge, that is where the normal force of the floor is exerted. This force is labeled FN on the diagram below. The force of friction is denoted by f, the applied force by F, and the force of gravity by W.

## Instructor’s Solution Manual for Fundamentals of Physics

Note that the force of gravity is applied at the center of the box. When the minimum force is applied the box does not accelerate, so the sum of the horizontal force components vanishes: Here L is the length of a side of the box and the origin was chosen to be at the lower right edge. In this case F is the weight of the object hung on the end: Thus, the shear stress is F mg kg 9. The y components of the forces must sum to zero since the knot is in equilibrium. This means TB cos With the pivot at the hinge, Eq.

When it is about to move, we are still able to apply the equilibrium conditions, but to obtain the critical condition we set static friction equal to its maximum value and picture f the normal force FN as a concentrated force upward at the bottom corner of the cube, f directly below the point O where P is being applied.

Thus, the line of action of FN passes through point O and exerts no torque about O of course, a similar observation applied to the pull P. We choose an axis through the top where the ladder comes into contact with the wall , perpendicular to the plane of the figure and take torques that would cause counterclockwise rotation as positive.

Note that the line of action of the applied force f F intersects the wall at a height of 8. Similarly, the moment arms for the x and y components of f the force at the ground Fg are 8. We solve for r: Values for Me, Ms, and d can be found in Appendix C. The initial kinetic energy is 1 2 mv 2. The particle just escapes if its kinetic energy is zero when it is infinitely far from the asteroid. The final potential and kinetic energies are both zero.

We solve for v: Suppose the particle is a distance h above the surface when it momentarily comes to rest. Write 1 2 mv 2f for the final kinetic energy. We use the principle of conservation of energy. The initial kinetic energy is zero since the stars are at rest. We write Mv2 for the final kinetic energy of the system. Let N be the number of stars in the galaxy, M be the mass of the Sun, and r be the radius of the galaxy.

The force points toward the galactic center. GT 2 M The period is 2. Here 6. From Fig. The radius of the orbit is twice the radius of Earth s orbit: We note that it is proportional to m and inversely proportional to r. Here R is the radius of the satellite and d is the distance from its surface to the center of the meteor. We apply the work-energy theorem to the object in question. It starts from a point at the surface of the Earth with zero initial speed and arrives at the center of the Earth with final speed vf.

T Plugging in for speed v, we arrive at an equation for period T: The pressure increase is the applied force divided by the area: This is equivalent to 1. The air inside pushes outward with a force given by piA, where pi is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by poA, where po is the pressure outside. Suppose h1 is greater than h2.

For this calculation, we take the density to be uniformly 1. Then, p 1. We use the equation of continuity. Let v1 be the speed of the water in the hose and v2 be its speed as it leaves one of the holes. Here A1 is the area of the pipe at the top and v1 is the speed of the water there; A2 is the area of the pipe at the bottom and v2 is the speed of the water there.

Since the tank is large we may neglect the water speed at the top; it is much smaller than the speed at the hole. The water is in free fall and we wish to know how far it has fallen when its speed is doubled to 4. We solve for v. Chapter 15 Student Solutions Manual 3. The surface cannot supply the larger force and the block slips. Here T is the period.

The arguments of the cosine functions are in radians. That is, particle 1 lags particle 2 by onetwelfth a period. We want the coordinates of the particles 0. When the block is at the end of its path and is momentarily stopped, its displacement is equal to the amplitude and all the energy is potential in nature.

We use the answer from part b to do part a , so it is best to look at the solution for part b first. We differentiate with respect to time to find the angular velocity: If the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod and if the torque tends to pull the rod toward its equilibrium orientation, then the rod will oscillate in simple harmonic motion.

This immediately gives the period in terms of given quantities.

## Instructor’s Solution Manual for Fundamentals of Physics - PDF Drive

This is also the equilibrium length of the spring. See Table The result is zero. The pendulum does not oscillate. It is started from rest 5. The linear mass density is the mass per unit length of rope: The plus sign is used since the wave is traveling in the negative x direction. It is about 5.

The solution is either 0. In the second case it has negative slope and does not match the graph. A plus sign appears in the argument of the trigonometric function because the wave is moving in the negative x direction. The phasor diagram is shown below: Suppose the lower frequency is associated with the integer n. The amplitude ym is half the maximum displacement of the standing wave, or 5. Different waves on different cords have the same ratio of speeds if they have the same amplitude and wavelength, regardless of the wave speeds, linear densities of the cords, and the tensions in the cords.

