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ENGINEERING ELECTROMAGNETICS HAYT PDF

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Engineering electromagnetics / William H. Hayt, Jr., John A. Buck. industry, Professor Hayt joined the faculty of Purdue University, where he served as. Hayt,Buck Engineering Electromagnetics 6e Pdf previous post Harrison Advanced Engineering Dynamics Pdf. next post Heat Exchanger. Engineering Electromagnetics 7th Edition William H. Hayt Solution Manual. The BookReader requires JavaScript to be enabled. Please check that your browser.


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The McGraw-Hill Companies. Engineering Electromagnetics. Sixth Edition. William H. Hayt, Jr.. John A. Buck. Textbook Table of Contents. The Textbook Table. Library of Congress Cataloging-in-Publication Data Hayt, William Hart, – Engineering electromagnetics / William H. Hayt, Jr., John A. Buck. — 8th ed. p. cm. Engineering Electromagnetics - 7th Edition - William H. Hayt - Solution Manual. Arsh Khan. CHAPTER 1 Given the vectors M = −10ax + 4ay − 8az and N.

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Find the voltmeter reading at: Have 0. First the flux through the loop is evaluated, where the unit normal to the loop is az. The rails in Fig. In this case, there will be a contribution to the current from the right loop, which is now closed.

Develop a function of time which expresses the ohmic power being delivered to the loop: First, since the field does not vary with y, the loop motion in the y direction does not produce any time-varying flux, and so this motion is immaterial.

Engineering Electromagnetics 8th Edition William H. Hayt Original

We can evaluate the flux at the original loop position to obtain: This will be Id 0. Then D 1. We set the given expression for Jd equal to the result of part c to obtain: Find the total displacement current through the dielectric and compare it with the source current as determined from the capacitance Sec. This we already found during the development in part a: We have. The parallel plate transmission line shown in Fig.

Neglect fields outside the dielectric. We evaluate the flux integral of Jd over the given cross section: Note that B as stated is constant with position, and so will have zero curl. The equation is thus not valid with these fields.

Nevertheless, we press on: We use the result of part a: The procedure here is similar to the development that leads to Eq. Begin by taking the curl of both sides of the Faraday law equation: In free space, the magnetic field of the uniform plane wave can be easily found using the intrinsic impedance: First, we note that if E at a given instant points in the negative x direction, while the wave propagates in the forward y direction, then H at that same position and time must point in the positive z direction.

We use the general formula, Eq. Using With Es in the positive y direction at a given time and propagating in the positive x direction, we would have a positive z component of Hs , at the same time. Find the distance a uniform plane wave can propagate through the material before: A 10 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region.

In a non-magnetic material, we would have: Find both the power factor and Q in terms of the loss tangent: First, the impedance will be: Then the power factor is P. Assume x-polarization for the electric field.

At this point a flaw becomes evident in the problem statement, since solving this part in two different ways gives results that are not the same. I will demonstrate: We apply our equation to the result of part a: Note that in Problem A positive y component of E requires a posi- tive z component of H for propagation in the forward x direction. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. From part a, we have 4. This will be 4. The external and internal regions are non-conducting.

We use J 1. Use The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respectively. The dielectric is lossless and the operating frequency is MHz. Calculate the resistance per meter length of the: Again, 70 applies but with a different conductor radius.

A hollow tubular conductor is constructed from a type of brass having a conductivity of 1. The inner and outer radii are 9 mm and 10 mm respectively. Calculate the resistance per meter length at a frequency of a dc: In this case the current density is uniform over the entire tube cross-section. We write: Now the skin effect will limit the effective cross-section. Therefore we can approximate the resistance using the formula: Most microwave ovens operate at 2. Since the conductivity is high, we.

A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0. Assuming the conductor is non-magnetic, determine the frequency and the conductivity: The outer conductor thickness is 0. Use information from Secs. The coax is air-filled. The result is squared, terms collected, and the square root taken. For a good dielectric Teflon we use the approximations: Consider a left-circularly polarized wave in free space that propagates in the forward z direction.

