Express your answer in spherical coordinates. To first order, the monopole term the potential vanishes, since the total charge is zero. For each of the arrangements below, find i the monopole moment, ii the dipole moment, and iii the approximate potential in spherical coordinates at large r include both the monopole and dipole contributions.
The monopole moment is that total charge, which is again 2q. Find the approximate electric field at points far from the origin. Express your answer in spherical coordinates, and include the two lowest orders in the multipole expansion. Find the potential at the center of the sphere relative to infinity , if its radius is R and its dielectric constant is fr.
We need to find the electric field everywhere first. This polarization generates a field of its own, E1 Ex. Sum the series, and compare your answer with Eq. The field resulting from this polarization is given by equation 4. Find the energy of this configuration. We will use equation 4. Find all the bound charges, and check that they add up to zero. What is the total bound charge on the surface? Where is the compensating negative bound charge located? Find the force on a square loop side a , lying in the yz plane and centered at the origin, if it carries a current I, flowing counterclockwise, when you look down the x axis.
The force on the two horizontal segments in z field flips sign. How great would v have to be in order for the magnetic attraction to balance the electric repulsion? The inner solenoid radius a has n1 turns per unit length, and the outer one radius b has n2.
Find B in each of the three regions: i inside the inner solenoid, ii between them, and iii outside both. Outside, the field is zero. Thus, in region i , we have contributions from both the larger and the smaller solenoid. In region ii , the only contribution to the magnetic field is from the larger solenoid. The region is outside the smaller solenoid, so the magnetic field due to the smaller one is zero.
According to example 5. How is this derived? Thus, above and below the capacitor, the field is zero, since the surface current K is pointed in opposite direction due to the opposite charge on the plates. Thus, the forces would never balance. If the square loop is free to rotate, what will its equilibrium orientation be?
Why minus? Problem 6. Find the magnetic field due to M inside and outside the cylinder. Find the magnetic field due to M, for points inside and outside the cylinder. Since the surface current is Find the magnetic field inside and outside the cylinder by two different methods: a As in Sect. Notice that the second method is much faster, and avoids any explicit reference to the bound currents.
A current I flows down the inner conductor and returns along the outer one; in each case the current distributes itself uniformly over the surface Fig. Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that together, of course, with the free currents they generate the correct field. Find all the bound currents. What is the net bound Rcurrent flowing down the wire. A resistor R is connected across the rails and a uniform magnetic field B, pointing into the page, fills the entire region.
In what direction does it flow? If the bar is moving to the right, that means that the flux through the loop is increasing. In what direction? Since the current is up through the bar and the magnetic field is into the page, the force is to the left, as it must be. Otherwise, the bar would gain more and more energy as it sped up, without a source of that energy. The bar must slow down. Find the current induced in the loop, as a function of time.
Find the electric field magnitude and direction at a distance s from the axis both inside and outside the solenoid , in the quasi static approximation. A narrow gap in the wire, of width w c a, forms a parallel-plate capacitor, as shown in Fig.
This is a displacement current problem. The displacement current is given by the change of electric flux. For a more realistic model, imagine thin wires that connect to the centers of the plates Fig. Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w c a. Notice that the displacement current through this surface is zero, and there are two contributions to Ienc.
We know have a total enclosed current of i the current in the wire with radius s, and ii the displacement current from the donut region around the wire. The displacement current will be the old displacement current in equation , minus the current in the wire. R We already found the magnetic field in problem 7. Note especially the direction of S. Check that Eq. Calculate the total power flowing into the gap, by integrating the Poynting vector of the appropriate surface.
Check that the power input is equal to the rate of increase of energy in the gap Eq. Our instinct is to hand the students shovels and tell them to start digging. They may develop blisters at first, but we still think this is the most efficient and exciting way to learn. Several readers have noted that there are fewer worked examples in this book than is customary, and that some important material is relegated to the problems. This is no accident. This seemed like a good idea though, like the quality of a restaurant, the significance of a problem is partly a matter of taste ; we have adopted the following rating scheme: an essential problem that every reader should study; a somewhat more difficult or peripheral problem; an unusually challenging problem, that may take over an hour.
Most of the one-star problems appear at the end of the relevant section; most of the three-star problems are at the end of the chapter. If a computer is required, we put a mouse in the margin. A solution manual is available to instructors only from the publisher. In preparing this third edition we have tried to retain as much as possible the spirit of the first and second.
Schroeter brings the fresh perspective of a solid state theorist, and he is largely responsible for the new chapter on symmetries. We have added a number of problems, clarified many explanations, and revised the Afterword. But we were determined not to allow the book to grow fat, and for that reason we have eliminated the chapter on the adiabatic approximation significant insights from that chapter have been incorporated into Chapter 11 , and removed material from Chapter 5 on statistical mechanics which properly belongs in a book on thermal physics.
It goes without saying that instructors are welcome to cover such other topics as they see fit, but we want the textbook itself to represent the essential core of the subject. Do you like this book? Please share with your friends, let's read it!! Search Ebook here:.
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