# DESCARGAR LIBRO TEORIA ELECTROMAGNETICA WILLIAM HAYT PDF

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy. See our Privacy Policy and User Agreement for details. Author: Shajind Kikus Country: Liechtenstein Language: English (Spanish) Genre: Technology Published (Last): 1 January 2007 Pages: 330 PDF File Size: 9.95 Mb ePub File Size: 19.73 Mb ISBN: 265-5-42815-825-2 Downloads: 5638 Price: Free* [*Free Regsitration Required] Uploader: Maulkree For the G field in Problem 1. Plots are shown below. Given the points M 0. This is the equation of a cylinder, centered on the x axis, and of radius 2. Thus 1. Each does, as shown above. A fifth 10nC positive charge is located at a point 8cm distant from the other charges. By symmetry, the force on the fifth charge will be z-directed, and will be four times the z component of force produced by each of the four other charges.

To solve this problem, the z coordinate of the third charge is immaterial, so we can place it in the xy plane at coordinates x, y, 0. We take its magnitude to be Q3. Find the total force on the charge at A. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q2. Therefore, we look for P3 at coordinates x, 2. This expression simplifies to the following quadratic: 0.

Now, since the charge is at the origin, we expect to obtain only a radial component of EM. A uniform volume charge density of 0. What is the average volume charge density throughout this large region? Each cube will contain the equivalent of one little sphere. Neglecting the little sphere volume, the average density becomes 3. Uniform line charges of 0. What force per unit length does each line charge exert on the other?

The charges are parallel to the z axis and are separated by 0. This is evident just from the symmetry of the problem. For this reason, the net field magnitude will be the same everywhere, whereas the field direction will depend on which side of a given sheet one is positioned. This will be the magnitude at the other two points as well. A uniform surface charge density of 0.

Find E at the origin: Since each pair consists of equal and opposite charges, the effect at the origin is to double the field produce by one of each type. Taking the sum of the fields at the origin from the surface and line charges, respectively, we find: 0.

An empty metal paint can is placed on a marble table, the lid is removed, and both parts are discharged honorably by touching them to ground. An insulating nylon thread is glued to the center of the lid, and a penny, a nickel, and a dime are glued to the thread so that they are not touching each other. The assembly is lowered into the can so that the coins hang clear of all walls, and the lid is secured. The outside of the can is again touched momentarily to ground.

The device is carefully disassembled with insulating gloves and tools. Again, since the coins are insulated, they retain their original charges.

A point charge of 12 nC is located at the origin. This sphere encloses the point charge, so its flux of 12 nC is included. The flux from the line charges will equal the total line charge that lies within the sphere. We just integrate the charge density on that surface to find the flux that leaves it. This layer, being of uniform density, will not contribute to D at P. These fluxes will thus cancel.

The enclosed charge is the result of part a. Note also that the expression is valid for all x positive or negative values. Find D everywhere: Since the charge varies only with radius, and is in the form of a cylinder, symmetry tells us that the flux density will be radially-directed and will be constant over a cylindrical surface of a fixed radius. The set up is the same, except the upper limit of the above integral is 1 instead of r.

These are plotted on the next page. We note that there is no z component of D, so there will be no outward flux contributions from the top and bottom surfaces.

Evaluate both sides of Eq. Evaluate the partial derivatives at the center of the volume. A spherical surface of radius 3 mm is centered at P 4, 1, 5 in free space. Use the. A cube of volume a 3 has its faces parallel to the cartesian coordinate surfaces. Show that div D is zero everywhere except at the origin.

Using the formula for divergence in spherical coordinates see problem 3. We note that D has only a radial component, and so flux would leave only through the cylinder sides. The total surface charge should be equal and opposite to the total volume charge.

The plot is zero at larger radii. Note that since the x component of D does not vary with x, the outward fluxes from the front and back surfaces will cancel each other. The same is true for the left and right surfaces, since Dy does not vary with y.

We could just as well position the two points at the same z location and the problem would not change. Halfway along this line is a point of symmetry in the field make a sketch to see this. This means that when starting from either point, the initial force will be the same. This is also found by going through the same procedure as in part a, but with the direction roles of P and Q reversed.

Repeat Problem 4. A point charge Q1 is located at the origin in free space. Three point charges, 0. The sketch will show that V maximizes to a value of 8. Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them?

We keep in mind the definition of absolute potential as the work done in moving a unit positive charge from infinity to location r. This will be just How much charge lies within the cylinder? Two point charges, 1 nC at 0, 0, 0. From the above field expression, the radial component magnitude is twice that of the theta component.

Therefore, the given surface cannot be an equipotential the problem was ill-conceived. Only a surface of constant r could be an equipotential in this field. Four 0. Again find the total stored energy: This will be the energy found in part a plus the amount of work done in moving the fifth charge into position from infinity.

The latter is just the potential at the square center arising from the original four charges, times the new charge value, or 4. Make the assumption that the electrons are emitted continuously as a beam with a 0. In part c, the net outward flux was found to be zero, and in part b, the divergence of J was found to be zero as will be its volume integral. Therefore, the divergence theorem is satisfied.

Assuming that there is no transformation of mass to energy or vice-versa, it is possible to write a continuity equation for mass. If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center. The continuity equation for mass equates the divergence of the mass rate of flow mass per second per square meter to the negative of the density mass per cubic meter. After setting up a cartesian coordinate system inside a star, Captain Kirk and his intrepid crew make measurements over the faces of a cube centered at the origin with edges 40 km long and parallel to the coordinate axes.

Therefore, the rate of change of density at the origin will be just the negative. Thus 3. Let the total current carried by this hybrid conductor be 80 A dc. Find: a Jst. We begin with the fact that electric field must be the same in the aluminum and steel regions.

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