The electric field is given by the appropriate form of Eq. We find the two components of Hs separately, using the two components of Es. Specifically, the x component of Es is associated with a y component of Hs , and the y component of Es is associated with a negative x component of Hs.

With the Poynting vector in the positive x direction, a positive y component for E requires a positive z component for H. Similarly, a positive z component for E requires a negative y component for H. Therefore, 10! This is most clearly seen by first converting the given field to real instantaneous form: With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the output.

Suppose that the length of the medium of Problem Describe the polarization of the output wave in this case: With the wave propagating in the forward z direction, we find: Given the general elliptically-polarized wave as per Eq. What percentage of the incident power density is transmitted into the copper?

We need to find the reflection coefficient. We nevertheless proceed: A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2. There are two possible answers. This wave will experience loss in region 2, along with a different phase constant. First, using Eq. The field in region 2 is then constructed by using the resulting amplitude, along with the attenuation and phase constants that are appropriate for region 2.

Also, the intrinsic impedance We now find the input impedance: Calculate the ratio of the final power to the incident power after this round trip: Pi Try measuring that.

A MHz uniform plane wave in normally-incident from air onto a material whose intrinsic impedance is unknown. Measurements yield a standing wave ratio of 3 and the appearance of an electric field minimum at 0.

Determine the impedance of the unknown material: A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean. First we find the loss tangent: Within the limits of our good conductor approximation loss tangent greater than about ten , the reflected power fraction, using the formula derived in part a, is found to decrease with increasing frequency.

The transmitted power fraction thus increases.

Calculate the fractions of the incident power that are reflected and trans- mitted. The total electric field in the plane of the interface must rotate in the same direction as the incident field, in order to continually satisfy the boundary condition of tangential electric field continuity across the interface.

Therefore, the reflected wave will have to be left circularly polarized in order to make this happen. The transmitted field will be right circularly polarized as the incident field for the same reasons. A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect conductor. Assume positive z travel for the incident electric field.

This is a standing wave exhibiting circular polarization in time. Determine the standing wave ratio in front of the plate. Repeat Problem A uniform plane wave is normally incident from the left, as shown. Plot a curve of the standing wave ratio, s, in the region to the left: Thus, at 2. In this case we use 2. MathCad was used in both cases. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally-incident.

The slabs are to be arranged such that the air spaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness. Specify an arrangement of slabs and air spaces such that a the wave is totally transmitted through the stack: In this case, we look for a combination of half- wave sections.

Let the inter-slab distances be d1 , d2 , and d3 from left to right. Two possibilities are i. Thus every thickness is one-quarter wavelength. The impedances transform as follows: The reflection coefficient for waves incident on the front slab thus gets close to unity, and approaches 1 as the number of slabs approaches infinity. The 50MHz plane wave of Problem Determine the fractions of the incident power that are reflected and transmitted for a s polarization: To review Problem 12, we first we find the loss tangent: Therefore, for s polarization,.

The fraction transmitted is then 0. Again, with the refracted angle close to zero, the relection coefficient for p polar- ization is. Since the wave is circularly-polarized, the s-polarized component represents one-half the total incident wave power, and so the fraction of the total power that is reflected is. Since all the p-polarized com- ponent is transmitted, the reflected wave will be entirely s-polarized linear. The transmitted wave, while having all the incident p-polarized power, will have a reduced s-component, and so this wave will be right-elliptically polarized.

A dielectric waveguide is shown in Fig. All subsequent reflections from the upper an lower boundaries will be total as well, and so the light is confined to the guide. Find n0 in terms of n1 and n2: A Brewster prism is designed to pass p-polarized light without any reflective loss.

The prism of Fig. In the Brewster prism of Fig. The light is incident from air, and the returning beam also in air may be displaced sideways from the incident beam. More than one design is possible here. Using the result of Example For this to work, the Brewster angle must be greater than or equal to the critical angle.

Using Eq. Assume conductivity does not vary with frequency: In a good conductor: Over a certain frequency range, the refractive index of a certain material varies approximately linearly. The pulse will broaden and will acquire a frequency sweep chirp that is precisely linear with time.

