# Eletromagnetismo - Hayt - 7ª Ed - Soluções - chapter02 7th solution

CHAPTER 2

2.1. Four 10nC positive charges are located in the z =0 plane at the corners of a square 8cm on a side. A ﬁfth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the magnitude of the total force on this ﬁfth charge for = 0:

Arrange the charges in the xy plane at locations (4,4), (4,-4), (-4,4), and (-4,-4). Then the ﬁfth charge will be on the z axis at location z =4 √ 2, which puts it at 8cm distance from the other four. 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.

2.2. Two point charges of Q1 coulombs each are located at (0,0,1) and (0,0,-1). (a) Determine the locus of the possible positions of a third charge Q2 where Q2 may be any positive or negative value, such that the total ﬁeld E =0 at (0,1,0):

The total ﬁeld at (0,1,0) from the two Q1 charges (where both are positive) will be

where R = √ 2. To cancel this ﬁeld, Q2 must be placed on the y axis at positions y> 1i f Q2 > 0, and at positions y< 1i f Q2 < 0. In either case the ﬁeld from Q2 will be

and the total ﬁeld is then

Therefore

where the plus sign is used if Q2 > 0, and the minus sign is used if Q2 < 0.

(b) What is the locus if the two original charges are Q1 and −Q1?

located at the positive z (= 1) value. We now need Q2 to lie along the line x =0 , y =1 in order to cancel the ﬁeld from the positive and negative Q1 charges. Assuming Q2 is located at (0,1,z ), the total ﬁeld is now

The force will be:

and |RBA| =2 . Substituting these leads to where distances are in meters.

2.4. Eight identical point charges of Q C each are located at the corners of a cube of side length a, with one charge at the origin, and with the three nearest charges at (a,0,0), (0,a,0), and (0,0,a). Find an expression for the total vector force on the charge at P(a,a,a), assuming free space:

The total electric ﬁeld at P(a,a,a) that produces a force on the charge there will be the sum of the ﬁelds from the other seven charges. This is written below, where the charge locations associated with each term are indicated:

Enet(a,a,a)= q

The force is now the product of this ﬁeld and the charge at (a,a,a). Simplifying, we obtain

Now the x component of E at the new P3 will be:

To obtain Ex =0 ,w e require the expression in the large brackets to be zero. This expression simpliﬁes to the following quadratic:

2.6. Three point charges, each 5 × 10−9 C, are located on the x axis at x = −1, 0, and 1 in free space. a) Find E at x =5 :A ta general location, x,

b) Determine the value and location of the equivalent single point charge that would produce the same ﬁeld at very large distances: For x> > 1, the above general ﬁeld in part a becomes

2.8. 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. The uncharged spheres are centered at x =0 and x = d, the latter ﬁxed. If the spheres are given equal and opposite charges of Q coulombs:

a) Obtain the expression by which Q may be found as a function of x: The spheres will attract, and so the movable sphere at x =0 will move toward the other until the spring and Coulomb forces balance. This will occur at location x for the movable sphere. With equal and opposite forces, we have Q2

b) Determine the maximum charge that can be measured in terms of 0, k, and d, and state the separation of the spheres then: With increasing charge, the spheres move toward each other until they just touch at xmax = d − 2a. Using the part a result, we ﬁnd the maximum measurable charge: Qmax =4 a√ π 0k(d − 2a). Presumably some form of stop mechanism is placed at x = x−max to prevent the spheres from actually touching. c) What happens if a larger charge is applied? No further motion is possible, so nothing happens.

