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CHAPTER 1
1.1. Given the vectors M = − 10 ax + 4ay − 8 az and N = 8ax + 7ay − 2 az , find: a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4 ay + 8az + 16ax + 14ay − 4 az = (26, 10 , 4) Thus
a =
b) the magnitude of 5ax + N − 3 M:
(5, 0 , 0) + (8, 7 , −2) − (− 30 , 12 , −24) = (43, − 5 , 22), and |(43, − 5 , 22)| = 48.6.
c) |M|| 2 N|(M + N): |(− 10 , 4 , −8)||(16, 14 , −4)|(− 2 , 11 , −10) = (13.4)(21.6)(− 2 , 11 , −10) = (− 580. 5 , 3193 , −2902)
1.2. The three vertices of a triangle are located at A(− 1 , 2 , 5), B(− 4 , − 2 , −3), and C(1, 3 , −2). a) Find the length of the perimeter of the triangle: Begin with AB = (− 3 , − 4 , −8), BC = (5, 5 , 1), and CA = (− 2 , − 1 , 7). Then the perimeter will be = |AB| + |BC| + |CA| =
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side BC: The vector from the origin to the midpoint of AB is MAB = 12 (A + B) = 12 (− 5 ax + 2az ). The vector from the origin to the midpoint of BC is MBC = 12 (B + C) = 12 (− 3 ax + ay − 5 az ). The vector from midpoint to midpoint is now MAB − MBC = 12 (− 2 ax − ay + 7az ). The unit vector is therefore
aM M =
MAB − MBC
|MAB − MBC |
(− 2 ax − ay + 7az )
- 35
= − 0. 27 ax − 0. 14 ay + 0. 95 az
where factors of 1/2 have cancelled. c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the unit vector is therefore parallel to AC. First we find AC = 2ax + ay − 7 az , which we recognize as − 7. 35 aM M. The vectors are thus parallel (but oppositely-directed).
1.3. The vector from the origin to the point A is given as (6, − 2 , −4), and the unit vector directed from the origin toward point B is (2, − 2 , 1)/3. If points A and B are ten units apart, find the coordinates of point B. With A = (6, − 2 , −4) and B = 13 B(2, − 2 , 1), we use the fact that |B − A| = 10, or |(6 − 23 B)ax − (2 − 23 B)ay − (4 + 13 B)az | = 10 Expanding, obtain 36 − 8 B + 49 B^2 + 4 − 83 B + 49 B^2 + 16 + 83 B + 19 B^2 = 100 or B^2 − 8 B − 44 = 0. Thus B = 8 ±
√ 64 − 176 2 = 11.75 (taking positive option) and so
B =
(11.75)ax −
(11.75)ay +
(11.75)az = 7. 83 ax − 7. 83 ay + 3. 92 az
1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit vector in rectangular components that lies in the xy plane, is tangent to the circle at (
3 , 1 , 0), and is in the general direction of increasing values of y: A unit vector tangent to this circle in the general increasing y direction is t = aφ. Its x and y components are tx = aφ · ax = − sin φ, and ty = aφ · ay = cos φ. At the point (
3 , 1), φ = 30◦, and so t = − sin 30◦ax + cos 30◦ay = 0.5(−ax +
3 ay ).
1.5. A vector field is specified as G = 24xyax + 12(x^2 + 2)ay + 18z^2 az. Given two points, P (1, 2 , −1) and Q(− 2 , 1 , 3), find: a) G at P : G(1, 2 , −1) = (48, 36 , 18) b) a unit vector in the direction of G at Q: G(− 2 , 1 , 3) = (− 48 , 72 , 162), so
aG =
c) a unit vector directed from Q toward P :
aQP =
P − Q
|P − Q|
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x^2 + 2), 18 z^2 )|, or 10 = |(4xy, 2 x^2 + 4, 3 z^2 )|, so the equation is
100 = 16x^2 y^2 + 4x^4 + 16x^2 + 16 + 9z^4
1.6. If a is a unit vector in a given direction, B is a scalar constant, and r = xax + yay + zaz , describe the surface r · a = B. What is the relation between the the unit vector a and the scalar B to this surface? (HINT: Consider first a simple example with a = ax and B = 1, and then consider any a and B.): We could consider a general unit vector, a = A 1 ax + A 2 ay + A 3 az , where A^21 + A^22 + A^23 = 1. Then r · a = A 1 x + A 2 y + A 3 z = f (x, y, z) = B. This is the equation of a planar surface, where f = B. The relation of a to the surface becomes clear in the special case in which a = ax. We obtain r · a = f (x) = x = B, where it is evident that a is a unit normal vector to the surface (as a look ahead (Chapter 4), note that taking the gradient of f gives a).