Additionally, a pulse of a given bandwidth will broaden by the same amount, regardless of what carrier frequency is used. Describe the pulse at the output of the second channel and give a physical explanation for what hap- pened.

In fact, we may write in general: Physically, the pulse acquires a positive linear chirp frequency increases with time over the pulse envelope during the first half of the channel. The pulse, if originally transform-limited at input, will emerge, again transform-limited, at its original width. We use the expression for input impedance Eq.

Find and s: If the velocity on the line is 2. I will do this using two different methods: The Hard Way: Find L, C, R, and G for the line: If the inner radius of the outer conductor is 4 mm, find the radius of the inner conductor so that assuming a lossless line: Two aluminum-clad steel conductors are used to construct a two-wire transmission line.

The radius of the steel wire is 0. The dielectric is air, and the center-to-center wire separation is 4 in. The first question is whether we are in the high frequency or low frequency regime. Furthermore, the skin depth is considerably less than the aluminum layer thickness, so the bulk of the current resides in the aluminum, and we may neglect the steel.

Each conductor of a two-wire transmission line has a radius of 0. Pertinent dimensions for the transmission line shown in Fig. The conductors and the dielectric are non-magnetic. A transmission line constructed from perfect conductors and an air dielectric is to have a maximum dimension of 8mm for its cross-section. The line is to be used at high frequencies. Specify its dimensions if it is: With the maximum dimension of 8mm, we have, using Find c L: Coaxial lines 1 and 2 have the following parameters: The reflection coefficient encountered by waves incident on ZL1 from line 1 can now be found, along with the standing wave ratio: The line 1 length now has a load impedance of We need to find its input impedance.

In line 1, having a dielectric constant of 2. For the transmission line represented in Fig. A ohm transmission line is 0. The line is operating in air with a wavelength of 0. Determine the average power absorbed by each resistor in Fig. The next step is to determine the input impedance of the 2. The power dissipated by the ohm resistor is now 1 V 2 1 Find s on both sections 1 and 2: For section 2, we consider the propagation of one forward and one backward wave, comprising the superposition of all reflected waves from both ends of the section.

We first need the input impedance of the. A lossless transmission line is 50 cm in length and operating at a frequency of MHz. We then find the input impedance to the shorted line section of length 20 cm putting this impedance at the location of ZL , so we can combine them: This problem was originally posed incorrectly.

The corrected version should have an inductor in the input circuit instead of a capacitor. I will proceed with this replacement understood, and will change the wording as appropriate in parts c and d: Note that the dielectric is air: With the length of the line at 2. To achieve this, the imaginary part of the total impedance of part c must be reduced to zero so we need an inductor. Continuing, for this value of L, calculate the average power: The power delivered to the load will be the same as the power delivered to the input impedance.

Engineering Electromagnetics 8th Edition William H. Hayt Original

Use analytic methods or the Smith chart or both to find: I will first use the analytic approach. Using normalized impedances, Eq. A line drawn from the origin through this point intersects the outer chart boundary at the position 0. With a wavelength of 1. On the WTL scale, we add 0. A straight line is now drawn from the origin though the 0. A compass is then used to measure the distance between the origin and zin. With this distance set, the compass is then used to scribe off the same distance from the origin to the load impedance, along the line between the origin and the 0.

The difference in imaginary parts arises from uncertainty in reading the chart in that region. This is close to the value of the VSWR, as we found earlier. Next, yL is inverted to find zL by transforming the point halfway around the chart, using the compass and a straight edge. This is close to the computed inverse of yL , which is 1. Now, the position of zL is read on the outer edge of the chart as 0. The point is now transformed through the line length distance of 1.

The final reading on the WTG scale after the transformation is found through 0. Drawing a line between this mark on the WTG scale and the chart center, and scribing the compass arc length on this line, yields the normalized input impedance. I will specify answers in terms of wavelength. Make use of the Smith chart to find: Referring to the figure below, we start by marking the given zL on the chart and drawing a line from the origin through this point to the outer boundary.