2.9. A 100 nC point charge is located at A(−1,1,3) in free space. a) Find the locus of all points P(x,y,z)a t which Ex = 500 V/m: The total ﬁeld at P will be:

b) Find y1 if P(−2,y1,3) lies on that locus: At point P, the condition of part a becomes

2.10. Ap ositive test charge is used to explore the ﬁeld of a single positive point charge Q at P(a,b,c). If the test charge is placed at the origin, the force on it is in the direction 0.5ax − 0.5√ 3ay, and when

We ﬁrst construct the ﬁeld using the form of Eq. (12). We identify r = xax + yay + zaz and r′ = aax + bay + caz. Then

Using (1), we can write the two force directions at the two test charge positions as follows:

a2 + b2 =2 a. Using this information in (3), we write for the x component:

2.1. Ac harge Q0 located at the origin in free space produces a ﬁeld for which Ez =1 kV/m at point P(−2,1,−1).

a) Find Q0: The ﬁeld at P will be

obtain only a radial component of EM. This will be:

2.12. Electrons are in random motion in a ﬁxed region in space. During any 1µsi nterval, the probability of ﬁnding an electron in a subregion of volume 10−15 m2 is 0.27. 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. With e = −1.602 × 10−19 C, the density becomes

2.13. A uniform volume charge density of 0.2 µC/m3 is present throughout the spherical shell extending from r =3 cm to r =5 cm. If ρv =0 elsewhere:

a) ﬁnd the total charge present throughout the shell: This will be

b) ﬁnd r1 if half the total charge is located in the region 3cm <r <r 1:I f the integral over r in part a is taken to r1,w ew ould obtain[

Thus

2.14. The charge density varies with radius in a cylindrical coordinate system as ρv = ρ0/(ρ2 +a2)2 C/m3. Within what distance from the z axis does half the total charge lie?

Choosing a unit length in z, the charge contained up to radius ρ is

The total charge is found when ρ →∞ ,o r Qnet = πρ0/a2.I ti s seen from the Q(ρ) expression that half of this occurs when ρ = a.

2.15. A spherical volume having a 2 µm radius contains a uniform volume charge density of 1015 C/m3.

b) Now assume that a large region contains one of these little spheres at every corner of a cubical grid 3mm on a side, and that there is no charge between spheres. 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

2.16. Within a region of free space, charge density is given as ρv = ρ0r/a C/m3, where ρ0 and a are constants. Find the total charge lying within:

a) the sphere, r ≤ a: This will be a r2 sinθdrdθdφ =4 π

b) Find E at that point in the z =0 plane where the direction of E is given by (1/3)ay − (2/3)az: With z =0 , the general ﬁeld will be

2.18. An inﬁnite uniform line charge ρL =2 nC/m lies along the x axis in free space, while point charges of 8 nC each are located at (0,0,1) and (0,0,-1).

a) Find E at (2,3,-4).

The net electric ﬁeld from the line charge, the point charge at z =1 , and the point charge at z = −1 will be (in that order):

b) To what value should ρL be changed to cause E to be zero at (0,0,3)? In this case, we only need scalar addition to ﬁnd the net ﬁeld:

Therefore q

2.19. A uniform line charge of 2 µC/m is located on the z axis. Find E in cartesian coordinates at P(1,2,3) if the charge extends from a) −∞ <z < ∞: 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). The ﬁeld from an inﬁnite line on the z axis is generally E =[ ρl/(2π 0ρ)]aρ. Therefore, at point P:

RzP

∫ ρldz

where r = ax +2 ay +3 az and r′ = zaz.S o the integral becomes

Using integral tables, we obtain:

The student is invited to verify that when evaluating the above expression over the limits −∞ < z< ∞, the z component vanishes and the x and y components become those found in part a.

