**UFRGS**

# Engineering Electromagneti...- W. Hayt, solution manual - hayt8e sm ch3

(Parte **1** de 2)

CHAPTER 3

3.1. Suppose that the Faraday concentric sphere experiment is performed in free space using a central charge at the origin, Q1 , and with hemispheres of radius a. A second charge Q2

(this time a point charge) is located at distance R from Q1 , where R >> a.

a) What is the force on the point charge before the hemispheres are assembled around Q1 ?

This will be simply the force beween two point charges, or a r b) What is the force on the point charge after the hemispheres are assembled but before they are discharged? The answer will be the same as in part a because induced charge

Q1 now resides as a surface charge layer on the sphere exterior. This produces the same electric field at the Q2 location as before, and so the force acting on Q2 is the same.

c) What is the force on the point charge after the hemispheres are assembled and after they are discharged? Discharging the hemispheres (connecting them to ground) neutralizes the positive outside surface charge layer, thus zeroing the net field outside the sphere. The force on Q2 is now zero.

d) Qualitatively, describe what happens as Q2 is moved toward the sphere assembly to the extent that the condition R >> a is no longer valid. Q2 itself begins to induce negative surface charge on the sphere. An attractive force thus begins to strengthen as the charge moves closer. The point charge field approximation used in parts a through c is no longer valid.

3.2. An electric field in free space is E = (5z2

V/m. Find the total charge contained within a cube, centered at the origin, of 4-m side length, in which all sides are parallel to coordinate axes (and therefore each side intersects an axis at ±2.

E = 5z2 az . As D is z-directed only, it will intersect only the top and bottom surfaces (both parallel to the x-y plane). From Gauss’ law, the charge in the cube is equal to the net outward flux of D, which in this case is

Qencl az · az dxdy + where the first and second integrals on the far right are over the top and bottom surfaces respectively.

3.3. The cylindrical surface ρ = 8 cm contains the surface charge density, ρs = 5e a) What is the total amount of charge present? We integrate over the surface to find:

integrate the charge density on that surface to find the flux that leaves it.

3.4. An electric field in free space is E = (5z3

V/m. Find the total charge contained within a sphere of 3-m radius, centered at the origin. Using Gauss’ law, we set up the integral in free space over the sphere surface, whose outward unit normal is ar :

az · ar

(3)2 sinθ dθ dφ where in this case z = 3cosθ and (in all cases) az · ar

= cosθ. These are substituted to yield

C/m2 and evaluate surface integrals to find the total charge enclosed in the rectangular parallelepiped 0 < x < 2, 0 < y < 3, 0 < z < 5 m: Of the 6 surfaces to consider, only 2 will contribute to the net outward flux. Why? First consider the planes at y = 0 and 3. The y component of D will penetrate those surfaces, but will be inward at y = 0 and outward at y = 3, while having the same magnitude in both cases. These fluxes will thus cancel. At the x = 0 plane, Dx = 0 and at the z = 0 plane, Dz

= 0, so there will be no flux contributions from these surfaces. This leaves the 2 remaining surfaces at x = 2 and z = 5. The net outward flux becomes:

· az dxdy

3.6. In free space, volume charge of constant density ρv = ρ0

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. For finding the field inside the charge, an appropriate Gaussian surface will be that which encloses a rectangular region defined by −1 < x < 1, −1 < y < 1, and |z| < d/2. The outward flux from this surface will be limited to that through the two parallel surfaces at ±z:

Φin

Dz dxdy = Qencl

Z z ρ0 dxdydz where the factor of 2 in the second integral account for the equal fluxes through the two surfaces. The above readily simplifies, as both Dz and ρ0 are constants, leading to

Din = ρ0 z az

Outside the charge, the Gaussian surface is the same, except that the parallel boundaries at ±z occur at |z| > d/2. As a result, the calculation is nearly the same as before, with the only change being the limits on the total charge integral:

Φout

Dz dxdy = Qencl ρ0 dxdydz

Solve for Dz to find the constant values:

3.7. Volume charge density is located in free space as ρv = 2e a) Find the total charge enclosed by the spherical surface r = 1 m: To find the charge we integrate:

2e −1000rr2 sinθ drdθ dφ

Integration over the angles gives a factor of 4π. The radial integration we evaluate using tables; we obtain

b) By using Gauss’s law, calculate the value of Dr on the surface r = 1 m: The gaussian surface is a spherical shell of radius 1 m. The enclosed charge is the result of part a.

We thus write 4πr2

3.8. Use Gauss’s law in integral form to show that an inverse distance field in spherical coordinates,

D = Aar /r, where A is a constant, requires every spherical shell of 1 m thickness to contain

4πA coulombs of charge. Does this indicate a continuous charge distribution? If so, find the charge density variation with r.