1.7. Given the vector field E = 4zy^2 cos 2xax + 2zy sin 2xay + y^2 sin 2xaz for the region |x|, |y|, and |z| less than 2, find: a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0, with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2, |z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2. b) the region in which Ey = Ez : This occurs when 2zy sin 2x = y^2 sin 2x, or on the plane 2z = y, with |x| < 2, |y| < 2, |z| < 1. c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy^2 cos 2x = zy sin 2x = y^2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.
1.11. Given the points M (0. 1 , − 0. 2 , − 0 .1), N (− 0. 2 , 0. 1 , 0 .3), and P (0. 4 , 0 , 0 .1), find:
a) the vector RM N : RM N = (− 0. 2 , 0. 1 , 0 .3) − (0. 1 , − 0. 2 , − 0 .1) = (− 0. 3 , 0. 3 , 0 .4). b) the dot product RM N · RM P : RM P = (0. 4 , 0 , 0 .1) − (0. 1 , − 0. 2 , − 0 .1) = (0. 3 , 0. 2 , 0 .2). RM N · RM P = (− 0. 3 , 0. 3 , 0 .4) · (0. 3 , 0. 2 , 0 .2) = − 0 .09 + 0.06 + 0.08 = 0.05. c) the scalar projection of RM N on RM P :
RM N · aRM P = (− 0. 3 , 0. 3 , 0 .4) ·
d) the angle between RM N and RM P :
θM = cos−^1
RM N · RM P
|RM N ||RM P |
= cos−^1
1.12. Show that the vector fields A = ρ cos φ aρ + ρ sin φ aφ + ρ az and B = ρ cos φ aρ + ρ sin φ aφ − ρ az are everywhere perpendicular to each other: We find A · B = ρ^2 (sin^2 φ + cos^2 φ) − ρ^2 = 0 = |A||B| cos θ. Therefore cos θ = 0 or θ = 90◦.
1.13. a) Find the vector component of F = (10, − 6 , 5) that is parallel to G = (0. 1 , 0. 2 , 0 .3):
F||G =
F · G
|G|^2
G =
b) Find the vector component of F that is perpendicular to G:
FpG = F − F||G = (10, − 6 , 5) − (0. 93 , 1. 86 , 2 .79) = (9. 07 , − 7. 86 , 2 .21)
c) Find the vector component of G that is perpendicular to F:
GpF = G − G||F = G −
G · F
|F|^2
F = (0. 1 , 0. 2 , 0 .3) −
1.14. Show that the vector fields A = ar (sin 2θ)/r^2 +2aθ (sin θ)/r^2 and B = r cos θ ar +r aθ are everywhere parallel to each other: Using the definition of the cross product, we find
A × B =
sin 2θ r
2 sin θ cos θ r
aφ = 0 = |A||B| sin θ n
Identify n = aφ, and so sin θ = 0, and therefore θ = 0 (they’re parallel).