On the WTG scale, we read the zL location as 0. Moving from here toward the generator, we cross the positive R axis at which the impedance is purely real and greater than 1 at 0.

The distance is then 0. Using a compass, we set its radius at the distance between the origin and zL. What is s on the remainder of the line?

Electromagnetics pdf engineering hayt

This will be just s for the line as it was before. This would return us to the original point, requiring a complete circle around the chart one-half wavelength distance.

The distance from the resistor will therefore be: With the aid of the Smith chart, plot a curve of Zin vs. Then, using a compass, draw a circle beginning at zL and progressing clockwise to the positive real axis. On the chart, radial lines are drawn at positions corresponding to. The intersections of the lines and the circle give a total of 11 zin values.

Electromagnetics pdf engineering hayt

The table below summarizes the results. A fairly good comparison is obtained. Use the Smith chart to find ZL with the short circuit replaced by the load if the voltage readings are: We mark this on the positive real axis of the chart see next page.

The load position is now 0. A line is drawn from the origin through this point on the chart, as shown. We then scribe this same distance along the line drawn through the.

A line is drawn from the origin through this location on the chart. Use the Smith chart to find: Drawing a line from the chart center through this point yields its location at 0.

Alternately, use the s scale at the bottom of the chart, setting the compass point at the center, and scribing the distance on the scale to the left. This distance is found by transforming the load impedance clockwise around the chart until the negative real axis is reached. This distance in wavelengths is just the load position on the WTL scale, since the starting point for this scale is the negative real axis.

So the distance is 0. Transforming the load through this distance toward the generator involves revolution once around the chart 0. A line is drawn between this point and the chart center.

This is plotted on the Smith chart below. We then set on the compass the distance between yL and the origin. The same distance is then scribed along the positive real axis, and the value of s is read as 2.

First we draw a line from the origin through zL and note its intersection with the WTG scale on the chart outer boundary. We note a reading on that scale of about 0.

To this we add 0. A line drawn from the 0. On the admittance chart, the Vmax position is on the negative r axis. This is at the zero position on the WTL scale.

The load is at the approximate 0.

Hayt pdf electromagnetics engineering

The wavelength on a certain lossless line is 10cm. We begin by marking zin on the chart see below , and setting the compass at its distance from the origin. First, use a straight edge to draw a line from the origin through zin , and through the outer scale. We read the input location as slightly more than 0. The line length of 12cm corresponds to 1. Thus, to transform to the load, we go counter-clockwise twice around the chart, plus 0. A line is drawn to the origin from that position, and the compass with its previous setting is scribed through the line.

A standing wave ratio of 2. Probe measurements locate a voltage minimum on the line whose location is marked by a small scratch on the line.

When the load is replaced by a short circuit, the minima are 25 cm apart, and one minimum is located at a point 7 cm toward the source from the scratch. Find ZL: Suppose that the scratch locates the first voltage minimum. With the short in place, the first minimum occurs at the load, and the second at 25 cm in front of the load. The effect of replacing the short with the load is to move the minimum at 25 cm to a new location 7 cm toward the load, or at 18 cm.

This is a possible location for the scratch, which would otherwise occur at multiples of a half-wavelength farther away from that point, toward the generator. Using the Smith chart see below we first draw a line from the origin through the 0.

As a check, I will do the problem analytically. A 2-wire line, constructed of lossless wire of circular cross-section is gradually flared into a coupling loop that looks like an egg beater.

At the point X, indicated by the arrow in Fig. A probe is moved along the line and indicates that the first voltage minimum to the left of X is 16cm from X. With the short circuit removed, a voltage minimum is found 5cm to the left of X, and a voltage maximum is located that is 3 times voltage of the minimum. Use the Smith chart to determine: No Smith chart is needed to find f , since we know that the first voltage minimum in front of a short circuit is one-half wavelength away.

Again, no Smith chart is needed, since s is the ratio of the maximum to the minimum voltage amplitudes. Now we need the chart. This point is then transformed, using the compass, to the negative real axis, which corresponds to the location of a voltage minimum. On the chart, we now move this distance from the Vmin location toward the load, using the WTL scale. A line is drawn from the origin through the 0. We begin with the general phasor voltage in the line: We use the same equation for V z , which in this case reads: With a short circuit replacing the load, a minimum is found at a point on the line marked by a small spot of puce paint.