Thus, 5. Chapter 17 Student Solutions Manual 5. Let tf be the time for the stone to fall to the water and ts be the time for the sound of the splash to travel from the water to the top of the well. This is a quadratic equation for d. If d is the distance from the location of the earthquake to the seismograph and vs is the speed of the S waves then the time for these waves to reach the seismograph is ts.

The frequency is the same for air and tissue. Let L1 be the distance from the closer speaker to the listener. This means n ranges from 0 to Thus, the factor is 3. Thus, the factor is 5. Thus, the factor is 2. The intensity is the rate of energy flow per unit area perpendicular to the flow. The rate at which energy flow across every sphere centered at the source is the same, regardless of the sphere radius, and is the same as the power output of the source.

The intensity is increased by a factor of 1. The wavelength is different because the wave speed is different. The results are 2. Thus, there are 3 frequencies. There are no resonant frequencies between so you know that the integers associated with the given frequencies differ by 1.

All speeds are relative to the air. Since the locomotive is moving away from the uncle the frequency decreases and we use the plus sign in the denominator. This tends to increase the frequency and we use the plus sign in the numerator. This tends to decrease the frequency and we use the plus sign in the denominator. Use the plus sign in the denominator. Use the plus sign in the numerator. Use the plus signs in both the numerator and the denominator.

The siren is between you and the cliff, moving away from you and towards the cliff. Since the interference pattern changes from a minimum to the next maximum, this distance must be half a wavelength of the sound. At the minimum, interference is destructive and the displacement amplitude is the difference in the amplitudes of the individual waves: At the maximum, the waves interfere constructively and the displacement amplitude is the sum of the amplitudes of the individual waves: Those losses are greater on path B since it is longer than path A.

Chapter 18 Student Solutions Manual 8. See Eq. Note that the original volume of glycerin is the same as the original volume of the cup. Consider half the bar.

The melting point of silver is K, so the temperature of the silver must first be raised from Now the silver at its melting point must be melted. The total heat required is 2. There are three possibilities: None of the ice melts and the water-ice system reaches thermal equilibrium at a temperature that is at or below the melting point of ice. The system reaches thermal equilibrium at the melting point of ice, with some of the ice melted.

All of the ice melts and the system reaches thermal equilibrium at a temperature at or above the melting point of ice. First, suppose that no ice melts. Since no energy is lost to the environment, these two heats in absolute value must be the same.

That is, the calculation just completed does not take into account the melting of the ice and is in error. Consequently, we start with a new assumption: This is then the solution: All the ice melts and the final temperature is above the melting point of ice.

Over a cycle, the internal energy is the same at the beginning and end, so the heat Q absorbed equals the work done: The CA portion of the cycle is at constant volume, so no work is done.

This means the gas loses 30 J of energy in the form of heat. The thermal conductivity is found in Table of the text. Recall that a change in Kelvin temperature is numerically equivalent to a change on the Celsius scale. Let h be the thickness of the slab and A be its area. Thus, 1. This is similar to Sample Problem As discussed in part a of that sample problem, there are three steps to the total process: And we convert the volume to SI units: Now, according to the ideal gas law, 1. Suppose the gas expands from volume Vi to volume Vf during the isothermal portion of the process.

The final pressure is atmospheric pressure: The gas starts in a state with pressure pf, so this is the pressure throughout this portion of the process. We also note that the volume decreases from Vf to Vi. According to Table , the molar mass of molecular hydrogen is 2.

This value can be used in Eq. Here we illustrate the latter approach, using v for vrms: Thus Evaluate the integral by calculating the area under the curve in Fig. The area of the triangular portion is half the product of the base and altitude, or 12 av0. The area of the rectangular portion is the product of the sides, or av0. Thus the number of particles with speeds in the given range is N a 2. Two formulas other than the first law of thermodynamics will be of use to us.

It is straightforward to show, from Eq. Further, Eq. In this case, the work done consists of that done during the constant pressure part the horizontal line in the graph plus that done during the constant volume part the vertical line: In this solution we will use non-standard notation: This is very similar to Sample Problem and we use similar notation here except for the use of Eq.