1.15. Three vectors extending from the origin are given as r 1 = (7, 3 , −2), r 2 = (− 2 , 7 , −3), and r 3 = (0, 2 , 3). Find: a) a unit vector perpendicular to both r 1 and r 2 :
ap 12 =
r 1 × r 2 |r 1 × r 2 |
b) a unit vector perpendicular to the vectors r 1 − r 2 and r 2 − r 3 : r 1 − r 2 = (9, − 4 , 1) and r 2 − r 3 = (− 2 , 5 , −6). So r 1 − r 2 × r 2 − r 3 = (19, 52 , 32). Then
ap =
c) the area of the triangle defined by r 1 and r 2 :
Area =
|r 1 × r 2 | = 30. 3
d) the area of the triangle defined by the heads of r 1 , r 2 , and r 3 :
Area =
|(r 2 − r 1 ) × (r 2 − r 3 )| =
|(− 9 , 4 , −1) × (− 2 , 5 , −6)| = 32. 0
1.16. The vector field E = (B/ρ) aρ, where B is a constant, is to be translated such that it originates at the line, x = 2, y = 0. Write the translated form of E in rectangular components: First, transform the given field to rectangular components:
Ex =
B
ρ
aρ · ax =
B
x^2 + y^2
cos φ =
B
x^2 + y^2
x √ x^2 + y^2
Bx x^2 + y^2
Using similar reasoning:
Ey =
B
ρ
aρ · ay =
B
x^2 + y^2
sin φ =
By x^2 + y^2
We then translate the two components to x = 2, y = 0, to obtain the final result:
E(x, y) =
B [(x − 2) ax + y ay ] (x − 2)^2 + y^2
1.17. Point A(− 4 , 2 , 5) and the two vectors, RAM = (20, 18 , −10) and RAN = (− 10 , 8 , 15), define a triangle. a) Find a unit vector perpendicular to the triangle: Use
ap =
RAM × RAN
|RAM × RAN |
The vector in the opposite direction to this one is also a valid answer.
1.19b) Evaluate D at the point where ρ = 2, φ = 0. 2 π, and z = 5, expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0. 5 aρ. To express this in cartesian, we use
D = 0.5(aρ · ax)ax + 0.5(aρ · ay )ay = 0.5 cos 36◦ax + 0.5 sin 36◦ay = 0. 41 ax + 0. 29 ay
1.20. A cylinder of radius a, centered on the z axis, rotates about the z axis at angular velocity Ω rad/s. The rotation direction is counter-clockwise when looking in the positive z direction. a) Using cylindrical components, write an expression for the velocity field, v, that gives the tan- gential velocity at any point within the cylinder: Tangential velocity is angular velocity times the perpendicular distance from the rotation axis. With counter-clockwise rotation, we therefore find v(ρ) = −Ωρ aφ (ρ < a). b) Convert your result from part a to spherical components: In spherical, the component direction, aφ, is the same. We obtain
v(r, θ) = −Ωr sin θ aφ (r sin θ < a)
c) Convert to rectangular components:
vx = −Ωρaφ · ax = −Ω(x^2 + y^2 )^1 /^2 (− sin φ) = −Ω(x^2 + y^2 )^1 /^2
−y (x^2 + y^2 )^1 /^2
= Ωy
Similarly
vy = −Ωρaφ · ay = −Ω(x^2 + y^2 )^1 /^2 (cos φ) = −Ω(x^2 + y^2 )^1 /^2
x (x^2 + y^2 )^1 /^2
= −Ωx
Finally v(x, y) = Ω [y ax − x ay ], where (x^2 + y^2 )^1 /^2 < a.
1.21. Express in cylindrical components: a) the vector from C(3, 2 , −7) to D(− 1 , − 4 , 2): C(3, 2 , −7) → C(ρ = 3. 61 , φ = 33. 7 ◦, z = −7) and D(− 1 , − 4 , 2) → D(ρ = 4. 12 , φ = − 104. 0 ◦, z = 2). Now RCD = (− 4 , − 6 , 9) and Rρ = RCD · aρ = −4 cos(33.7) − 6 sin(33.7) = − 6 .66. Then Rφ = RCD · aφ = 4 sin(33.7) − 6 cos(33.7) = − 2 .77. So RCD = − 6. 66 aρ − 2. 77 aφ + 9az b) a unit vector at D directed toward C: RCD = (4, 6 , −9) and Rρ = RDC · aρ = 4 cos(− 104 .0) + 6 sin(− 104 .0) = − 6 .79. Then Rφ = RDC · aφ = 4[− sin(− 104 .0)] + 6 cos(− 104 .0) = 2.43. So RDC = − 6. 79 aρ + 2. 43 aφ − 9 az Thus aDC = − 0. 59 aρ + 0. 21 aφ − 0. 78 az c) a unit vector at D directed toward the origin: Start with rD = (− 1 , − 4 , 2), and so the vector toward the origin will be −rD = (1, 4 , −2). Thus in cartesian the unit vector is a = (0. 22 , 0. 87 , − 0 .44). Convert to cylindrical: aρ = (0. 22 , 0. 87 , − 0 .44) · aρ = 0.22 cos(− 104 .0) + 0.87 sin(− 104 .0) = − 0 .90, and aφ = (0. 22 , 0. 87 , − 0 .44) · aφ = 0.22[− sin(− 104 .0)] + 0.87 cos(− 104 .0) = 0, so that finally, a = − 0. 90 aρ − 0. 44 az.