The net outward flux of this field through a spherical surface of radius r is

ar · ar r2 sinθdθdφ = 4πAr = Q encl

We see from this that with every increase in r by one m, the enclosed charge increases by 4πA (done). It is evident that the charge density is continuous, and we can find the density indirectly by constructing the integral for the enclosed charge, in which we already found the latter from Gauss’s law:

Z r

Z r

To obtain the correct enclosed charge, the integrand must be ρ(r) = A/r2 .

3.9. A uniform volume charge density of 80µC/m3 is present throughout the region 8mm < r < a) Find the total charge inside the spherical surface r = 10 m: This will be

b) Find Dr at r = 10 m: Using a spherical gaussian surface at r = 10, Gauss’ law is c) If there is no charge for r > 10 m, find Dr at r = 20 m: This will be the same computation as in part b, except the gaussian surface now lies at 20 m. Thus

3.10. An infinitely long cylindrical dielectric of radius b contains charge within its volume of density

= aρ2 , where a is a constant. Find the electric field strength, E, both inside and outside the cylinder.

Inside, we note from symmetry that D will be radially-directed, in the manner of a line charge field. So we apply Gauss’ law to a cylindrical surface of radius ρ, concentric with the charge distribution, having unit length in z, and where ρ < b. The outward normal to the surface is aρ .

Dρaρ · aρ ρdφdz = Qencl in which the dummy variable ρ0 must be used in the far-right integral because the upper radial limit is ρ. Dρ is constant over the surface and can be factored outside the integral.

Evaluating both integrals leads to or Ein =

To find the field outside the cylinder, we apply Gauss’ law to a cylinder of radius ρ > b. The setup now changes only by the upper radius limit for the charge integral, which is now the charge radius, b:

Dρaρ · aρ ρdφdz = Qencl

Z b aρ2 ρdρdφdz where the dummy variable is no longer needed. Evaluating as before, the result is ab 4 or Eout = ab 4

3.1. In cylindrical coordinates, let ρv = 0 for ρ < 1 m, ρv 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. Gauss’ law applied to such a surface of unit length in z gives:

a) for ρ < 1 m, Dρ = 0, since no charge is enclosed by a cylindrical surface whose radius lies within this range.

3.11c) for ρ > 1.5mm, the gaussian cylinder now lies at radius ρ outside the charge distribution, so the integral that evaluates the enclosed charge now includes the entire charge distribution. To accomplish this, we change the upper limit of the integral of part b from ρ to 1.5 m, finally obtaining:

3.12. The sun radiates a total power of about 3.86×1026 watts (W). If we imagine the sun’s surface to be marked oﬀ in latitude and longitude and assume uniform radiation, a) What power is radiated by the region lying between latitude 50◦

N and 60◦ N and longi-

12◦ and 27◦ correspond directly to the limits on φ. Since the sun for our purposes is

The required power is now found through ar · ar r2 sinθ dθ dφ b) What is the power density on a spherical surface 93,0,0 miles from the sun in W/m2 ?

First, 93,0,0 miles = 15,0,0 km = 1.5 × 1011 m. Use this distance in the flux density expression above to obtain a) Find D at r = 1, 3and5 m: Noting that the charges are spherically-symmetric, we ascertain that D will be radially-directed and will vary only with radius. Thus, we apply Gauss’ law to spherical shells in the following regions: r < 2: Here, no charge is enclosed, b) Determine ρ s0 such that D = 0 at r = 7 m. Since fields will decrease as 1/r2 , the question could be re-phrased to ask for ρ s0 such that D = 0 at all points where r > 6 m. In this region, the total field will be

3.14. A certain light-emitting diode (LED) is centered at the origin with its surface in the xy plane.

At far distances, the LED appears as a point, but the glowing surface geometry produces a far-field radiation pattern that follows a raised cosine law: That is, the optical power (flux) density in Watts/m2 is given in spherical coordinates by ar Watts/m where θ is the angle measured with respect to the normal to the LED surface (in this case, the z axis), and r is the radial distance from the origin at which the power is detected.

a) Find, in terms of P0 , the total power in Watts emitted in the upper half-space by the

LED: We evaluate the surface integral of the power density over a hemispherical surface of radius r:

ar · ar r2 sinθdθdφ = − b) Find the cone angle, θ1 , within which half the total power is radiated; i.e., within the range 0 < θ < θ1 : We perform the same integral as in part a except the upper limit for θ is now θ1 . The result must be one-half that of part a, so we write:

P t c) An optical detector, having a 1 mm2 cross-sectional area, is positioned at r = 1 m and at θ = 45◦ , such that it faces the LED. If one nanowatt (stated in error as 1mW) is measured by the detector, what (to a very good estimate) is the value of P0 ? Start with a r

Then the detected power in a 1-mm2 area at r = 1 m approximates as

If the originally stated 1mW value is used for the detected power, the answer would have been 4π kW (!).