The negative value implies that the heat transfer is from the sample to its environment. Chapter 20 Student Solutions Manual 5. Now the internal energy of an ideal gas depends only on the temperature and not on the pressure and volume. The two energies are the same in magnitude since no energy is lost.

Since the temperatures in the equation must be in Kelvins, the temperature in the denominator is converted to the Kelvin scale. Vb and pb are given. We need to find pa. Now pa is the same as pc and points c and b are connected by an adiabatic process.

The coefficient of performance is the energy QL drawn from the cold reservoir as heat divided by the work done: Here QH is the energy ejected to the hot reservoir as heat. There are N! These rearrangements do not produce a new configuration. WB 33! Except for the phase change which just uses Eq. In this case, using Eq. Thus, 3! Thus, 5! Let q1 and q2 be the original charges and choose the coordinate system so the force on q2 is positive if it is repelled by q1. Take the distance between the charges to be r.

After the wire is connected, the spheres, being identical, have the same charge. Since charge is conserved, the total charge is the same as it was originally. Since the spheres are identical, the solutions are essentially the same: Let the charge on the third particle be q0.

Thus the y coordinate of the particle must be zero. Suppose the third particle is a distance x from the particle with charge q, as shown on the diagram to the right. Solve this equation for x.

The force on the particle with charge 4: Solve for the charge: Each force is a force of attraction and is directed toward the cesium ion that exerts it, along the body diagonal of the cube. We can pair every cesium ion with another, diametrically positioned at the opposite corner of the cube.

Since the two ions in such a pair exert forces that have the same magnitude but are oppositely directed, the two forces sum to zero and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero. This neutralizes the ion and, as far as the electrical force on the chlorine ion is concerned, it is equivalent to removing the ion.

The forces of the eight cesium ions at the cube corners sum to zero, so the only force on the chlorine ion is the force of the added charge. The chlorine ion is pulled away from the site of the missing cesium ion. Atomic numbers numbers of protons and numbers of electrons and molar masses combined numbers of protons and neutrons can be found in Appendix F of the text. One of the neutrons is freed in the reaction.

Equate these forces to each other and solve for d. The result is s s 1: Particles 2 and 3 repel each other. The later angle is associated with a vector that has negative x and y components and so is the correct angle. Since electrostatic forces are along the lines that join the particles, particle 3 must be on the x axis. Its y coordinate is zero. Particle 3 is repelled by one of the other charges and attracted by the other.

As a result, particle 3 cannot be between the other two particles and must be either to the left of particle 1 or to the right of particle 2.

Since the magnitude of q1 is greater than the magnitude of q2 , particle 3 must Chapter 21 be closer to particle 2 than to particle 1 and so must be to the right of particle 2. Let x be the coordinate of particle 3. At the center of the square, the electric fields produced by the particles at the lower left and upper right corners are both along the x axis and each points away from the center and toward the particle that producespit.

Ex It is upward in the diagram, from the center of the square toward the center of the upper side. The moments point in opposite directions and produce fields in opposite directions at points on the quadrupole axis.

Consider the point P on the axis, a distance z to the right of the quadrupole center and take a rightward pointing field to be positive. Thus F 3: The force is downward. Such a line has slope 3: Since the field is downward, the flux through the upper face is negative and the flux through the lower face is positive. Neglect fringing. Symmetry can be used to show that the electric field is radial, both between the rod and the shell and outside the shell.

It is zero, of course, inside the rod and inside the shell since they are conductors. The flux through the ends is zero. The negative sign indicates that the field points inward. The charge enclosed by the Gaussian surface is only the charge Q1 on the conducting rod. The electric field is zero at all points on the curved surface and is parallel to the ends, so the total electric flux through the Gaussian surface is zero and the net charge within it is zero.

The charge is distributed uniformly over both sides of the original plate, with half being on the side near the field point. Thus 8: You bet! Just post a question you need help with, and one of our experts will provide a custom solution. You can also find solutions immediately by searching the millions of fully answered study questions in our archive. You can download our homework help app on iOS or Android to access solutions manuals on your mobile device. Asking a study question in a snap - just take a pic.

Textbook Solutions. Get access now with. Get Started. Select your edition Below by. David Halliday. Fundamentals of Physics 10th Edition. How is Chegg Study better than a printed Fundamentals of Physics student solution manual from the bookstore?

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