1.22. A sphere of radius a, centered at the origin, rotates about the z axis at angular velocity Ω rad/s. The rotation direction is clockwise when one is looking in the positive z direction. a) Using spherical components, write an expression for the velocity field, v, which gives the tan- gential velocity at any point within the sphere: As in problem 1.20, we find the tangential velocity as the product of the angular velocity and the perperdicular distance from the rotation axis. With clockwise rotation, we obtain
v(r, θ) = Ωr sin θ aφ (r < a)
b) Convert to rectangular components: From here, the problem is the same as part c in Problem 1.20, except the rotation direction is reversed. The answer is v(x, y) = Ω [−y ax + x ay ], where (x^2 + y^2 + z^2 )^1 /^2 < a.
1.23. The surfaces ρ = 3, ρ = 5, φ = 100◦, φ = 130◦, z = 3, and z = 4.5 define a closed surface. a) Find the enclosed volume:
Vol =
3
100 ◦
3
ρ dρ dφ dz = 6. 28
NOTE: The limits on the φ integration must be converted to radians (as was done here, but not shown).
b) Find the total area of the enclosing surface:
Area = 2
100 ◦
3
ρ dρ dφ +
3
100 ◦
3 dφ dz
3
100 ◦
5 dφ dz + 2
3
3
dρ dz = 20. 7
c) Find the total length of the twelve edges of the surfaces:
Length = 4 × 1 .5 + 4 × 2 + 2 ×
[
360 ◦^
× 2 π × 3 +
360 ◦^
× 2 π × 5
]
d) Find the length of the longest straight line that lies entirely within the volume: This will be between the points A(ρ = 3, φ = 100◦, z = 3) and B(ρ = 5, φ = 130◦, z = 4.5). Performing point transformations to cartesian coordinates, these become A(x = − 0 .52, y = 2.95, z = 3) and B(x = − 3 .21, y = 3.83, z = 4.5). Taking A and B as vectors directed from the origin, the requested length is Length = |B − A| = |(− 2. 69 , 0. 88 , 1 .5)| = 3. 21
1.27. The surfaces r = 2 and 4, θ = 30◦^ and 50◦, and φ = 20◦^ and 60◦^ identify a closed surface. a) Find the enclosed volume: This will be
Vol =
20 ◦
30 ◦
2
r^2 sin θdrdθdφ = 2. 91
where degrees have been converted to radians. b) Find the total area of the enclosing surface:
Area =
20 ◦
30 ◦
(4^2 + 2^2 ) sin θdθdφ +
2
20 ◦
r(sin 30◦^ + sin 50◦)drdφ
30 ◦
2
rdrdθ = 12. 61
c) Find the total length of the twelve edges of the surface:
Length = 4
2
dr + 2
30 ◦
(4 + 2)dθ +
20 ◦
(4 sin 50◦^ + 4 sin 30◦^ + 2 sin 50◦^ + 2 sin 30◦)dφ
= 17. 49
d) Find the length of the longest straight line that lies entirely within the surface: This will be from A(r = 2, θ = 50◦, φ = 20◦) to B(r = 4, θ = 30◦, φ = 60◦) or
A(x = 2 sin 50◦^ cos 20◦, y = 2 sin 50◦^ sin 20◦, z = 2 cos 50◦)
to B(x = 4 sin 30◦^ cos 60◦, y = 4 sin 30◦^ sin 60◦, z = 4 cos 30◦)
or finally A(1. 44 , 0. 52 , 1 .29) to B(1. 00 , 1. 73 , 3 .46). Thus B − A = (− 0. 44 , 1. 21 , 2 .18) and