3.15. Volume charge density is located as follows: ρv = 0 for ρ < 1 m and for ρ > 2 m,

We find,

4ρρdρdφdz = 8πL b) Use Gauss’ law to determine Dρ at ρ = ρ1 : Gauss’ law states that 2πρ1LDρ

Q is the result of part a. So, with ρ1 in meters,

At ρ = 2.4, we evaluate the charge integral of part a from .001 to .002, and Gauss’ law is written as

3.16. An electric flux density is given by D = D0 aρ , where D0 is a given constant.

a) What charge density generates this field? Charge density is found by taking the divergence: With radial D only, we have

1ρ d dρ b) For the specified field, what total charge is contained within a cylinder of radius a and height b, where the cylinder axis is the z axis? We can either integrate the charge density over the specified volume, or integrate D over the surface that contains the specified volume:

Z b ρ ρdρdφdz =

Z b a) apply Gauss’ law to find the total flux leaving the closed surface of the cube. We call the surfaces at x = 1.2 and x = 1 the front and back surfaces respectively, those at y = 1.2 and y = 1 the right and left surfaces, and those at z = 1.2 and z = 1 the top and bottom surfaces. To evaluate the total charge, we integrate D · n over all six surfaces and sum the results. We note that there is no z component of D, so there will be no outward flux contributions from the top and bottom surfaces. The fluxes through the remaining four are

| {z } left b) evaluate ∇ · D at the center of the cube: This is c) Estimate the total charge enclosed within the cube by using Eq. (8): This is center

3.18. State whether the divergence of the following vector fields is positive, negative, or zero:

a) the thermal energy flow in J/(m2 − s) at any point in a freezing ice cube: One way to visualize this is to consider that heat is escaping through the surface of the ice cube as it freezes. Therefore the net outward flux of thermal energy through the surface is positive. Calling the thermal flux density F, the divergence theorem says and so if we identify the left integral as positive, the right integral (and its integrand) must also be positive. Answer: positive.

b) the current density in A/m2 in a bus bar carrying direct current: In this case, we have no accumulation or dissipation of charge within any small volume, since current is dc; this also means that the net outward current flux through the surface that surrounds any small volume is zero. Therefore the divergence must be zero.

c) the mass flow rate in kg/(m2 − s) below the surface of water in a basin, in which the water is circulating clockwise as viewed from above: Here again, taking any small volume in the water, the net outward flow through the surface that surrounds the small volume is zero; i.e., there is no accumulation or dissipation of mass that would result in a change in density at any point. Divergence is therefore zero.

3.19. A spherical surface of radius 3 m is centered at P(4,1,5) in free space. Let D = xax

Use the results of Sec. 3.4 to estimate the net electric flux leaving the spherical surface: We

3.20. A radial electric field distribution in free space is given in spherical coordinates as:

where ρ0 , a, and b are constants.

a) Determine the volume charge density in the entire region (0 ≤ r ≤ ∞) by appropriate use of ∇ · D = ρv . We find ρv by taking the divergence of D in all three regions, where

D = ≤0 E. As D has only a radial component, the divergences become:

3.20b) Find, in terms of given parameters, the total charge, Q, within a sphere of radius r where r > b. We integrate the charge densities (piecewise) over the spherical volume of radius b:

r2 sinθdrdθdφ −

Z b r2 sinθdrdθdφ =

3.21. Calculate the divergence of D at the point specified if aρ + 10ρzaz

c) D = 2rsinθsinφar + rcosθsinφaθ

+ rcosφaφ nates, we have

r sinθ cos2θ sinφ

sinθ − sinφ sinθ

3.2. (a) A flux density field is given as F1 = 5az . Evaluate the outward flux of F1 through the

The flux integral is hem.

cosθ b) What simple observation would have saved a lot of work in part a? The field is constant, and so the inward flux through the base of the hemisphere (of area πa2 ) would be equal in magnitude to the outward flux through the upper surface (the flux through the base is a much easier calculation).

c) Now suppose the field is given by F2 = 5zaz

. Using the appropriate surface integrals, evaluate the net outward flux of F2 through the closed surface consisting of the hemisphere of part a and its circular base in the xy plane:

Note that the integral over the base is zero, since F2 = 0 there. The remaining flux integral is that over the hemisphere:

5z az · ar a2 sinθdθdφ =

(Parte **1** de 2)