2.20. The portion of the z axis for which |z| < 2 carries a nonuniform line charge density of 10|z| nC/m, and ρL =0 elsewhere. Determine E in free space at:

a) (0,0,4): The general form for the diﬀerential ﬁeld at (0,0,4) is

b) (0,4,0): In this case, r =4 ay and r′ = z az as before. The ﬁeld at (0,4,0) is then

Note the symmetric limits on the integral. As the z component of the integrand changes sign at z =0 ,i t will contribute equal and opposite portions to the overall integral, which will cancel completely (the z component integral has odd parity). This leaves only the y component integrand, which has even parity. The integral therefore simpliﬁes to

2.21. Two identical uniform line charges with ρl =7 5 nC/m are located in free space at x =0 , y = ±0.4 m. 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 0.8m .T hus the ﬁeld from the charge at y = −0.4e valuated at the of the line at the positive y location is dF = dqE = ρldzE.T hus the force per unit length acting on the line at postive y arising from the charge at negative y is

The force on the line at negative y is of course the same, but with −ay.

2.2. Two identical uniform sheet charges with ρs = 100 nC/m2 are located in free space at z = ±2.0 cm. What force per unit area does each sheet exert on the other?

The ﬁeld from the top sheet is E = −ρs/(2 0)az V/m. The diﬀerential force produced by this ﬁeld on the bottom sheet is the charge density on the bottom sheet times the diﬀerential area

2.23. Given the surface charge density, ρs =2 µC/m2,i n the region ρ< 0.2m , z =0 , and is zero elsewhere, ﬁnd E at:

a) PA(ρ =0 ,z =0 .5): First, we recognize from symmetry that only a z component of E will be present. Considering a general point z on the z axis, we have r = zaz. Then, with r′ = ρaρ,w e obtain r − r′ = zaz − ρaρ. The superposition integral for the z component of E will be:

Ez,PA = ρs zρ dρdφ

With z =0 .5m , the above evaluates as Ez,PA =8 .1k V/m.

b) With z at −0.5m ,w e valuate the expression for Ez to obtain Ez,PB = −8.1k V/m.

2.24. For the charged disk of Problem 2.23, show that: a) the ﬁeld along the z axis reduces to that of an inﬁnite sheet charge at small values of z:I n general, the ﬁeld can be expressed as

At small z, this reduces to Ez .= ρs/2 0, which is the inﬁnite sheet charge ﬁeld.

b) the z axis ﬁeld reduces to that of a point charge at large values of z: The development is as follows:

where the last approximation is valid if z> > .04. Continuing:

This the point charge ﬁeld, where we identify q = π(0.2)2ρs as the total charge on the disk (which now looks like a point).

2.25. Find E at the origin if the following charge distributions are present in free space: point charge, 12nC at P(2,0,6); uniform line charge density, 3nC/ma t x = −2, y =3 ; uniform surface charge density, 0.2nC/m2 at x =2 . The sum of the ﬁelds at the origin from each charge in order is:

2.26. An electric dipole (discussed in detail in Sec. 4.7) consists of two point charges of equal and opposite magnitude ±Q spaced by distance d. With the charges along the z axis at positions z = ±d/2 (with the positive charge at the positive z location), the electric ﬁeld in spherical coordinates is given by

expressions for the vector force on a point charge of magnitude q:

dx = Ey

Thus 2(xdy + ydx)= ydy − xdx

2.28 A ﬁeld is given as E =2 xz2ax +2 z(x2 +1 )az. Find the equation of the streamline passing through the point (1,3,-1):

dx = Ez

At (1,3,-1), the expression is satisﬁed if C =0 . Therefore, the equation for the streamline is z2 = x2 +2 lnx.

b) a unit vector in the direction of EP: The unit vector associated with E is (cos5xax − sin5xay), which evaluated at P becomes aE = −0.87ax − 0.50ay.

c) the equation of the direction line passing through P: Use

Thus y = 15 lncos5x + C.E valuating at P,w eﬁ nd C =0 .13, and so

2.30. For ﬁelds that do not vary with z in cylindrical coordinates, the equations of the streamlines are obtained by solving the diﬀerential equation Eρ/Eφ = dρ(ρdφ). Find the equation of the line passing through the point (2,30◦,0) for the ﬁeld E = ρcos2φaρ − ρsin2φaφ:

= dρ

Integrate to obtain

2lnρ =l nsin2φ +l nC =l n