The 1m distance is therefore 3. Therefore, with the actual load installed, the Vmin position as stated would be 3. This being the case, the normalized load impedance will lie on the positive real axis of the Smith chart, and will be equal to the standing wave ratio. The Smith chart construction is shown on the next page. This point is to be transformed to a location at which the real part of the normalized admittance is unity.

The stub is connected at either of these two points. The stub input admittance must cancel the imaginary part of the line admittance at that point.

Pdf engineering electromagnetics hayt

This point is marked on the outer circle and occurs at 0. The length of the stub is found by computing the distance between its input, found above, and the short-circuit position stub load end , marked as Psc. The length of the main line between its load and the stub attachment point is found on the chart by measuring the distance between yL and yin2 , in moving clockwise toward generator. In this case, everything is the same, except for the load-end position of the stub, which now occurs at the Poc point on the chart.

We find the stub length by moving from Poc to the point at which the admittance is j 0. This occurs at 0. The attachment point is found by transforming yL to yin1 , where the former point is located at 0. The lossless line shown in Fig. For the line to be matched, it is required that the sum of the normalized input admittances of the shorted stub and the main line at the point where the stub is connected be unity.

So the input susceptances of the two lines must cancel. To find the stub input susceptance, use the Smith chart to transform the short circuit point 0. This line is one-quarter wavelength long, so the normalized load impedance is equal to the normalized input admittance. The Smith chart construction is shown below. To cancel the input normalized susceptance of We therefore write 2. The two-wire lines shown in Fig. In this case, we have a series combination of the loaded line section and the shorted stub, so we use impedances and the Smith chart as an impedance diagram.

The requirement for matching is that the total normalized impedance at the junction consisting of the sum of the input impedances to the stub and main loaded section is unity. In the transmission line of Fig. Determine and plot the voltage at the load resistor and the current in the battery as functions of time by constructing appropriate voltage and current reflection diagrams: Referring to the figure, closing the switch launches a voltage wave whose value is given by Eq.

So the voltage wave traverses the line and does not reflect. The voltage reflection diagram would be that shown in Fig. Likewise, the current reflection diagram is that of Fig. Using these values, voltage and current reflection diagrams are constructed, and are shown below: First, the load voltage is found by adding voltages along the right side of the voltage diagram at the indicated times.

Second, the current through the battery is found by adding currents along the left side of the current reflection diagram. Both plots are shown below, where currents and voltages are expressed to three significant figures. Note also that when the switch is opened, the reflection coefficient at the generator end of the line becomes unity.

The reflection diagram is now constructed in the usual manner, and is shown on the next page. The load voltage as a function of time is found by accumulating voltage values as they are read moving up along the right hand boundary of the chart. The resulting function, plotted just below the reflection diagram, is found to be a sequence of pulses that alternate signs.

The pulse amplitudes are calculated as follows: In the charged line of Fig. This problem accompanies Example Plots of the voltage and current at the resistor are then found by accumulating values from the left sides of the two charts, producing the plots as shown.

Plot the load resistor voltage as a function of time: With the left half of the line charged to V0 , closing the switch initiates at the switch location two voltage waves: No reflection occurs at the load end, since the load is matched to the line. The reflection diagram and load voltage plot are shown below. The results are summarized as follows: A simple frozen wave generator is shown in Fig. Determine and plot the load voltage as a function of time: 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 Q 3. The force on the third charge is now. Find the total force on the charge at A. This force will be. This force in general will be:.

Solutions of engineering electromagnetics 6th edition william h. hayt, john a. echecs16.info - Docsity

Note, however, that all three charges must lie in a straight line, and the location of Q 3 will be along the vector R 12 extended past Q 2. Therefore, we look for P 3 at coordinates x, 2. With this restriction, the force becomes:.