Length = |B − A| = 2. 53
1.28. Express the vector field, G = 8 sin φ aθ in
a) rectangular components:
Gx = 8 sin φ aθ · ax = 8 sin φ cos θ cos φ =
8 y √ x^2 + y^2
z √ x^2 + y^2 + z^2
x √ x^2 + y^2
=
8 xyz (x^2 + y^2 )
x^2 + y^2 + z^2
Gy = 8 sin φ aθ · ay = 8 sin φ cos θ sin φ =
8 y √ x^2 + y^2
z √ x^2 + y^2 + z^2
y √ x^2 + y^2
=
8 y^2 z (x^2 + y^2 )
x^2 + y^2 + z^2
1.28a) (continued)
Gz = 8 sin φ aθ · az = 8 sin φ(− sin θ) =
− 8 y √ x^2 + y^2
x^2 + y^2 √ x^2 + y^2 + z^2
=
− 8 y √ x^2 + y^2 + z^2 Finally, G(x, y, z) =
8 y √ x^2 + y^2 + z^2
[
xz x^2 + y^2
ax +
yz x^2 + y^2
ay − az
]
b) cylindrical components: The aθ direction will transform to cylindrical components in the aρ and az directions only, where
Gρ = 8 sin φ aθ · aρ = 8 sin φ cos θ = 8 sin φ
z √ ρ^2 + z^2 The z component will be the same as found in part a, so we finally obtain
G(ρ, z) =
8 ρ sin φ √ ρ^2 + z^2
[
z ρ
aρ − az
]
1.29. Express the unit vector ax in spherical components at the point: a) r = 2, θ = 1 rad, φ = 0.8 rad: Use ax = (ax · ar )ar + (ax · aθ )aθ + (ax · aφ)aφ = sin(1) cos(0.8)ar + cos(1) cos(0.8)aθ + (− sin(0.8))aφ = 0. 59 ar + 0. 38 aθ − 0. 72 aφ
b) x = 3, y = 2, z = −1: First, transform the point to spherical coordinates. Have r =
θ = cos−^1 (− 1 /
- = 105. 5 ◦, and φ = tan−^1 (2/3) = 33. 7 ◦. Then ax = sin(105. 5 ◦) cos(33. 7 ◦)ar + cos(105. 5 ◦) cos(33. 7 ◦)aθ + (− sin(33. 7 ◦))aφ = 0. 80 ar − 0. 22 aθ − 0. 55 aφ
c) ρ = 2.5, φ = 0.7 rad, z = 1.5: Again, convert the point to spherical coordinates. r =
√ ρ^2 +^ z^2 = 8 .5, θ = cos−^1 (z/r) = cos−^1 (1. 5 /
8 .5) = 59. 0 ◦, and φ = 0.7 rad = 40. 1 ◦. Now ax = sin(59◦) cos(40. 1 ◦)ar + cos(59◦) cos(40. 1 ◦)aθ + (− sin(40. 1 ◦))aφ = 0. 66 ar + 0. 39 aθ − 0. 64 aφ
1.30. At point B(5, 120 ◦, 75 ◦) a vector field has the value A = − 12 ar − 5 aθ + 15 aφ. Find the vector component of A that is: a) normal to the surface r = 5: This will just be the radial component, or − 12 ar. b) tangent to the surface r = 5: This will be the remaining components of A that are not normal, or − 5 aθ + 15 aφ. c) tangent to the cone θ = 120◦: The unit vector normal to the cone is aθ , so the remaining components are tangent: − 12 ar + 15 aφ. d) Find a unit vector that is perpendicular to A and tangent to the cone θ = 120◦: Call this vector b = br ar + bφ aφ, where b^2 r + b^2 φ = 1. We then require that A · b = 0 = − 12 br + 15bφ, and therefore bφ = (4/5)br. Now b^2 r [1 + (16/25)] = 1, so br = 5/
- Then bφ = 4/
- Finally, b =
(5 ar + 4 aφ)
2.3. Point charges of 50nC each are located at A(1, 0 , 0), B(− 1 , 0 , 0), C(0, 1 , 0), and D(0, − 1 , 0) in free space. Find the total force on the charge at A.