The coordinates of P 3 are thus P 3 This field will be. This expression simplifies to the following quadratic:. The field will take the general form:. The total field at P will be:. The x component of the field will be. At point P , the condition of part a becomes. Determine E at P 0 , y, 0: The field will be. This field will be:. Now, since the charge is at the origin, we expect to obtain only a radial component of E M.

This will be:. Calculate the total charge present: A uniform volume charge density of 0. If the integral over r in part a is taken to r 1, we would obtain[. With the limits thus changed, the integral for the charge becomes:. 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. Find the charge within the region: The integral that gives the charge will be.

Uniform line charges of 0. This field will in general be:. Find E in cartesian coordinates at P 1 , 2 , 3 if the charge extends from. With the infinite line, we know that the field will have only a radial component in cylindrical coordinates or x and y components in cartesian.

Therefore, at point P:. So the integral becomes. Since all line charges are infinitely-long, we can write:. Substituting these into the expression for E P gives. 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. Thus the force per unit length acting on the line at postive y arising from the charge at negative y is.

The integral becomes:. Since the integration limits are symmetric about the origin, and since the y and z components of the integrand exhibit odd parity change sign when crossing the origin, but otherwise symmetric , these will integrate to zero, leaving only the x component.

A crude device for measuring charge consists of two small insulating spheres of radius a, one of which is fixed in position. The other is movable along the x axis, and is subject to a restraining force kx, where k is a spring constant. If the spheres are given equal and opposite charges of Q coulombs: This will occur at location x for the movable sphere. Using the part a result, we find the maximum measurable charge: No further motion is possible, so nothing happens. The total field at P will be: At point P, the condition of part a becomes 3.

A positive test charge is used to explore the field of a single positive point charge Q at P a, b, c. Find a, b, and c: We first construct the field using the form of Eq. Using this information in 3 , we write for the x component: So the two possible P coordinate sets are 0. This field will be: Now, since the charge is at the origin, we expect to obtain only a radial component of EM. This will be: Electrons are in random motion in a fixed region in space.

What volume charge density, appropriate for such time durations, should be assigned to that subregion? The finite probabilty effectively reduces the net charge quantity by the probability fraction. A uniform volume charge density of 0. Within what distance from the z axis does half the total charge lie? What is the average volume charge density throughout this large region? Each cube will contain the equivalent of one little sphere.

Find the total charge lying within: In this case, we only need scalar addition to find the net field: With the infinite line, we know that the field will have only a radial component in cylindrical coordinates or x and y components in cartesian. Therefore, at point P: Determine E in free space at: This leaves only the y component integrand, which has even parity.

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. What force per unit area does each sheet exert on the other? The differential force produced by this field on the bottom sheet is the charge density on the bottom sheet times the differential area there, multiplied by the electric field from the top sheet: First, we recognize from symmetry that only a z component of E will be present.

The superposition integral for the z component of E will be: For the charged disk of Problem 2. The development is as follows: Find E at the origin if the following charge distributions are present in free space: The sum of the fields at the origin from each charge in order is: An electric dipole discussed in detail in Sec.

Find the equation of the streamline passing through the point 1,3, 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. All coins were insulated during the entire procedure, so they will retain their original charges: Again, since the coins are insulated, they retain their original charges.

First, from part b, the point charge will now lie inside.

We integrate over the surface to find: We just integrate the charge density on that surface to find the flux that leaves it. Determine the total flux leaving: The total flux through the cylindrical surface and the two end caps are, in this order: Of the 6 surfaces to consider, only 2 will contribute to the net outward flux. These fluxes 28 The net outward flux becomes: Find D and E everywhere. From the symmetry of the configuration, we surmise that the field will be everywhere z-directed, and will be uniform with x and y at fixed z.

As a result, the calculation is nearly the same as before, with the only change being the limits on the total charge integral: To find the charge we integrate: The gaussian surface is a spherical shell of radius 1 mm.

The enclosed charge is the result of part a. Does this indicate a continuous charge distribution? If so, find the charge density variation with r. This will be the same computation as in part b, except the gaussian surface now lies at 20 mm. 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.