The force will be: F =
(50 × 10 −^9 )^2
4 π 0
[
RCA
|RCA|^3
RDA
|RDA|^3
RBA
|RBA|^3
]
where RCA = ax − ay , RDA = ax + ay , and RBA = 2ax. The magnitudes are |RCA| = |RDA| =
and |RBA| = 2. Substituting these leads to
F =
(50 × 10 −^9 )^2
4 π 0
[
]
ax = 21. 5 ax μN
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 4 π 0 a^2
ax + ay + az 3
︸ ︷︷^3 ︸
(0, 0 ,0)
ay + az 2
︸ ︷︷^2 ︸
(a, 0 ,0)
ax + az 2
︸ ︷︷^2 ︸
(0,a,0)
ax + ay 2
︸ ︷︷^2 ︸
(0, 0 ,a)
(^) ︸︷︷︸ax (0,a,a)
ay ︸︷︷︸ (a, 0 ,a)
(^) ︸︷︷︸az (a,a,0)
The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtain
F(a, a, a) = qEnet(a, a, a) =
q^2 4 π 0 a^2
[
]
(ax + ay + az ) =
- 90 q^2 4 π 0 a^2
(ax + ay + az )
in which the magnitude is |F| = 3. 29 q^2 /(4π 0 a^2 ).
2.5. Let a point charge Q 1 25 nC be located at P 1 (4, − 2 , 7) and a charge Q 2 = 60 nC be at P 2 (− 3 , 4 , −2).
a) If = 0 , find E at P 3 (1, 2 , 3): This field will be
E =
10 −^9
4 π 0
[
25 R 13
|R 13 |^3
60 R 23
|R 23 |^3
]
where R 13 = − 3 ax + 4ay − 4 az and R 23 = 4ax − 2 ay + 5az. Also, |R 13 | =
41 and |R 23 | =
So E =
10 −^9
4 π 0
[
25 × (− 3 ax + 4ay − 4 az ) (41)^1.^5
60 × (4ax − 2 ay + 5az ) (45)^1.^5
]
= 4. 58 ax − 0. 15 ay + 5. 51 az
b) At what point on the y axis is Ex = 0? P 3 is now at (0, y, 0), so R 13 = − 4 ax + (y + 2)ay − 7 az and R 23 = 3ax + (y − 4)ay + 2az. Also, |R 13 | =
65 + (y + 2)^2 and |R 23 | =
13 + (y − 4)^2. Now the x component of E at the new P 3 will be:
Ex =
10 −^9
4 π 0
[
25 × (−4)
[65 + (y + 2)^2 ]^1.^5
60 × 3
[13 + (y − 4)^2 ]^1.^5
]
To obtain Ex = 0, we require the expression in the large brackets to be zero. This expression simplifies to the following quadratic:
- 48 y^2 + 13. 92 y + 73.10 = 0
which yields the two values: y = − 6. 89 , − 22. 11
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: At a general location, x,
E(x) =
q 4 π 0
[
(x + 1)^2
x^2
(x − 1)^2
]
ax
At x = 5, and with q = 5 × 10 −^9 C, this becomes E(x = 5) = 5. 8 ax V/m. 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
E(x >> 1) =.
3 q 4 π 0 x^2
ax
Therefore, the equivalent charge will have value 3q = 1. 5 × 10 −^8 C, and will be at location x = 0. c) Determine E at x = 5, using the approximation of (b). Using 3q = 1. 5 × 10 −^8 C and x = 5 in the part b result gives E(x = 5)
= 5. 4 ax V/m, or about 7% lower than the exact result.
2.7. A 2 μC point charge is located at A(4, 3 , 5) in free space. Find Eρ, Eφ, and Ez at P (8, 12 , 2). Have
EP =
2 × 10 −^6
4 π 0
RAP
|RAP |^3
2 × 10 −^6
4 π 0
[
4 ax + 9ay − 3 az (106)^1.^5
]
= 65. 9 ax + 148. 3 ay − 49. 4 az
Then, at point P , ρ =
82 + 12^2 = 14.4, φ = tan−^1 (12/8) = 56. 3 ◦, and z = z. Now,
Eρ = Ep · aρ = 65.9(ax · aρ) + 148.3(ay · aρ) = 65.9 cos(56. 3 ◦) + 148.3 sin(56. 3 ◦) = 159. 7
and
Eφ = Ep · aφ = 65.9(ax · aφ) + 148.3(ay · aφ) = − 65 .9 sin(56. 3 ◦) + 148.3 cos(56. 3 ◦) = 27. 4
Finally, Ez = − 49 .4 V/m
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 Q^2 4 π 0 (d − x)^2
= kx
We observe immediately that c = 0. Also, from (2) we find that b = −a
√ 3, and therefore a^2 + b^2 = 2a. Using this information in (3), we write for the x component:
1 − a √ (1 − a)^2 + b^2
1 − a √ 1 − 2 a + 4a^2
or 0. 44 a^2 + 1. 28 a − 0 .64 = 0, so that
a =
(1.28)^2 + 4(0.44)(0.64)
= 0.435 or − 3. 344
The corresponding b values are respectively − 0 .753 and 5.793. So the two possible P coordinate sets are (0. 435 , − 0. 753 , 0) and (− 3. 344 , 5. 793 , 0). By direct substitution, however, it is found that only one possibility is entirely consistent with both (2) and (3), and this is
P (a, b, c) = (− 3. 344 , 5. 793 , 0)
2.11. A charge Q 0 located at the origin in free space produces a field for which Ez = 1 kV/m at point P (− 2 , 1 , −1).
a) Find Q 0 : The field at P will be
EP =
Q 0
4 π 0
[
− 2 ax + ay − az 61.^5
]
Since the z component is of value 1 kV/m, we find Q 0 = − 4 π 061.^5 × 103 = − 1. 63 μC.
b) Find E at M (1, 6 , 5) in cartesian coordinates: This field will be:
EM =
− 1. 63 × 10 −^6
4 π 0
[
ax + 6ay + 5az [1 + 36 + 25]^1.^5
]
or EM = − 30. 11 ax − 180. 63 ay − 150. 53 az.
c) Find E at M (1, 6 , 5) in cylindrical coordinates: At M , ρ =
1 + 36 = 6.08, φ = tan−^1 (6/1) =
- 54 ◦, and z = 5. Now
Eρ = EM · aρ = − 30 .11 cos φ − 180 .63 sin φ = − 183. 12
Eφ = EM · aφ = − 30 .11(− sin φ) − 180 .63 cos φ = 0 (as expected)
so that EM = − 183. 12 aρ − 150. 53 az.
d) Find E at M (1, 6 , 5) in spherical coordinates: At M , r =
1 + 36 + 25 = 7.87, φ = 80. 54 ◦^ (as before), and θ = cos−^1 (5/ 7 .87) = 50. 58 ◦. Now, since the charge is at the origin, we expect to obtain only a radial component of EM. This will be:
Er = EM · ar = − 30 .11 sin θ cos φ − 180 .63 sin θ sin φ − 150 .53 cos θ = − 237. 1
2.12. Electrons are in random motion in a fixed region in space. During any 1μs interval, the probability of finding an electron in a subregion of volume 10−^15 m^2 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
ρv = −
0. 27 × 1. 602 × 10 −^19
10 −^15
= − 43. 3 μC/m^3
2.13. A uniform volume charge density of 0.2 μC/m^3 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
Q =
∫ (^2) π
0
∫ (^) π
0
. 03
- 2 r^2 sin θ dr dθ dφ =
[
4 π(0.2)
r^3 3
]. 05
. 03
= 8. 21 × 10 −^5 μC = 82.1 pC
b) find r 1 if half the total charge is located in the region 3 cm < r < r 1 : If the integral over r in part a is taken to r 1 , we would obtain [ 4 π(0.2)
r^3 3
]r 1
. 03
= 4. 105 × 10 −^5
Thus
r 1 =
[
3 × 4. 105 × 10 −^5
- 2 × 4 π
+ (.03)^3
] 1 / 3
= 4.24 cm
2.14. The charge density varies with radius in a cylindrical coordinate system as ρv = ρ 0 /(ρ^2 + a^2 )^2 C/m^3. 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
Q(ρ) =
0
∫ (^2) π
0
∫ (^) ρ
0
ρ 0 (ρ′^2 + a^2 )^2
ρ′dρ′dφdz = 2πρ 0
[
2(a^2 + ρ′2)
]ρ
0
πρ 0 a^2
[
1 + ρ^2 /a^2
]
The total charge is found when ρ → ∞, or Qnet = πρ 0 /a^2. It is 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 10^15 C/m^3.
a) What total charge is enclosed in the spherical volume? This will be Q = (4/3)π(2 × 10 −^6 )^3 × 1015 = 3. 35 × 10 −^2 C.
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
ρv,avg =
3. 35 × 10 −^2
(0.003)^3
= 1. 24 × 106 C/m^3
Then, with the given values of ρL and q, the field evaluates as
Etot = 2. 0 ax + 7. 3 ay − 9. 4 az V/m
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:
E(0, 0 , 3) =
ρL 2 π 0 (3)
q 4 π 0 (2)^2
q 4 π 0 (4)^2
Therefore q
[
]
2 ρL 3
⇒ ρL = −
q = − 0. 47 q = − 3 .75 nC/m
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 0 ρ)]aρ. Therefore, at point P :
EP =
ρl 2 π 0
RzP |RzP |^2
(2 × 10 −^6 )
2 π 0
ax + 2ay 5
= 7. 2 ax + 14. 4 ay kV/m
where RzP is the vector that extends from the line charge to point P , and is perpendicular to the z axis; i.e., RzP = (1, 2 , 3) − (0, 0 , 3) = (1, 2 , 0).
b) − 4 ≤ z ≤ 4: Here we use the general relation
EP =
ρldz 4 π 0
r − r′ |r − r′|^3
where r = ax + 2ay + 3az and r′^ = zaz. So the integral becomes
EP =
(2 × 10 −^6 )
4 π 0
− 4
ax + 2ay + (3 − z)az [5 + (3 − z)^2 ]^1.^5
dz
Using integral tables, we obtain:
EP = 3597
[
(ax + 2ay )(z − 3) + 5az (z^2 − 6 z + 14)
] 4
− 4
V/m = 4. 9 ax + 9. 8 ay + 4. 9 az kV/m
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
dE =
ρL dz (r − r′) 4 π 0 |r − r′|^3
where r = 4 az and r′^ = z az. Therefore, r − r′^ = (4 − z) az and |r − r′| = 4 − z. Substituting ρL = 10|z| nC/m, the total field is
E(0, 0 , 4) =
− 2
10 −^8 |z| dz az 4 π 0 (4 − z)^2
0
10 −^8 z dz az 4 π 0 (4 − z)^2
− 2
10 −^8 z dz az 4 π 0 (4 − z)^2
10 −^8
4 π × 8. 854 × 10 −^12
{[
ln(4 − z) +
4 − z
] 2
0
[
ln(4 − z) +
4 − z
] 0
− 2
az
= 34. 0 az V/m
b) (0,4,0): In this case, r = 4 ay and r′^ = z az as before. The field at (0,4,0) is then
E(0, 4 , 0) =
− 2
10 −^8 |z| dz (4 ay − z az ) 4 π 0 (16 + z^2 )^3 /^2
Note the symmetric limits on the integral. As the z component of the integrand changes sign at z = 0, it will contribute equal and opposite portions to the overall integral, which will can- cel completely (the z component integral has odd parity). This leaves only the y component integrand, which has even parity. The integral therefore simplifies to
E(0, 4 , 0) = 2
0
4 × 10 −^8 z dz ay 4 π 0 (16 + z^2 )^3 /^2
− 2 × 10 −^8 ay π × 8. 854 × 10 −^12
[
16 + z^2
] 2
0
= 18. 98 ay V/m
2.21. Two identical uniform line charges with ρl = 75 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 0.8 m. Thus the field from the charge at y = − 0 .4 evaluated at the location of the charge at y = +0.4 will be E = [ρl/(2π 0 (0.8))]ay. The force on a differential length of the line at the positive y location is dF = dqE = ρldzE. Thus the force per unit length acting on the line at postive y arising from the charge at negative y is
F =
0
ρ^2 l dz 2 π 0 (0.8)
ay = 1. 26 × 10 −^4 ay N/m = 126 ay μN/m
The force on the line at negative y is of course the same, but with −ay.
2.22. Two identical uniform sheet charges with ρs = 100 nC/m^2 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 there, multiplied by the electric field from the top sheet: dF = ρsdaE. The force per unit area is then just F = ρsE = (100 × 10 −^9 )(− 100 × 10 −^9 )/(2 0 ) az = − 5. 6 × 10 −^4 az N/m^2.
